Quay lại đáp án

	7	bớt 46 ký tự trong đáp án		Sửa 18-02-13 02:25 PM hoàng anh thọ 17 2
BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} - x >{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} - x >{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} - x >{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$

	6	thêm 33 ký tự trong đáp án		Sửa 18-02-13 02:24 PM hoàng anh thọ 17 2
BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì $4^x \leq 3.2^{\sqrt{x}+x+4x BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} - x >{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì4x≤3.2x+x+4x BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} - x >{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì $4^x \leq 3.2^{\sqrt{x}+x+4x BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} - x >{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$

	5	bớt 2 ký tự trong đáp án		Sửa 18-02-13 02:21 PM Đỗ Quang Chính 15
BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} - x >{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} - x >{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} - x >{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$

	4	thêm 2 ký tự trong đáp án		Sửa 18-02-13 02:21 PM Đỗ Quang Chính 15
BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì 4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} - x >{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$$\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì 4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} -x>{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì 4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} - x >{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$$\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$

	3	thêm 2 ký tự trong đáp án		Sửa 18-02-13 02:17 PM hoàng anh thọ 17 2
BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì 4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} -x>{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$$\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì 4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x}-x>{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4.}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì 4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x} -x>{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$$\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4}$

	2	sửa đáp án		Sửa 18-02-13 11:43 AM Trần Nhật Tân 1
BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì 4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x}-x>{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4.}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì 4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x}-x>{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4.}$ BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì 4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x}-x>{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4.}$

	1	tạo mới		Trả lời 18-02-13 11:38 AM Trần Nhật Tân 1
BPT $\Leftrightarrow 2^{2x}< 3.2^{\sqrt{x}+x}+2^{2\sqrt{x}+2}$ $\Leftrightarrow 1< 3.2^{\sqrt{x}-x}+2^{2\sqrt{x}-2x+2}$ $\Leftrightarrow 3.2^{\sqrt{x}-x}+4.2^{2(\sqrt{x}-x)}-1>0$ Đặt $t=2^{\sqrt{x}-x}>0$ thì 4x≤3.2x√+x+41+x√ BPT $\Leftrightarrow 4t^2+3t-1>0\Leftrightarrow (t+1)(4t-1)>0\Leftrightarrow t>2^{-2}\Leftrightarrow 2^{\sqrt{x}-x}>2^{-2}$ $\Leftrightarrow \sqrt{x}-x>{-2}\Leftrightarrow x-\sqrt{x}-2<0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}+1)<0$ $\Leftrightarrow \sqrt{x}<2\Leftrightarrow \boxed{0 \le x <4.}$

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Giấy xác nhận cung cấp dịch vụ mạng xã hội trực tuyến số 97/GXN-TTĐT cấp ngày 03/08/2012