\(\frac{150}{x-1}-\frac{140}{x-1}=5\left(ĐK:x\ne1\right)\)

\(\Leftrightarrow\frac{10}{x-1}=5\)

\(\Leftrightarrow x-1=2\)

\(\Leftrightarrow x=3\)

ĐKXĐ: x-1\(\ne\)0=> x\(\ne\)1

=> \(\frac{150-140}{x-1}\)=5

=> \(\frac{10}{x-1}\)=5

=> 10= 5(x-1)=> x-1=2=> x=1(ko thỏa mã ĐKXĐ x\(\ne\)1)

phương trình này vô nghiệm.

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

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

\(\frac{10}{x-1}=5\)

\(\frac{10}{x-1}=\frac{5x-5}{x-1}\)

10 = 5x - 5 

-5x = -10 - 5

-5x = -15

x = -15 : -5

x = 3

Điều kiện xác định của pt : \(x\ge1\)

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

\(\Leftrightarrow x+1=x^2-2x+1\)

\(\Leftrightarrow x^2-3x=0\)

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(\text{loại}\right)\\x=3\left(\text{nhận}\right)\end{cases}}\)

Vậy phương trình có nghiệm x = 3

ĐKXĐ \(x+1\ge0\Leftrightarrow x\ge-1\)

Ta có \(\sqrt{x+1}=x-1\)

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

  \(\Leftrightarrow\)\(x+1=x^2-2x+1\)

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

  \(\Leftrightarrow\)\(x^2-3x=0\)

  \(\Leftrightarrow\)\(x\left(x-3\right)=0\)

  \(\Leftrightarrow\)\(\hept{\begin{cases}x=0\\x-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\left(nhận\right)\\x=3\left(nhận\right)\end{cases}}}\)

Vậy tập nghiệm của phương trình \(S=\left\{0;3\right\}\)

BÌnh 2 vê lên ta có:

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

\(\Leftrightarrow x+1=x^2-2x+1\)

\(\Leftrightarrow3x-x^2=0\)

\(\Leftrightarrow x\left(3-x\right)=0\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}\left(TM\right)}\)

a) \(3-2x>4\)

\(\Leftrightarrow-2x>1\)

\(\Leftrightarrow x< \frac{-1}{2}\)

b) \(\frac{2}{3-x}-\frac{9}{3+x}=\frac{1}{2}\)ĐKXĐ : \(x\pm3\)

\(\Leftrightarrow\frac{-4\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}-\frac{18\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-3\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow-4x-13-18x+54=x^2-9\)

\(\Leftrightarrow x^2+22x-50=0\)

\(\Leftrightarrow x^2+2\cdot x\cdot11+11^2-171=0\)

\(\Leftrightarrow\left(x+11\right)^2=\left(\pm\sqrt{171}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{171}-11\\x=-\sqrt{171}-11\end{cases}}\)( thỏa )

Vậy....

\(a,\)\(3-2x>4\)

\(\Rightarrow-2x>1\)

\(\Rightarrow x< \frac{-1}{2}\)

\(a,\)Để pt \(x^2+\left(2m+1\right)x+m\left(m-1\right)=0\) có nghiệm kép thì \(\Delta=0\)

\(\Leftrightarrow\left(2m+1\right)^2-4\left(m^2-m\right)=0\)

\(\Leftrightarrow4m^2+4m+1-4m^2+4m=0\)

\(\Leftrightarrow8m+1=0\)

\(\Leftrightarrow m=-\dfrac{1}{8}\)

Thay \(m=-\dfrac{1}{8}\) vào pt 

\(\Rightarrow x^2+\left[2.\left(-\dfrac{1}{8}\right)+1\right]x-\dfrac{1}{8}\left(-\dfrac{1}{8}-1\right)=0\)

\(\Rightarrow x^2+\dfrac{3}{4}x+\dfrac{9}{64}=0\)

\(\Rightarrow x=-\dfrac{3}{8}\)

\(b,\) Thay \(m=1\) vào pt :

\(\Rightarrow x^2+\left(2.1+1\right)x+1\left(1-1\right)=0\)

\(\Rightarrow x^2+3x=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

\(x^2+3x-10=0\Leftrightarrow x^2+5x-2x-10=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

Tập nghiệm của phương trình: \(S=\left\{2;-5\right\}\)

\(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

Vậy: \(S=\left\{-5;2\right\}\)

=.= hk tốt!!

Lời giải:

Để pt có 2 nghiệm phân biệt thì:

$\Delta'=m^2-(2m-4)=m^2-2m+4>0$

$\Leftrightarrow (m-1)^2+3>0$

$\Leftrightarrow m\in\mathbb{R}$
Áp dụng định lý Viet:

$x_1+x_2=2m$

$x_1x_2=2m-4$

Khi đó:
$x_1+2x_2=8$

$\Leftrightarrow 2m+x_2=8$

$\Leftrightarrow x_2=8-2m$

$\Leftrightarrow x_1=2m-x_2=2m-(8-2m)=4m-8$

$2m-4=x_1x_2=(4m-8)(8-2m)$

$\Leftrightarrow m-2=(2m-4)(8-2m)=2(m-2)(8-2m)$

$\Leftrightarrow (m-2)[2(8-2m)-1]=0$

$\Leftrightarrow (m-2)(15-4m)=0$

$\Leftrightarrow m=2$ hoặc $m=\frac{15}{4}$