\(ĐKXĐ:x\ne\pm5\)

 \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)

\(\Rightarrow\frac{3\left(x+5\right)}{4\left(x-5\right)\left(x+5\right)}+\frac{30}{4\left(25-x^2\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)

\(\Rightarrow\frac{3x+15}{4\left(x-5\right)\left(x+5\right)}+\frac{-30}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)

\(\Rightarrow\frac{3x+15-30}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)

\(\Rightarrow\frac{3x-15}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)

\(\Rightarrow\frac{3\left(x-5\right)}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)

\(\Rightarrow\frac{3}{4\left(x+5\right)}=\frac{-7}{6\left(x+5\right)}\)

\(\Rightarrow18\left(x+5\right)=-28\left(x+5\right)\)

\(\Rightarrow18\left(x+5\right)+28\left(x+5\right)=0\)

\(\Rightarrow46\left(x+5\right)=0\Leftrightarrow x+5=0\Leftrightarrow x=-5\)(ktm)

Vậy pt vô nghiệm

Từ 1 ta có 2x²-  2Xy +x-y=0

2x(x-y) + x-y=0

(2x+1)(x-y) =0

Th1 x-y=0 thay vao phương trình 2

th2 thay 2x+1=0 vao phương trình 2

Tự giải tiếp nhé.... 

Mãi không thấy ai sol nên mình làm bạn xem nhé ^_^

a)

Để căn bậc 2 có nghĩa tức là \(\left(3-x\right)\left(x+1\right)\ge0\Leftrightarrow-1\le x\le3\)

b)

Để căn bậc 2 có nghĩa tức là \(\frac{2-x}{x-1}\ge0\) mặt khác cũng cần có điều kiện \(x-1\ne0\)

\(\Rightarrow1< x\le2\)

Đặt a=2x-1; b=3x+2

=>a+b=2x-1+3x+2=5x+1

\(\left(2x-1\right)^3+\left(3x+2\right)^3-\left(5x+1\right)^3=0\)

=>\(a^3+b^3-\left(a+b\right)^3=0\)

=>\(\left(a+b\right)^3-3ab\left(a+b\right)-\left(a+b\right)^3=0\)

=>-3ab(a+b)=0

=>(2x-1)(5x+1)(3x+2)=0

=>\(\left[\begin{array}{l}2x-1=0\\ 5x+1=0\\ 3x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=-\frac15\\ x=-\frac23\end{array}\right.\)

Khi \(x=4>3\Rightarrow f\left(x\right)=2x-3\)

\(\Rightarrow f\left(4\right)=2.4-3=5\)

Cảm ơn ạ