ĐKXĐ \(2x^2-1\ge0\)

PT <=> \(10x^2+3x-6=2\left(3x+1\right)\sqrt{2x^2-1}\)

<=> \(\left(3x+1\right)\left(x+2-2\sqrt{2x^2-1}\right)+7x^2-4x-8=0\)

<=> \(\left(3x+1\right).\frac{\left(x+2\right)^2-4\left(2x^2-1\right)}{x+2+2\sqrt{2x^2-1}}+7x^2-4x-8=0\)

<=> \(\left(3x+1\right).\frac{-7x^2+4x+8}{x+2+2\sqrt{2x^2-1}}+7x^2-4x-8=0\)

<=> \(\orbr{\begin{cases}-7x^2+4x+8=0\left(1\right)\\3x+1=x+2+2\sqrt{2x^2-1}\left(2\right)\end{cases}}\)

Giải (2)

\(2x-1=2\sqrt{2x^2-1}\)

<=> \(4x^2+4x-5=0\)với \(x\ge\frac{1}{2}\)

=> \(x=\frac{-1+\sqrt{6}}{2}\)

GIải (1)

\(x=\frac{2\pm2\sqrt{15}}{7}\)thỏa mãn ĐKXĐ

Vậy \(S=\left\{\frac{2\pm2\sqrt{15}}{7},\frac{-1+\sqrt{6}}{2}\right\}\)

ĐKXĐ \(x\ge\frac{1}{2}\)

Đặt \(\sqrt{x^2+2x}=a,\sqrt{2x-1}=b\left(a,b\ge0\right)\)

=> \(3a^2-b^2=3x^2+4x+1\)

Khi đó PT <=> 

\(a+b=\sqrt{3a^2-b^2}\)

=> \(a^2+2ab+b^2=3a^2-b^2\)

=> \(a^2-ab-b^2=0\)

=> \(a=\frac{1+\sqrt{5}}{2}.b\)

=> \(x^2+2x=\frac{6+2\sqrt{5}}{4}.\left(2x-1\right)\)

=> \(x=\frac{1+\sqrt{5}}{2}\)thỏa mãn ĐKXĐ

Vậy \(x=\frac{1+\sqrt{5}}{2}\)

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Học tốt!!!!!!!!!!

P/s : có hình

cám ơn bạn

Áp dụng bđt Mincopxki \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\) ta được

\(VT\ge\sqrt{\left(x+y\right)^2+\left(\frac{1}{x}+\frac{1}{y}\right)^2}+\sqrt{z^2+\frac{1}{z^2}}\)

        \(\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)

Áp dụng bđt Cô-si có

\(\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\ge9\sqrt[3]{\left(xyz\right)^2}+\frac{9}{\sqrt[3]{\left(xyz\right)^2}}\)

Đặt \(\sqrt[3]{\left(xyz\right)^2}=t\)

\(\Rightarrow0\le t=\sqrt[3]{\left(xyz\right)^2}\le\left(\frac{x+y+z}{3}\right)^2=\frac{1}{4}\)

Khi đó \(VT\ge\sqrt{9t+\frac{9}{t}}=\sqrt{3\left(48t+\frac{3}{t}-45t\right)}\ge\sqrt{3\left(2.\sqrt{3.48}-\frac{45}{4}\right)}=\frac{3\sqrt{17}}{2}\)

Nếu không dùng bđt đó làm ra ko bạn

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

Đề có nhầm không????

Học tốt!!!!!!!!!!!!!!

đề đúng mà

em mới lớp 6 nên chưa biết

\(B=\sqrt{x-4\sqrt{x}+4}+\sqrt{x-6\sqrt{x}+9}\)

    \(=\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}\)

    \(=\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|\ge\left|\sqrt{x}-2+3-\sqrt{x}\right|=1\)

Dấu "=" <=> 4 < x < 9

a, \(A=x^2-x\sqrt{y}-2x\sqrt{y}+2y\)

\(=x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)

\(=\left(x-2\sqrt{y}\right)\left(x-\sqrt{y}\right)\)

\(a,\)\(A=x^2-3x\sqrt{y}+2y\)

\(=x^2-2x\sqrt{y}-x\sqrt{y}+2y\)

\(=x\left(x-2\sqrt{y}\right)-\sqrt{y}\left(x-2\sqrt{y}\right)\)

\(=\left(x-\sqrt{y}\right)\left(x-2\sqrt{y}\right)\)

\(b,\)Ta có : \(x=\frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\)

\(y=\frac{1}{9+4\sqrt{5}}=\frac{9-4\sqrt{5}}{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}=\frac{9-4\sqrt{5}}{81-80}=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)

\(\Rightarrow A=\left[\sqrt{5}+2-\sqrt{\left(\sqrt{5}-2\right)^2}\right]\left[\sqrt{5}+2-2\sqrt{\left(\sqrt{5}-2\right)^2}\right]\)

\(=\left(\sqrt{5}+2-\sqrt{5}-2\right)\left(\sqrt{5}+2-2\sqrt{5}+4\right)\)

\(=4\left(6-\sqrt{5}\right)\)

\(=24-4\sqrt{5}\)