trần đắc lợi lần sau nhớ gõ latex nha bạn, như này người làm dễ bị sai đề lắm

\(a\sqrt{b-1}+b\sqrt{a-1}\)

Áp dụng AM-GM :

\(a\sqrt{b-1}+b\sqrt{a-1}\)

\(=a\sqrt{1\cdot\left(b-1\right)}+b\sqrt{1\cdot\left(a-1\right)}\le a\cdot\frac{1+b-1}{2}+b\cdot\frac{1+a-1}{2}\)

\(=\frac{ab}{2}+\frac{ab}{2}=ab\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=2\)

cảm ơn nha

\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\)\(\Rightarrow\frac{|4x+3|}{|x+1|}=3\)

\(\Rightarrow|4x+3|=3|x+1|\)

TH1 : \(4x+3=-3\left(x+1\right)\)

\(\Rightarrow4x+3=-3x-3\)

\(\Rightarrow7x=-6\Leftrightarrow x=-\frac{6}{7}\)

Th2 : \(4x+3=3\left(x+1\right)\)

\(\Rightarrow4x+3=3x+3\)

\(\Rightarrow x=0\)

Vậy \(x\in\left\{0;-\frac{6}{7}\right\}\)

\(\frac{\sqrt{4x+3}}{\sqrt{x+1}}=3\Rightarrow\frac{|4x+3|}{|x+1|}=9\)

\(\Rightarrow|4x+3|=9|x+1|\)

\(TH1:|4x+3|=-9|x+1|\)

\(\Rightarrow4x+3=-9x-9\Rightarrow13x=-12\Rightarrow x=-\frac{12}{13}\)

\(TH2:|4x+3|=9|x+1|\)

\(\Rightarrow4x+3=9x+9\)

\(\Rightarrow5x=-6\Rightarrow x=\frac{-6}{5}\)

\(A=x^2-8x+3\)

\(=x^2-8x+16-13\)

\(=\left(x-4\right)^2-13\)

\(A_{min}=-13\Leftrightarrow\left(x-4\right)^2=0\)

\(\Rightarrow x-4=0\Leftrightarrow x=4\)

Vậy \(A_{min}=-13\Leftrightarrow x=4\)

Ta có:

   A = x² - 8x + 3 = (x² - 8x + 16) - 13 = (x - 4)² - 13

Ta luôn có: (x - 4)² \(\ge\)0 \(\forall\)x

=> (x - 4)² - 13 \(\ge\)-13 \(\forall\)x

hay A \(\ge\)-13 \(\forall\)x

Dấu "=" xảy ra khi : x - 4 = 0 <=> x = 4

Vậy Min A = -13 tại x = 4

\(Pt\Leftrightarrow\left(x^3+4x^2+3x\right)+3\left(x+2-\sqrt[3]{2x^2+9x+8}\right)=0.\)

\(\Leftrightarrow\left(x^3+4x^2+3x\right)+3.\frac{\left(x+2\right)^3-2x^2-9x-8}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{2x^2+9x+8}+\sqrt[3]{\left(2x^2+9x+8\right)^2}}=0\)

\(\Leftrightarrow\left(x^3+4x^2+3x\right)+3.\frac{x^3+4x^2+3x}{MS}=0\Leftrightarrow\left(x^3+4x^2+3x\right)\left(1+\frac{3}{MS}\right)=0\)

Dễ thấy MS >0 \(\Rightarrow PT\Leftrightarrow x^3+4x^2+x=0\Leftrightarrow x\left(x^2+4x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+4x+3=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\end{cases}}\)

\(\Rightarrow Pt\Leftrightarrow x^3+4x^2+3x=0\Leftrightarrow x\left(x^2+4x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+4x+3=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\end{cases}}\)\(\Rightarrow PT\Leftrightarrow x^3+4x^2+3x=0\)<=>\(x\in\left\{-3;-1;0\right\}\)

\(A=15x^2+\left(x^2-4x+4\right).\)

\(=15x^2+\left(x-2\right)^2\)

Vì \(15x^2\ge0\)với \(\forall x\)và \(\left(x-2\right)^2\ge0\)với \(\forall x\)

\(\Rightarrow A>0\)với mọi x 

\(B=x^2\left(x^2+6x+9\right)\)

\(=x^2\left(x+3\right)^2\)

Vì \(x^2\ge0\)với mọi x và \(\left(x+3\right)^2\ge0\)với mọi x 

\(\Rightarrow x^2\left(x+3\right)^2>0\)với mọi x

\(A=15x^2+\left(x^2-4x+4\right)\)

\(A=15x^2+x^2-4x+4\)

\(A=16x^2-4x+4\)

\(A=16x^2-4x+\frac{1}{4}+\frac{15}{4}\)

\(A=\left(16x^2-4x+\frac{1}{4}\right)+\frac{15}{4}\)

\(A=\left(4x+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)

Vậy A không âm

\(x^2+9x-400=0\)

\(\Leftrightarrow x^2-16x+25x-400=0\)

\(\Leftrightarrow x\left(x-16\right)+25\left(x-16\right)=0\)

\(\Leftrightarrow\left(x-16\right)\left(x+25\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=16\\x=-25\end{cases}}\)

\(a=1;b=9;c=-400\)

\(\Delta=b^2-4ac=9^2-4.1.\left(-400\right)=1681>0\)

Phương trình có 2 nghiệm phân biệt 

\(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-9+\sqrt{1681}}{2.1}=16\)

\(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-9-\sqrt{1681}}{2.1}=-25\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{x\sqrt{x}-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)}{\sqrt{x}^3-8}-\frac{\left(x-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}^3-8}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right)\)\(:\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\frac{\sqrt{x}^3+2x+4\sqrt{x}-\sqrt{x}^3+2x+3\sqrt{x}-6-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}.\frac{\left(x+2\sqrt{x}+4\right)}{\sqrt{x}+7}\)

\(=\)\(\frac{\left(4x-16\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}=\frac{4\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

Sai đề không ?

A= \(\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)-\left(x-3\right)\left(\sqrt{x}-2\right)-7\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\right)\)     .  \(\frac{x+2\sqrt{x}+4}{\sqrt{x}+7}\)

= \(\frac{x\sqrt{x}+2x+4\sqrt{x}-x\sqrt{x}+3\sqrt{x}-6+2x-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

= \(\frac{4x-16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

=\(\frac{4\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

= \(\frac{4\left(\sqrt{x}+2\right)}{\sqrt{x}+7}\)

= \(\frac{4\sqrt{x}+8}{\sqrt{x}+7}\)

#mã mã#

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

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

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

b, \(A>A^2\Rightarrow\frac{1}{2\sqrt{x}}>\left(\frac{1}{2\sqrt{x}}\right)^2\Rightarrow\frac{1}{2\sqrt{x}}>\frac{1}{4x}\Rightarrow\frac{1}{2\sqrt{x}}-\frac{1}{4x}>0\Rightarrow\frac{2\sqrt{x}-1}{4x}>0\)

\(2\sqrt{x}-1>0\);\(4x>0\)

\(\Rightarrow x>0\)thì \(A>A^2\)