Kiểm tra lại đề bài đi em, chỗ CMR đó

Đặt \(\sqrt[3]{x^2}=m\Leftrightarrow x^2=m^3;\sqrt[3]{y^2}=n\Leftrightarrow y^2=n^3\)

Thay vào biểu thức:

\(\Leftrightarrow\sqrt{m^3+m^2n}+\sqrt{n^3+mn^2}=a\\ \Leftrightarrow m^3+n^3+mn\left(m+n\right)+2\sqrt{\left(m^3+m^2n\right)\left(n^3+mn^2\right)}=a^2\\ \Leftrightarrow m^3+n^3+mn\left(m+n\right)+2\sqrt{m^2n^2\left(m+n\right)}=a^2\\ \Leftrightarrow m^3+n^3+3mn\left(m+n\right)=a^2\\ \Leftrightarrow\left(m+n\right)^3=a^2\\ \Leftrightarrow m+n=\sqrt[3]{a^2}\\ \Leftrightarrow\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)

Em chắc chắn là đề bài đúng chứ? Trước khi nhìn kĩ lại?

1.

\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)

hết tối giải rồi

Đặt \(\left(\sqrt[3]{x^2};\sqrt[3]{y^2}\right)=\left(X;Y\right)\ge0\)

\(\Rightarrow\sqrt{X^3+X^2Y}+\sqrt{Y^3+XY^2}=a\)

\(\Leftrightarrow X\sqrt{X+Y}+Y\sqrt{X+Y}=a\)

\(\Leftrightarrow\left(X+Y\right)\sqrt{X+Y}=a\)

\(\Leftrightarrow\sqrt{\left(X+Y\right)^3}=a\)

\(\Rightarrow\left(X+Y\right)^3=a^2\Rightarrow X+Y=\sqrt[3]{a^2}\)

\(\Rightarrow\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}\)

Bài này đề CSP năm ngoái hay sao á

a, (3 ; -3)

a, Với y >= 0 

hpt có dạng \(\left\{{}\begin{matrix}2x+y=3\\x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=9\\y=x-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)(ktmđk)

Với y < 0 hpt có dạng 

\(\left\{{}\begin{matrix}2x-y=3\\x-y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-3-6=-9\end{matrix}\right.\)(tm) 

b, bạn tự làm 

c, đk : x>= 3 

\(\left\{{}\begin{matrix}2\sqrt{x+3}+\left|y-2\right|=2\\\sqrt{x+3}-3\left|y-2\right|=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x+3}+\left|y-2\right|=2\\2\sqrt{x+3}-6\left|y-2\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7\left|y-2\right|=1\\2\sqrt{x+3}+\left|y-2\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y-2=\dfrac{1}{7}\\y-2=-\dfrac{1}{7}\end{matrix}\right.\\2\sqrt{x+3}+\left|y-2\right|=2\end{matrix}\right.\)

bạn tự giải nốt nhé 

a) \(x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-7x^2-9x+4+x^3+3x^2+4x+2=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-\left(7x^2+9x-4\right)+\left(x+1\right)^3+x+1=\sqrt[3]{7x^2+9x-4}\) (*)

Đặt \(\sqrt[3]{7x^2+9x-4}=a;x+1=b\)

Khi đó (*) \(\Leftrightarrow-a^3+b^3+b=a\)

\(\Leftrightarrow\left(b-a\right).\left(b^2+ab+a^2+1\right)=0\)

\(\Leftrightarrow b=a\)

Hay \(x+1=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow\left(x+1\right)^3=7x^2+9x-4\)

\(\Leftrightarrow x^3-4x^2-6x+5=0\)

\(\Leftrightarrow x^3-4x^2-5x-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-1\pm\sqrt{5}}{2}\end{matrix}\right.\)