\(A=\dfrac{\left(2-\sqrt[3]{x}\right)\left(4+2\sqrt[3]{x}+\sqrt[3]{x^2}\right)}{2+\sqrt[3]{x}}:\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}+\dfrac{\sqrt[3]{x^2}-2\sqrt[3]{x}+2\sqrt[3]{x}}{\sqrt[3]{x}-2}.\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x}+2\right)}{\sqrt[3]{x}\left(\sqrt[3]{x}+2\right)}\)

\(=\dfrac{\left(2-\sqrt[3]{x}\right)\left(4+2\sqrt[3]{x}+\sqrt[3]{x^2}\right)}{2+\sqrt[3]{x}}.\dfrac{2+\sqrt[3]{x}}{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}+\dfrac{\sqrt[3]{x}.\sqrt[3]{x}}{\sqrt[3]{x}-2}.\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x}+2\right)}{\sqrt[3]{x}\left(\sqrt[3]{x}+2\right)}\)

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

\(\dfrac{8-x}{2+\sqrt[3]{x}}:\left(2+\dfrac{\sqrt[3]{x^2}}{2+\sqrt[3]{x}}\right)+\left(\sqrt[3]{x}+\dfrac{2\sqrt[3]{x}}{\sqrt[3]{x}-2}\right).\left(\dfrac{\sqrt[3]{x^2}-4}{\sqrt[3]{x^2}+2\sqrt[3]{x}}\right)\)

\(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{2+\sqrt[3]{x}}:\left(\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}\right)+\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x}-2}.\left(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x}+2\right)}{\sqrt[3]{x}\left(\sqrt[3]{x}+2\right)}\right)\)

\(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{2+\sqrt[3]{x}}.\dfrac{2+\sqrt[3]{x}}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x}-2}.\dfrac{\sqrt[3]{x}-2}{\sqrt[3]{x}}\)

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

4) Ta có: \(\left(x+3\right)\cdot\sqrt{10-x^2}=x^2-x-12\)

\(\Leftrightarrow\left(x+3\right)\cdot\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\10-x^2=x^2-8x+16\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-8x+16-10+x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x^2-8x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\2\left(x^2-4x+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\\x=3\end{matrix}\right.\)

a: =>\(x^2\cdot2\sqrt{2}+x\left(2+2\sqrt{2}\right)+4=0\)

\(\text{Δ}=\left(2\sqrt{2}+2\right)^2-4\cdot2\sqrt{2}\cdot4=12-24\sqrt{2}< 0\)

=>PTVN

b: 

\(\Leftrightarrow2x^2+2x+\sqrt{3}-x^2+2\sqrt{3}x+\sqrt{3}=0\)

=>\(x^2+x\left(2\sqrt{3}+2\right)+2\sqrt{3}=0\)

\(\text{Δ}=\left(2\sqrt{3}+2\right)^2-4\cdot2\sqrt{3}=16>0\)

PT có hai nghiệm là;

\(\left\{{}\begin{matrix}x_1=\dfrac{-2\sqrt{3}-2-4}{2}=-\sqrt{3}-3\\x=\dfrac{-2\sqrt{3}-2+4}{2}=-\sqrt{3}+1\end{matrix}\right.\)

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge2\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=2\)

3.

ĐKXĐ: \(x\ge-1\)

\(x^2+x-12+12\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\dfrac{12\left(x-3\right)}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4+\dfrac{12}{\sqrt{x+1}+2}\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

từ dòng cuối là sai rồi bạn à

Bạn bỏ dòng cuối đi còn lại đúng rồi

Ở tử đặt nhân tử chung căn x chung  rồi lại đặt căn x +1 chung

Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra 

rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)

cảm ơn bạn nha 

\(=\dfrac{\left(2-\sqrt[3]{x}\right)\left(4+2\sqrt[3]{x}+\sqrt[3]{x^2}\right)}{2-\sqrt[3]{x^2}}:\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}+\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x}-2}\cdot\dfrac{\sqrt[3]{x^2}-4}{\sqrt[3]{x^2}+2\sqrt[3]{x}}\)

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

\(=1+\sqrt[3]{x}\)

a.

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

ĐKXĐ: \(x\ge1\)

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

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

\(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)

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

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\)

b.

\(\sqrt{4x^2-20x+25}=5-2x\)

\(\Leftrightarrow\sqrt{\left(2x-5\right)^2}=5-2x\)

\(\Leftrightarrow\left|5-2x\right|=5-2x\)

\(\Leftrightarrow5-2x\ge0\)

\(\Leftrightarrow x\le\dfrac{5}{2}\)

c.

ĐKXĐ: \(x\ge3\)

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

\(\Rightarrow x^2-x-6=x-3\)

\(\Leftrightarrow x^2-2x-3=0\Rightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=3\end{matrix}\right.\)

d.

ĐKXĐ: \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)

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

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

\(\Leftrightarrow x^2=3\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\) (thỏa mãn)

a.

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x-2}=b\end{matrix}\right.\) ta được:

\(2a^2-b^2=ab\)

\(\Leftrightarrow\left(a-b\right)\left(2a+b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=-b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\8a^3=-b^3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(vô-nghiệm\right)\\8\left(x+2\right)=-\left(x-2\right)\end{matrix}\right.\)

\(\Leftrightarrow x=-\dfrac{14}{9}\)

b.

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\)

\(\Rightarrow a^2+4b^2=5ab\)

\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\a^3=64b^3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}65+x=65-x\\65+x=64\left(65-x\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

c.

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x+1}=b\end{matrix}\right.\)

\(\Rightarrow a+b=1+ab\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=1\\b^3=1\end{matrix}\right.\)

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

\(\Leftrightarrow...\)