Question

\(\frac{150}{x-1}-\frac{140}{x}=5\)

Answer 1

       \(\frac{150}{x-1}-\frac{140}{x}=5\)

\(ĐKXĐ:x\ne1;x\ne0\)

\(MTC:x\left(x-1\right)\) 

 \(\Leftrightarrow\frac{150x}{x\left(x-1\right)}-\frac{140\left(x-1\right)}{x\left(x-1\right)}=\frac{5x\left(x-1\right)}{x\left(x-1\right)}\)

\(\Rightarrow150x-140\left(x-1\right)=5x\left(x-1\right)\)

\(\Leftrightarrow150x-140x+140=5x^2-5x\)

\(\Leftrightarrow150x-140x+140-5x^2+5x=0\)

\(\Leftrightarrow-5x^2+15x+140=0\)

\(\Leftrightarrow-5x^2-20x+35x+140=0\)

\(\Leftrightarrow\left(-5x^2+35x\right)+\left(-20x+140\right)=0\)

\(\Leftrightarrow-5x\left(x-7\right)-20\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(-5x-20\right)=0\)

HOẶC \(x-7=0\Leftrightarrow x=7\)(nhận)

HOẶC\(-5x-20=0\Leftrightarrow x=-4\)(nhận)

VẬY TẬP NGHIỆM CỦA PT LÀ \(S=\left\{7;-4\right\}\)

Answer 2

\(\frac{150}{x-1}-\frac{140}{x}=5\)

\(\Leftrightarrow\frac{150x}{x\left(x-1\right)}-\frac{140\left(x-1\right)}{x\left(x-1\right)}=5\)

\(\Leftrightarrow\frac{150x-140\left(x-1\right)}{x\left(x-1\right)}=5\)

\(\Leftrightarrow\frac{150x-140x+140}{x\left(x-1\right)}=5\)

\(\Leftrightarrow\frac{10x+140}{x\left(x-1\right)}=5\)

\(\Leftrightarrow10x+140=5x\left(x-1\right)\)

\(\Leftrightarrow5\left(2x+28\right)=5x\left(x-1\right)\)

\(\Leftrightarrow2x+28=x\left(x-1\right)\)

\(\Leftrightarrow28=x\left(x-1\right)-2x\)

\(\Leftrightarrow28=x\left(x-1-2\right)\)

\(\Leftrightarrow28=x\left(x-3\right)\)

\(\Leftrightarrow x\left(x-3\right)=7.4=\left(-4\right)\left(-7\right)\)

\(\Leftrightarrow x\in\left\{7;-4\right\}\)

Answer 3

ĐKXĐ: \(x\ne0;1\)

Ta có: \(\frac{150}{x-1}-\frac{140}{x}=5\Leftrightarrow\frac{150x}{x\left(x-1\right)}-\frac{140\left(x-1\right)}{x\left(x-1\right)}=\frac{150x-140x+140}{x\left(x-1\right)}=5\)

\(\Leftrightarrow\frac{10x+140}{x\left(x-1\right)}=5\Leftrightarrow10x+140=5x\left(x-1\right)\Leftrightarrow10x+140=5x^2-5x\)

\(\Leftrightarrow5x^2-5x-10x-140=0\Leftrightarrow5x^2-15x-140=0\)

\(\Leftrightarrow x^2-3x-28=0\Leftrightarrow x^2-7x+4x-28=0\Leftrightarrow x\left(x-7\right)+4\left(x-7\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-7\right)=0\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=7\end{cases}}}\)

Vậy...