kho..................lam............................tich,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,minh..........................troi........................ret............................wa.................ung ho minh.................hu....................hu..............hu................hat..............hat....................s

Mình cũng thắc mắc câu này ;-;

Ta có:

\(\left|x-\frac{3}{4}\right|+\left|x+\frac{9}{7}\right|=\left|\frac{3}{4}-x\right|+\left|x+\frac{9}{7}\right|\ge\left|\frac{3}{4}-x+x+\frac{9}{7}\right|=\frac{57}{28}\)

=> \(28\cdot\left(\left|x-\frac{3}{4}\right|+\left|x+\frac{9}{7}\right|\right)\ge57\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(\frac{3}{4}-x\right)\left(x+\frac{9}{7}\right)\ge0\Rightarrow-\frac{9}{7}\le x\le\frac{3}{4}\)

Vậy \(Min=28\Leftrightarrow-\frac{9}{7}\le x\le\frac{3}{4}\)

Đặt \(A=\left|x-\frac{3}{4}\right|+\left|x+\frac{9}{7}\right|\)

\(\Rightarrow A=\left|\frac{3}{4}-x\right|+\left|x+\frac{9}{7}\right|\ge\left|\frac{3}{4}-x+x+\frac{9}{7}\right|=\left|\frac{57}{28}\right|=\frac{57}{28}\)

Dấu " = " xảy ra \(\Leftrightarrow\left(\frac{3}{4}-x\right)\left(x+\frac{9}{7}\right)\ge0\)

TH1: \(\hept{\begin{cases}\frac{3}{4}-x\le0\\x+\frac{9}{7}\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{3}{4}\le x\\x\le\frac{-9}{7}\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{4}\\x\le\frac{-9}{7}\end{cases}}\)( vô lý )

TH2: \(\hept{\begin{cases}\frac{3}{4}-x\ge0\\x+\frac{9}{7}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{3}{4}\ge x\\x\ge\frac{-9}{7}\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{4}\\x\ge\frac{-9}{7}\end{cases}}\Leftrightarrow\frac{-9}{7}\le x\le\frac{3}{4}\)

\(\Rightarrow28.\left(\left|x-\frac{3}{4}\right|+\left|x+\frac{9}{7}\right|\right)\ge28.\frac{57}{28}=57\)

Dấu " = " xảy ra \(\Leftrightarrow-\frac{9}{7}\le x\le\frac{3}{4}\)

Vậy GTNN của biểu thức đã cho là \(57\)\(\Leftrightarrow-\frac{9}{7}\le x\le\frac{3}{4}\)

a

\(ĐKXĐ:x\in R\)

\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\left(x^4+\frac{1-x^4}{1+x^2}\right)\)

\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\left(x^4-x^2+1\right)\)

\(=\frac{\left(x^2-1\right)\left(x^4-x^2+1\right)}{x^4-x^2+1}-\frac{x^4-x^2+1}{x^2+1}\)

\(=x^2-1-\frac{x^4-x^2+1}{x^2+1}\)

\(=-1+\frac{x^4+x^2-x^4+x^2+1}{x^2+1}\)

\(=\frac{2x^2+1}{x^2+1}-1=\frac{2x^2+1-x^2-1}{x^2+1}=\frac{x^2}{x^2+1}\)

b

Xét \(x>0\Rightarrow M>0\)

Xét \(x=0\Rightarrow M=0\)

Xét \(x< 0\Rightarrow M>0\)

Vậy \(M_{min}=0\) tại \(x=0\)

\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|\)

\(=\left|x+\dfrac{1}{2}\right|+\left|-x-\dfrac{1}{4}\right|+\left|x+\dfrac{1}{3}\right|\)

\(\ge\left|x+\dfrac{1}{2}-x-\dfrac{1}{4}\right|+0=\dfrac{1}{4}\)

Dấu = xảy ra khi \(x=-\dfrac{1}{3}\)