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

Ta có: \(2-\sqrt{3}=\frac{1}{2+\sqrt{3}}\)

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

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

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

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

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

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

Ta có pt ẩn t: \(1+t^2=4t\)

<=> \(t^2-4t+1=0\Leftrightarrow\orbr{\begin{cases}t=2-\sqrt{3}\\t=2+\sqrt{3}\end{cases}}\)

+) Với \(t=2+\sqrt{3}\), ta có: 

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

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

<=> x=2 

Trường hợp còn lại em làm tương tự

bạn ơi khu 7−4√3=(2+√3)2 nó phải là 7−4√3=(2-√3)2 mới đúng chứ?

\(a,9\sqrt{5}+3\sqrt{20}-7\sqrt{45}=9\sqrt{5}+6\sqrt{5}-21\sqrt{5}=-6\sqrt{5}\\ b,\dfrac{2\sqrt{6}+\sqrt{40}}{\sqrt{3}+\sqrt{5}}=\dfrac{2\sqrt{6}+2\sqrt{10}}{\sqrt{3}+\sqrt{5}}\\ =\dfrac{2\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{2\sqrt{2}\left(5-3\right)}{5-3}=2\sqrt{2}\)

Sửa đề `->sqrt{7+4sqrt3}`

`=sqrt{7+4sqrt3}`

`=sqrt{4+2.2.sqrt3+3}`

`=sqrt{(2+sqrt3)^2}`

`=|2+sqrt3|`

`=2+sqrt3`

Đề như trên không đừng à bạn

\(\sqrt{2-\sqrt{3}}=\frac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{2.\left(2-\sqrt{3}\right)}}{\sqrt{2}}=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\frac{\left|\sqrt{3}-1\right|}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{\sqrt{2}.\left(\sqrt{3}-1\right)}{\sqrt{2}.\sqrt{2}}=\frac{\sqrt{2}.\sqrt{3}-\sqrt{2}.1}{\sqrt{2.2}}=\frac{\sqrt{2.3}-\sqrt{2}}{\sqrt{4}}=\frac{\sqrt{6}-\sqrt{2}}{2}\)

`Answer:`

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

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

\(=2\sqrt{3}+2-\sqrt{3}\) (Do `2>\sqrt{3}`)

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

ĐKXĐ: `x>=0;x\ne9`

`(x^2-3)/(sqrtx-3)=((x-sqrt3)(x+sqrt3))/(x+sqrt3)=x-sqrt3`

`a)(x^2-3)/(x+\sqrt3)`

`->` ĐKXĐ : `x>=0;x\ne9`

`=((x-\sqrt3)(x+\sqrt3))/(x+\sqrt3)`

`=(x-\sqrt3)/1`

`=x-\sqrt3`

\(A=2\sqrt{3}+\sqrt{\left(2+\sqrt{3}\right)^2}\\ =2\sqrt{3}+\left|2+\sqrt{3}\right|\\ =2\sqrt{3}+2+\sqrt{3}\\ =3\sqrt{3}+2\)