a) Ta có : \(A=\left|x+1\right|+\left|y-2\right|\)

\(\ge\left|x+1+y-2\right|\)

\(=\left|x+y-1\right|=\left|5-1\right|=\left|4\right|=4\)

Dấu "=" xảy ra <=> (x + 1)(y - 2) \(\ge\)0

Vậy Min A = 4 <=>  (x + 1)(y - 2) \(\ge\)0

\(C=\left(2x+1\right)\left(x-5\right)\)

\(=2x^2-9x-5\)

=\(2.\left(x^2-\frac{9}{2}x-\frac{5}{2}\right)\)

\(=2.\left(x^2-2.x.\frac{9}{4}+\frac{81}{16}-\frac{81}{16}\right)\)

\(=2.\left[\left(x-\frac{9}{4}\right)^2-\frac{81}{16}\right]\)

\(=2.\left(x-\frac{9}{4}\right)^2-\frac{81}{8}\)

Ta có: \(2\left(x-\frac{9}{4}\right)^2\ge0\) với mọi x

\(2\left(x-\frac{9}{4}\right)^2-\frac{81}{8}\ge\frac{-81}{8}\)

hay \(C\ge\frac{-81}{8}\)

- Dấu " = " xảy ra khi và chỉ khi: \(x-\frac{9}{4}=0\Leftrightarrow x=\frac{9}{4}\)

Vậy GTNN của \(C=\frac{-81}{4}\)<=> \(x=\frac{9}{4}\)

\(C=\dfrac{5}{3-\left(4x+1\right)^2}\)

Điều kiện xác định khi 

\(3-\left(4x+1\right)^2\ne0\Leftrightarrow\left[{}\begin{matrix}4x+1\ne\sqrt[]{3}\\4x+1\ne-\sqrt[]{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{\sqrt[]{3}-1}{4}\\x\ne\dfrac{-\sqrt[]{3}-1}{4}\end{matrix}\right.\)

Ta có :

\(\left(4x+1\right)^2\ge0,\forall x\)

\(\Leftrightarrow3-\left(4x+1\right)^2\le3\)

\(\Leftrightarrow C=\dfrac{5}{3-\left(4x+1\right)^2}\ge\dfrac{5}{3}\)

Vậy \(GTNN\left(C\right)=\dfrac{5}{3}\left(tạix=-\dfrac{1}{4}\right)\)

\(B=\left(2x\right)^2+2\left(y-1\right)^2-5\)

vì \(\left\{{}\begin{matrix}\left(2x\right)^2\ge0,\forall x\\2\left(y-1\right)^2\ge0,\forall y\end{matrix}\right.\)

\(\Rightarrow B=\left(2x\right)^2+2\left(y-1\right)^2-5\ge-5\)

Dấu "=" xảy tại khi

\(\left\{{}\begin{matrix}2x=0\\2\left(y-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

Vậy \(GTNN\left(B\right)=-5\left(tạix=0;y=1\right)\)

a) x² - 2x + 5 = (x - 1)² + 4 >= 4

Min là 4 khi x = 1

Câu 1 :

\(B=\left|3x-5\right|+\left|2-3x\right|\ge\left|3x-5+2-3x\right|=\left|-3\right|=3\)

Dấu "=" xảy ra  

TH1: \(\Leftrightarrow\hept{\begin{cases}3x-5>0\\2-3x>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>\frac{5}{3}\\x< \frac{2}{3}\end{cases}\Rightarrow}\frac{5}{3}< x< \frac{2}{3}\left(\text{loại}\right)}\)

TH2: \(\Leftrightarrow\hept{\begin{cases}3x-5< 0\\2-3x< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< \frac{5}{3}\\x>\frac{2}{3}\end{cases}\Rightarrow}\frac{2}{3}< x< \frac{5}{3}\left(\text{thỏa mãn}\right)}\)

Vậy Bmin = 3 <=> 2/3 < x < 5/3 

Câu 2 :

\(C=\left|2x-20\right|-\left|2x+3\right|\le\left|2x-20-2x-3\right|=\left|-23\right|=23\)

Dấu "=" xảy ra 

TH1 : \(\Leftrightarrow\hept{\begin{cases}2x-20>0\\2x+3>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>10\\x>\frac{-3}{2}\end{cases}}\Rightarrow x>10\)

TH2: \(\Leftrightarrow\hept{\begin{cases}2x-20< 0\\2x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 10\\x< \frac{-3}{2}\end{cases}\Rightarrow}}x< \frac{-3}{2}\)

Vậy Cmax = 23 <=> 2 t/h ( ko chắc )

\(B=\left|3x-5\right|+\left|2-3x\right|\ge\left|3x-5+2-3x\right|=\left|-5+2\right|=3\)

Dấu "=" xảy ra \(\Leftrightarrow\left(3x-5\right)\left(2-3x\right)\ge0\)

                         \(\Leftrightarrow\hept{\begin{cases}3x-5\ge0\\2-3x\le0\end{cases}}\) hoặc   \(\hept{\begin{cases}3x-5\le0\\2-3x\ge0\end{cases}}\)

 Giải ra ta được: \(\Leftrightarrow\frac{2}{3}\le x\le\frac{5}{3}\)

Vậy B_min = 3 khi và chỉ khi \(\frac{2}{3}\le x\le\frac{5}{3}\)

_{\(C=\left|2x-20\right|-\left|2x+3\right|\le\left|2x-20-2x-3\right|=\left|-20-3\right|=23\)}

Dấu "=" xảy ra <=> \(\orbr{\begin{cases}2x-20\ge2x+3\ge0\\2x-20\le2x+3\le0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\ge10;x\ge\frac{-3}{2}\\x\le10;x\le\frac{-3}{2}\end{cases}}\)

Vậy C_max = 17 khi và chỉ khi ....

tao chịu

mk cx chịu lun

\(C=\left(2x+1\right)\left(x-5\right)=2x^2-9x-5\)

\(=2\left(x^2-2.\frac{9}{4}x+\frac{9^2}{4^2}\right)-5-2.\frac{9^2}{4^2}\)

\(=2\left(x-\frac{9}{4}\right)^2-\frac{121}{8}\ge-\frac{121}{8}\)

Dấu bằng xảy ra khi x = 9/4.