\(x^4\ge0;3x^2\ge0=>x^4+3x^2+2\ge0+0+2=2=>A_{min}=2<=>x=0\)

\(x^4\ge0=>x^4+5\ge5=>\left(x^4+5\right)^2\ge5^2=25=>B_{min}=25<=>x=0\)

tick nhé

\(\left(x-1;y-1\right)=\left(a;b\right)\Rightarrow\left\{{}\begin{matrix}a;b>0\\a+b\le2\end{matrix}\right.\)

\(A=\dfrac{\left(a+1\right)^4}{b^2}+\dfrac{\left(b+1\right)^4}{a^2}\ge\dfrac{1}{2}\left[\dfrac{\left(a+1\right)^2}{b}+\dfrac{\left(b+1\right)^2}{a}\right]^2\)

\(A\ge\dfrac{1}{2}\left[\dfrac{\left(a+b+2\right)^2}{a+b}\right]^2\ge\dfrac{1}{2}\left[\dfrac{8\left(a+b\right)}{a+b}\right]^2=32\)

\(A=\left(x-1\right)^2-3\)

Nhận xét :\(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-1\right)^2-3\ge-3\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy \(minA=-3\Leftrightarrow x=1\)

Các câu còn lại làm tương tự nhé

\(đkxđ\Leftrightarrow x\ge4\)

\(P=\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{\frac{16}{x^2}-\frac{8}{x}+1}}\)

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

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

\(=\frac{|x-2|+|x-6|}{|\frac{4}{x}-1|}=\frac{x-2+|x-6|}{|\frac{4}{x}-1|}\)

Dùng bảng xét dấu nha

A = 3 x | 1 - 2x | - 5

Ta co : | 1 - 2x | \(\ge\)0 nen 3 x | 1 - 2x | \(\ge\)0

A = 3 x | 1 - 2x | - 5 \(\ge\)- 5

Vậy min A = -5 \(\Leftrightarrow\)x = \(\frac{1}{2}\)

1 bài thôi . còn lại tương tự

bài cuối dùng BĐT : | a | + | b | \(\ge\)| a + b | nhé

Vậy còn tìm max ạ???

max hả

ta có : | 2 - 5x | \(\ge\)0 nen -|2 - 5x | \(\le\)0

B = \(\frac{3}{4}-\left|2-5x\right|\le\frac{3}{4}\)

Vậy max B = \(\frac{3}{4}\)\(\Leftrightarrow x=\frac{2}{5}\)

\(A=x^4+3x^2+2\)

Mà \(x^4\ge0;3x^2\ge0\) 

Để A có GTNN thì x⁴ = 3x² = 0 => x=  0

Vậy A = 0 + 0 + 2 = 2

KL: A_min = 2 tại x = 0

B = (x⁴ + 5)² có GTNN

Mà x⁴ \(\ge\) 0 => x⁴ = 0 => x = 0

B = 5² = 25

Vậy B_MIN = 25 tại x = 0

C = (x - 1)² + (y + 2)² có GTNN

MÀ \(\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0\)

Nên (x - 1)² = (y + 2)² = 0

=> x = 1 ; y = -2

C = 0 + 0 = 0

Vậy C_MIN = 0 tại x = 1 ; y = -2 

câu 1:MIN=2 khi x=0

câu 2:MIN=25 khi x=0

câu 3 MIN=0 khi x=1 ; y=-2

diều kiện x >= 0

P=\(\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right).\frac{4\sqrt{x}}{3}\)

= \(\frac{x+2-x+\sqrt{x}-1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)

=\(\frac{\sqrt{x}+1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)=\(\frac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)

P=8/9

<=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)

<=> \(3\sqrt{x}=2x-2\sqrt{x}+1\)

<=> \(2x-5\sqrt{x}+2=0\)

<=> \(\left[\begin{array}{nghiempt}x=4\\x=\frac{1}{4}\end{array}\right.\)

vậy x=4 hoặc x=1/4 thì p=8/9

a) \(P=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\left(ĐK:x\ge0;x\ne-1\right)\)

\(=\left[\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]\cdot\frac{4\sqrt{x}}{3}\)

\(=\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)

\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)

\(=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

b) Để P=8/9

\(\Leftrightarrow\)\(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)

\(\Leftrightarrow24\left(x-\sqrt{x}+1\right)=36\sqrt{x}\)

\(\Leftrightarrow24x-24\sqrt{x}+24-36\sqrt{x}=0\)

\(\Leftrightarrow24x-60\sqrt{x}+24=0\)

\(\Leftrightarrow12\left(2x-5\sqrt{x}+2\right)=0\)

\(\Leftrightarrow\left(2x-\sqrt{x}\right)-\left(4\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)-2\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{array}\right.\)