c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)

\(2,\left\{{}\begin{matrix}x^3-2x^2y-15x=6y\left(2x-5-4y\right)\left(1\right)\\\frac{x^2}{8y}+\frac{2x}{3}=\sqrt{\frac{x^3}{3y}+\frac{x^2}{4}}-\frac{y}{2}\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(2y-x\right)\left(x^2-12y-15\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}2y=x\\y=\frac{x^2-15}{12}\end{matrix}\right.\)

Ta xét các trường hợp sau:

Trường hợp 1:

\(y=\frac{x^2-15}{12}\) thay vào phương trình \(\left(2\right)\) ta được:

\(\frac{3x^2}{2\left(x^2-15\right)}+\frac{2x}{3}=\sqrt{\frac{4x^3}{x^2-15}+\frac{x^2}{4}}-\frac{x^2-15}{24}\)

\(\Leftrightarrow\frac{36x^2}{x^2-15}-12\sqrt{\frac{x^2}{x^2-15}\left(x^2+16x-15\right)}+\left(x^2+16x-15\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\6\sqrt{\frac{x^2}{x^2-15}}=\sqrt{\left(x^2+16x-15\right)}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36\frac{x^2}{x^2-15}=x^2+16x-15\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\left(3\right)\end{matrix}\right.\)

Ta xét phương trình \(\left(3\right):36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\)

Vì: \(x=0\) Không phải là nghiệm. Ta chia cả hai vế p.trình cho \(x^2\) ta được:

\(36=\left(x-\frac{15}{x}\right)\left(x+16-\frac{15}{x}\right)\)

Đặt: \(x-\frac{15}{x}=t\Rightarrow t^2+16t-36=0\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-18\end{matrix}\right.\)

+ Nếu như:

\(t=2\Leftrightarrow x-\frac{15}{x}=2\Leftrightarrow x^2-2x-15=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)\(\Leftrightarrow x=5\)

+ Nếu như:

\(t=-18\Leftrightarrow x-\frac{15}{x}=-18\Leftrightarrow x^2+18x-15=0\Leftrightarrow\left[{}\begin{matrix}x=-9-4\sqrt{6}\\x=-9+4\sqrt{6}\end{matrix}\right.\Leftrightarrow x=-9-4\sqrt{6}\)

Trường hợp 2:

\(x=2y\) thay vào p.trình \(\left(2\right)\) ta được:

\(\Leftrightarrow\frac{x^2}{4x}+\frac{2x}{3}=\sqrt{\frac{2x^3}{3x}+\frac{x^2}{4}}-\frac{x}{4}\Leftrightarrow\frac{7}{6}x=\sqrt{\frac{11x^2}{12}}\Leftrightarrow x=0\left(ktmđk\right)\)

Vậy nghiệm của hệ đã cho là: \(\left(x,y\right)=\left(5;\frac{5}{6}\right),\left(-9-4\sqrt{6};\frac{27+12\sqrt{6}}{2}\right)\)

Năm mới chắc bị lag @@ tớ sửa luôn đề câu 3 nhé :v

3, \(\left\{{}\begin{matrix}8\left(x^2+y^2\right)+4xy+\frac{5}{\left(x+y\right)^2}=13\left(1\right)\\2xy+\frac{1}{x+y}=1\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow8\left[\left(x+y\right)^2-2xy\right]+4xy+\frac{5}{\left(x+y\right)^2}=13\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow8\left(a^2-2b\right)+4b+\frac{5}{a^2}=13\)

\(\Leftrightarrow8a^2-12b+\frac{5}{a^2}=13\)

Ta cũng có \(\left(2\right)\Leftrightarrow2b+\frac{1}{a}=1\)

\(\Leftrightarrow2b=1-\frac{1}{a}\)

Thay vào (1) ta được :

\(8a^2+\frac{5}{a^2}-6\cdot\left(1-\frac{1}{a}\right)=13\)

\(\Leftrightarrow8a^2+\frac{5}{a^2}-6+\frac{6}{a}=13\)

\(\Leftrightarrow8a^2+\frac{5}{a^2}+\frac{6}{a}=19\)

Giải pt được \(a=1\)

Khi đó \(b=\frac{1-\frac{1}{1}}{2}=0\)

Ta có hệ :

\(\left\{{}\begin{matrix}x+y=1\\xy=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)

Vậy...

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

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

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

Thay x=\(\frac{1}{2}\)

\(=\frac{2\left(\frac{1}{2}^2+1\right)\left(\frac{1}{2}-2\right)}{\left(\frac{1}{2}+2\right)\left(\frac{1}{2}+1\right)}\)

\(=-1\)

\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)

Vậy pt có vô số nghiệm

\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)

Mấy câu rút gọn bạn quy đồng nha

bạn có thể giải ra giúp mik đc ko?