Đặt \(a=\sqrt[3]{\frac{12+\sqrt{135}}{3}};b=\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)

=>\(a^3+b^3=\frac{12+\sqrt{135}+12-\sqrt{135}}{3}=\frac{24}{3}=8\) ; \(ab=\sqrt[3]{\frac{\left(12+\sqrt{135}\right)\left(12-\sqrt{135}\right)}{3\cdot3}}=\sqrt[3]{\frac{144-135}{9}}=1\)

\(x=\frac13\left(1+\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\right)\)

=>\(3x=1+\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)

=>\(3x-1=\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)

=>\(\left(3x-1\right)=a+b\)

=>\(\left(3x-1\right)^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)=a^3+b^3+3ab\left(3x-1\right)\)

=>(3x-1)^3=8+3*1*(3x-1)=8+3(3x-1)

=>\(27x^3-27x^2+9x-1=9x+5\)

=>\(27x^3-27x^2=6\)

=>\(9x^3-9x^2=2\)

\(M=\left(9x^3-9x^2-3\right)^2\)

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

Bo thi:>

+ đk x > 0 , x khác 1

Mọi người ơi, giúp em với ạ!

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}\)

a) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=-1+3\sqrt{5}\)

b) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}+1+\sqrt{3}=2-\sqrt{3}+1+\sqrt{3}=3\)

a: \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)

\(=3\sqrt{5}-1\)

b: \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=2-\sqrt{3}+\sqrt{3}+1\)

=3

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

\(=\frac{1}{2}\left(2+\sqrt{\left(\sqrt{3}+1\right)^2}\right)\left(2-\sqrt{\left(\sqrt{3}-1\right)^2}\right)\)

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

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

\(=>x^3=(\sqrt[3]{2\left(\sqrt{3}+1\right)}-\sqrt[3]{2\left(\sqrt{3}-1\right)})^3\)

\(x^3=2\left(\sqrt{3}+1\right)-3.\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]^2.\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]\)

+\(3\left[\sqrt[3]{2\left(\sqrt{3}-1\right)}\right]^2\left[\sqrt[3]{2\left(\sqrt{3}+1\right)}\right]-2\left(\sqrt{3}-1\right)\)

\(x^3=\)

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

\(x^3=4-3.\left[\sqrt[3]{4\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\right].\)\(x\)

\(x^3=4-3\left[\sqrt[3]{4\left(3-1\right)}\right].x\)

\(x^3=4-3.2x\)

\(x^3=4-6x\)

thay \(x^3=4-6x\) vào A=>\(A=\left(4-6x+6x-5\right)^{2009}=\left(-1\right)^{2009}=-1\)

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Câu 1,2 bạn đã đăng và có lời giải rồi

Câu 3:

\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)