Chương 4: BẤT ĐẲNG THỨC, BẤT PHƯƠNG TRÌNH

a: (2x-1)(x+3)<=0

nên -3<=x<=1/2

b: \(\left(2x-7\right)\left(4-5x\right)>=0\)

=>(2x-7)(5x-4)<=0

=>4/5<=x<=7/2

i: \(\Leftrightarrow\dfrac{3-x+2}{x-2}>0\)

\(\Leftrightarrow\dfrac{x-5}{x-2}< 0\)

=>2<x<5

1: \(\Leftrightarrow\left\{{}\begin{matrix}2x+3>=-8\\2x+3< =8\end{matrix}\right.\Leftrightarrow-\dfrac{11}{2}< =x< =\dfrac{5}{2}\)

2: \(\Leftrightarrow\left[{}\begin{matrix}-5x+3>1\\-5x+3< -1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x>-2\\-5x< -4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{2}{5}\\x>\dfrac{4}{5}\end{matrix}\right.\)

3: \(\Leftrightarrow\dfrac{x-1}{2x-3}< 0\)

hay 1<x<3/2

D

\(e,\dfrac{2x-3}{4+x}\ge1\Leftrightarrow\dfrac{2x-3}{4+x}-1\ge0\\ \Leftrightarrow\dfrac{2x-3-4-x}{4+x}\ge0\Leftrightarrow\dfrac{x-7}{4+x}\ge0\\ \Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge-4\end{matrix}\right.\\\left\{{}\begin{matrix}x\le7\\x\le4\end{matrix}\right.\end{matrix}\right.\) 

\(\Leftrightarrow\dfrac{2x-3}{4+x}< 2\Leftrightarrow\dfrac{2x-3}{4+x}-2< 0\\ \Leftrightarrow\dfrac{2x+3-8-2x}{4+x}< 0\\ \Leftrightarrow\dfrac{-5}{4+x}< 0\Leftrightarrow4+x>0\Leftrightarrow x>\left(-4\right)\) 

\(k,\left(2x+1\right)\left(x+3\right)< x^2+3\\ \Leftrightarrow2x^2+7x+3< x^2+3\\ \Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>\left(-7\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< \left(-7\right)\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow-7< x< 0\rightarrow x\in0\left(ko.có.nghiệm\right)\) 

\(-7< 0< 0\) 

\(l,\left(4x-1\right)\left(x+3\right)+9>x^2\\ \left\{{}\begin{matrix}\left\{{}\begin{matrix}x>-3\\x>-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3\\x< -\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\\ x< -3.hoặc.x>-\dfrac{2}{3}\\ \)

lm câu nèo đay?

Mình sửa câu e nghen, do mình thấy bài bạn Châu làm KTM á 

e. \(1\le\dfrac{2x-3}{4+x}\le2\)

\(\left\{{}\begin{matrix}\dfrac{2x-3}{4+x}-1\ge0\\\dfrac{2x-3}{4+x}-2\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-3-4-x}{4+x}\ge0\\\dfrac{2x-3-8-2x}{4+x}\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-7}{4+x}\ge0\\\dfrac{-11}{4+x}\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-7}{4+x}\ge0\\4+x>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-7\ge0\\4+x>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge7\\x>-4\end{matrix}\right.\)\(\Rightarrow x\ge7\)

Chọn B

Chọn D 

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Lời giải:

Bổ sung điều kiện $a,b$ là các số dương. Áp dụng BĐT Cô-si ta có:

$a+b\geq 2\sqrt{ab}$

$\frac{1}{a}+\frac{1}{b}\geq 2\sqrt{\frac{1}{ab}}$

$\Rightarrow (a+b)(\frac{1}{a}+\frac{1}{b})\geq 2\sqrt{ab}.2\sqrt{\frac{1}{ab}}=4$

Ta có đpcm 

Dấu "=" xảy ra khi $a=b$