Giải pt sau (bằng 3 cách TẠO LŨY THỪA DƯỚI DẤU CĂN, ĐẶT ẨN PHỤ, DÙNG BĐT): \(x^2+6x-3=4x\sqrt{2x-1}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Với x > 0 ; x \(\ne\)9
a, \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}+\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}\right):\left(\frac{3\sqrt{x}+1+2\left(\sqrt{x}-3\right)}{x-3\sqrt{x}}\right)\)
\(=\left(\frac{-3\sqrt{x}-9}{x-9}\right):\left(\frac{5\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)=\frac{-3}{\sqrt{x}-3}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{5\left(\sqrt{x}-1\right)}=\frac{-3\sqrt{x}}{5\left(\sqrt{x}-1\right)}\)
b, Ta có : \(B< 0\Rightarrow\frac{-3\sqrt{x}}{5\left(\sqrt{x}-1\right)}< 0\Rightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)
Kết hợp vói đk vậy x > 1 ; x \(\ne\)9
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3\sqrt{3}-8\sqrt{3}+15\sqrt{3}=10\sqrt{3}\)
\(B=\left|\sqrt{5}-2\right|-\left|\sqrt{5}-3\right|=\sqrt{5}-2-\left(\sqrt{5}-3\right)=\sqrt{5}-2-\sqrt{5}+3=1\)
\(C=\left|2\sqrt{3}+1\right|+\left|2\sqrt{3}-5\right|=2\sqrt{3}+1+5-2\sqrt{3}=6\)
\(D=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}-2\right|-\left|\sqrt{5}-3\right|=\sqrt{5}-2-\sqrt{5}+3=1\)
\(E=\frac{4\left(\sqrt{5}+2\right)}{5-4}-\frac{32\left(\sqrt{5}-1\right)}{5-1}=4\sqrt{5}+8-\frac{32\sqrt{5}-32}{4}=4\sqrt{5}+8-8\sqrt{5}+8=16-4\sqrt{5}\)
\(M=\frac{10\left(3\sqrt{2}+4\right)}{18-16}+\frac{28\left(3\sqrt{2}-2\right)}{18-4}=\frac{30\sqrt{2}+40}{2}+\frac{84\sqrt{2}-56}{14}=15\sqrt{2}+20+6\sqrt{2}-4=16+21\sqrt{2}\)
\(x^2+6x-3=4x\sqrt{2x-1}\left(1\right)\) ĐK: \(x\ge\frac{1}{2}\)
Đặt \(\sqrt{2x-1}=a\ge0\)
\(\Rightarrow6x-3=3a^2\)
=> (1) <=> x^2 +3a^2 = 4ax
<=> x^2 -4ax +3a^2 =0
<=> x^2 -ax - 3ax + 3a^2 =0
<=> x(x-a) -3a(x-a) =0
<=> (x-a) ( x-3a ) =0
\(\Leftrightarrow\orbr{\begin{cases}x=a\\x=3a\end{cases}}\)
TH1: x=a
\(\Rightarrow x=\sqrt{2x-1}\)\(\left(x\ge0\right)\)
\(\Leftrightarrow x^2=2x-1\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
<=> x=1 (tm)
TH2: x= 3a
\(\Rightarrow x=3\sqrt{2x-1}\left(x\ge0\right)\)
\(\Leftrightarrow x^2=18x-9\)
\(\Leftrightarrow x^2-18x+9=0\)
\(\Delta=288\)
=> pt có 2 nghiệm pb \(\orbr{\begin{cases}x=\frac{18+12\sqrt{2}}{2}=9+6\sqrt{2}\left(tm\right)\\x=\frac{18-12\sqrt{2}}{2}=9-6\sqrt{2}\left(tm\right)\end{cases}}\)
Vậy ...