\(\sqrt{2x+3}+\sqrt{x-3}=3x-6+2\sqrt{2x^2-3x-9}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(S=\left\{\frac{2\pm2\sqrt{15}}{7},\frac{-1+\sqrt{6}}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(x=\frac{1+\sqrt{5}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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P/s : có hình
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(VT\ge\sqrt{9t+\frac{9}{t}}=\sqrt{3\left(48t+\frac{3}{t}-45t\right)}\ge\sqrt{3\left(2.\sqrt{3.48}-\frac{45}{4}\right)}=\frac{3\sqrt{17}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(PT\Leftrightarrow x^2-\sqrt{2}\left(x^2+2x-1\right)=0\)
Đề có nhầm không????
Học tốt!!!!!!!!!!!!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\sqrt{x-4\sqrt{x}+4}+\sqrt{x-6\sqrt{x}+9}\)
\(=\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}\)
\(=\left|\sqrt{x}-2\right|+\left|3-\sqrt{x}\right|\ge\left|\sqrt{x}-2+3-\sqrt{x}\right|=1\)
Dấu "=" <=> 4 < x < 9
ĐKXĐ \(x\ge3\)
Đặt \(\sqrt{2x+3}+\sqrt{x-3}=t\left(t\ge0\right)\)
=> \(t^2=3x+2\sqrt{2x^2-3x-9}\left(1\right)\)
Khi đó Pt
<=> \(t=t^2-6\)
=> \(t^2-t-6=0\)
Mà \(t\ge0\)
=> \(t=3\)
(1) <=> \(9-3x=2\sqrt{2x^2-3x-9}\)
=> \(9-3x\ge0\Rightarrow x\le3\)
Kết hợp với ĐKXĐ
=> \(x=3\)
Vậy x=3