\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)

\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)

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

2.

ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)

\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)

\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)

\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)

\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)

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

\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)

\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)

\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

mong mọi người giải giúp em vs 

* Nếu \(x< 2\) thì phương trình trở thành:

\(\Leftrightarrow\left(2-x\right)+\left(3-x\right)+\left(8-2x\right)=9\)

\(\Leftrightarrow2-x+3-x+8-2x=9\)

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\) (nhận)

* Nếu \(2\le x\le3\) thì phương trình trở thành:

\(\Leftrightarrow\left(x-2\right)+\left(3-x\right)+\left(8-2x\right)=9\)

\(\Leftrightarrow x-2+3-x+8-2x=9\)

\(\Leftrightarrow-2x=0\)

\(\Leftrightarrow x=0\) (loại)

*Nếu \(3\le x\le4\) thì phương trình trở thành:

\(\Leftrightarrow\left(x-2\right)+\left(x-3\right)+\left(8-2x\right)=9\)

\(\Leftrightarrow x-2+x-3+8-2x=9\)

\(\Leftrightarrow3=9\) (vô lí)

* Nếu \(x>4\) thì phương trình trở thành:

\(\Leftrightarrow\left(x-2\right)+\left(x-3\right)+\left(2x-8\right)=9\)

\(\Leftrightarrow x-2+x-3+2x-8=9\)

\(\Leftrightarrow4x=21\)

\(\Leftrightarrow x=\dfrac{21}{4}\)(nhận)

Vậy phương trình có tập nghiệm \(S=\left\{1;\dfrac{21}{4}\right\}\)

      \(\left(4x-5\right)^2\left(2x-3\right)\left(x-1\right)=9\)

\(\Leftrightarrow\left(4x-5\right)^2\left(2x-3\right).2.\left(x-1\right).4=9.2.4\)

\(\Leftrightarrow\left(4x-5\right)^2\left(4x-6\right)\left(4x-4\right)=72\)(1)

Đặt \(4x-5=a\)

Khi đó (1) trở thành: 

      \(a^2\left(a-1\right)\left(a+1\right)=72\)

\(\Leftrightarrow a^2\left(a^2-1\right)=72\)    

\(\Leftrightarrow a^4-a^2-72=0\)

\(\Leftrightarrow a^4-9a^2+8a^2-72=0\)

\(\Leftrightarrow a^2\left(a^2-9\right)+8\left(a^2-9\right)=0\)

\(\Leftrightarrow\left(a^2-9\right)\left(a^2+8\right)=0\)

\(\Leftrightarrow a^2-9=0\) (vì \(a^2+8>0\forall a\) )

\(\Leftrightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)

- Với \(a=3\Rightarrow4x-5=3\Rightarrow x=2\)

-Với \(a=-3\Rightarrow4x-5=-3\Rightarrow x=\frac{1}{2}\)

Vậy \(x=2,x=\frac{1}{2}\)

Chúc bạn học tốt.

Câu hỏi của Do Xuan Dat - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo nhé! 

Ta có:

\(\left|x-1\right|+\left|x-5\right|=\left|x-1\right|+\left|5-x\right|\)

≥ \(\left|x-1+5-x\right|=4\)

mà \(4-\left|x-3\right|\)≤ 4

Dấu "="⇔ \(x=3\)