a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)

\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)

\(=4-3\cdot A\)

\(\Leftrightarrow A^3+3A-4=0\)

\(\Leftrightarrow A^3-A+4A-4=0\)

\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)

\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)

\(\Leftrightarrow A=1\)

Mn giúp e với ak

a) \(\sqrt{x^2-6x+9}\)

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

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

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

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

9.

\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\)

\(=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

\(=-\sqrt{5}\)

10.

\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}-\sqrt{5+\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2+\dfrac{2}{3}}+6\sqrt{3}\)

\(=11\sqrt{3}-\sqrt{\dfrac{16}{3}}+\dfrac{9}{2}\sqrt{\dfrac{8}{3}}\)

\(=11\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}\)

\(=\dfrac{29\sqrt{3}}{3}+3\sqrt{6}\)

\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\\ =2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}=\sqrt{5}\)

\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\\ =5\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}+6\sqrt{3}\\ =\dfrac{15\sqrt{3}-4\sqrt{3}+6\sqrt{6}+18\sqrt{3}}{3}\\ =\dfrac{29\sqrt{3}+6\sqrt{6}}{3}\)

\(A=\sqrt[3]{2^3+3.2^2.\sqrt{2}+3.2.\sqrt{2}^2+\sqrt{2}^3}+\sqrt[3]{\sqrt{2}^3-3.\sqrt{2}^2.2+3.\sqrt{2}.2^2-2^3}\)

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

\(A=2+\sqrt{2}+\sqrt{2}-2=2\sqrt{2}\)

\(X=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)

\(\Rightarrow X^3=\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)^3\)

\(\Rightarrow X^3=2+3\sqrt[3]{1-\frac{84}{81}}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)\)

\(\Rightarrow X^3=2-3\sqrt[3]{\frac{1}{27}}.X\)

\(\Rightarrow X^3=2-X\)

\(\Rightarrow X^3+X-2=0\)

\(\Rightarrow\left(X-1\right)\left(X^2+2X+2\right)=0\)

\(\Rightarrow X=1\) (do \(X^2+2X+2=\left(X+1\right)^2+1>0\) \(\forall X\))

a

\(\sqrt{9\left(2-3x\right)^2}=6\\ \Leftrightarrow3\left|2-3x\right|=6\\ \Leftrightarrow\left|2-3x\right|=2\)

Với \(x\le\dfrac{2}{3}\) thì PT trở thành:

\(2-3x=2\\ \Leftrightarrow3x=0\\ \Leftrightarrow x=0\left(nhận\right)\)

Với \(x>\dfrac{2}{3}\) thì PT trở thành:

\(3x-2=2\\ \Leftrightarrow3x=4\\ \Leftrightarrow x=\dfrac{4}{3}\left(nhận\right)\)

b

ĐK: \(x\ge-\dfrac{3}{2}\)

\(\sqrt{4x^2-9}=2\sqrt{2x+3}\\ \Leftrightarrow\sqrt{\left(2x\right)^2-3^2}=2\sqrt{2x+3}\\ \Leftrightarrow\sqrt{2x-3}.\sqrt{2x+3}-2\sqrt{2x+3}=0\\ \Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=0\\\sqrt{2x-3}-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\2x-3=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(nhận\right)\\x=\dfrac{7}{2}\left(nhận\right)\end{matrix}\right.\)

c

ĐK: \(x\ge3\)

\(\sqrt{10\left(x-3\right)}=\sqrt{20}\\ \Leftrightarrow10\left(x-3\right)=20\\ \Leftrightarrow x-3=2\\ \Leftrightarrow x=5\left(nhận\right)\)

d

\(\sqrt{x^2+6x+9}=3x-6\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-6\\ \Leftrightarrow\left|x+3\right|=3x-6\)

Với \(x\ge-3\) thì PT trở thành:

\(x+3=3x-6\\ \Leftrightarrow x+3-3x+6=0\\ \Leftrightarrow-2x+9=0\\ \Leftrightarrow x=\dfrac{9}{2}\left(nhận\right)\)

Với \(x< -3\) thì PT trở thành:

\(-x-3=3x-6\\ \Leftrightarrow-x-3-3x+6=0\\ \Leftrightarrow-2x+3=0\\ \Leftrightarrow x=\dfrac{3}{2}\left(loại\right)\)

a: \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)

\(\Leftrightarrow A^3=9+4\sqrt{5}+9-4\sqrt{5}+3\cdot A\)

=>A^3-3A-18=0

=>A=3

b: \(B=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

=>\(B^3=5\sqrt{2}+7-5\sqrt{2}+7+3B\)

=>B^3-3B-14=0

=>B=2,82

c: \(C^3=20+14\sqrt{2}-14\sqrt{2}+20-6C\)

=>C^3+6C-40=0

=>C=2,84

Em kiểm tra lại đề câu đầu.

b.

\(\sqrt{3x^2-6x+1}=1\)

\(\Leftrightarrow3x^2-6x+1=1\)

\(\Leftrightarrow3x\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

c.

\(\sqrt{4x^2-4x+1}=9\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=9\)

\(\Leftrightarrow\left|2x-1\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

\(x=\dfrac{3\sqrt[3]{8-3\sqrt{5}}}{\sqrt[3]{57}}.\sqrt[3]{8+3\sqrt{5}}=\dfrac{3\sqrt[3]{\left(8-3\sqrt{5}\right)\left(8+3\sqrt[]{5}\right)}}{\sqrt[3]{57}}=\sqrt[3]{\dfrac{19}{57}}=\dfrac{1}{\sqrt[3]{3}}\)

\(y=\dfrac{\left(\sqrt[3]{3}+\sqrt[4]{2}\right)\left(\sqrt[3]{3}-\sqrt[4]{2}\right)}{\sqrt[3]{3}+\sqrt[4]{2}}+\dfrac{\left(\sqrt[4]{2}-\sqrt[3]{81}\right)\left(\sqrt[4]{2}+\sqrt[3]{81}\right)}{\sqrt[4]{2}-\sqrt[3]{81}}\)

\(=\sqrt[3]{3}-\sqrt[4]{2}+\sqrt[4]{2}+\sqrt[3]{81}=\sqrt[3]{3}+3\sqrt[3]{3}=4\sqrt[3]{3}\)

\(T=xy=\dfrac{4\sqrt[3]{3}}{\sqrt[3]{3}}=4\)

a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4

=>2*căn(x+5)=4

=>căn (x+5)=2

=>x+5=4

=>x=-1

b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

=>2*căn x-1=16

=>x-1=64

=>x=65

c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)

TH₁: \(x\ge3\)

\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)

TH₂: \(2\le x< 3\)

\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH₃: \(0\le x< 2\)

\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

TH₄: \(x< 0\)

\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)