\(x^2+\frac{9x^2}{\left(x+3\right)^2}=27\)

\(x^2+\frac{3x}{x+3}=27\)

\(\frac{x^2\left(x+3\right)+3x}{x+3}=27\)

\(\frac{x^3+3x^2+3x}{x+3}=27\)

\(x^3+3x^2+3x=27x+81\)

\(x^3+3x^2+3x-27x-81=0\)

\(x^3+3x^2-24x-81=0\)

đến đây bạn có thể làm được rồi

ầy... bỏ zô máy tính giải nghiệm là nhanh nhứt ahihi:))))

\(x^3+9x^2+27x+27=x^3+3x^2+6x^2+18x+9x+27=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)

\(=\left(x^2+6x+9\right)\left(x+3\right)=\left(x+3\right)^2\left(x+3\right)=\left(x+3\right)^3=0\)

=>x+3=0=>x=-3

Đk:\(x\ge-\frac{10}{3}\)

\(pt\Leftrightarrow\left(x^2+6x+9\right)+\left(3x+9\right)-\left(2\sqrt{3x+10}-2\right)=0\)

\(\Leftrightarrow\left(x+3\right)^2+3\left(x+3\right)-2\frac{\left(3x+10\right)-1}{\sqrt{3x+10}+2}=0\)(do \(\sqrt{3x+10}+2>0\) )

\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)+3-2\frac{3}{\sqrt{3x+10}+2}\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)+3-\frac{6}{\sqrt{3x+10}+2}\right]=0\)

Do \(\sqrt{3x+10}+2\ge0\) với mọi x

\(\Rightarrow\frac{6}{\sqrt{3x+10}}+2\le3\)

\(\Rightarrow\left(x+3\right)+3-\frac{6}{\sqrt{3x+10}+2}>0\)(loại)

\(\Rightarrow x+3=0\Leftrightarrow x=-3\)(thỏa mãn)

Vậy pt có nghiệm duy nhất x=-3.

ĐKXĐ : \(x\ge3\)

\(\sqrt{9x-27}-2\sqrt{\frac{x-3}{4}}=1\)

\(\Leftrightarrow\)\(3\sqrt{x-3}-\sqrt{x-3}=1\)

\(\Leftrightarrow\)\(2\sqrt{x-3}=1\)

\(\Leftrightarrow\)\(\sqrt{x-3}=\frac{1}{2}\)

\(\Leftrightarrow\)\(x=\frac{13}{4}\) ( tm đkxđ ) 

...

hết cứu đi mà làm

ĐK: \(x\ne-2;-3;-4;-5;-6\)

\(\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\Leftrightarrow\left(x+2\right)\left(x+6\right)=32\)

\(\Leftrightarrow x^2+8x-20=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

\(...\Leftrightarrow\frac{1}{\left(x+2\right) \left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{18}\Leftrightarrow\frac{x+6}{\left(x+2\right)\left(x+6\right)}-\frac{x+2}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\Rightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{18}\)

\(\Rightarrow\left(x+2\right)\left(x+6\right)=72\)

=> \(x^2+8x-60=0\)

Phân tich đa thức thành nhân tử để tìm x

\(2\sqrt{2\left(x+2\right)}\)+4\(\sqrt{2-x}=\sqrt{9x^2+16}\)

=>\(x=\frac{4\sqrt{2}}{3}\)

`a)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}`

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrtx{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`

`a)A=\sqrt{4+2sqrt3}`

`=\sqrt{3+2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}`

`=sqrt3+1`

`B)1/(2-sqrt3)+1/(2+sqrt3)`

`=(2+sqrt3)/(4-3)+(2-sqrt3)/(4-3)`

`=2+sqrt3+2-sqrt3`

`=4`

`\sqrt{4x-12}+sqrt{x-3}-1/3sqrt{9x-27}=8`

`đk:x>=3`

`pt<=>2sqrt{x-3}+sqrt{x-3}-sqrt{x-3}=8`

`<=>2sqrt{x-3}=8`

`<=>sqrt{x-3}=4`

`<=>x-3=16`

`<=>x=19`

Vậy `S={19}`