a) Thay \(x=2\) vào phương trình, ta được:

 \(15\left(m+6\right)+12=80\) \(\Rightarrow m=-\dfrac{22}{15}\)

Vậy \(m=-\dfrac{22}{15}\)

b) Thay \(x=1\) vào phương trình, ta được:

  \(15\left(2+m\right)-32=43\) \(\Rightarrow m=3\)

Vậy \(m=3\)

a,\(11-2x=x-1\Leftrightarrow-2x-x=-1-11\Leftrightarrow-3x=-12\Leftrightarrow x=-4\)

b,\(\text{5(3x+2)=4x+1}\Leftrightarrow15x+10=4x+1\Leftrightarrow15x-4x=1-10\Leftrightarrow11x=-9\Leftrightarrow x=\dfrac{-9}{11}\)

c,\(x^2-4-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x+2\right)\left(x-2\right)-\left(x-2\right)\left(x-5\right)\Leftrightarrow\left(x-2\right)[\left(x+2\right)-\left(x-5\right)]\Leftrightarrow\left(x-2\right)\left[x+2-x+5\right]\Leftrightarrow\left(x-2\right)7\Leftrightarrow7x-14\)

Câu 1: ĐKXĐ: ...

\(\Leftrightarrow4x\left(3x-1\right)+x-1=4x\sqrt{3x+1}\)

\(\Leftrightarrow12x^2-3x-1-4x\sqrt{3x+1}=0\)

\(\Leftrightarrow16x^2-\left(4x^2+4x\sqrt{3x+1}+3x+1\right)=0\)

\(\Leftrightarrow16x^2-\left(2x+\sqrt{3x+1}\right)^2=0\)

\(\Leftrightarrow\left(2x-\sqrt{3x+1}\right)\left(6x+\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow...\)

Câu 2:

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2-4\right)=y^3+2y\\x^2-4=-3y^2\end{matrix}\right.\)

\(\Leftrightarrow x\left(-3y^2\right)=y^3+2y\)

\(\Leftrightarrow y\left(y^2+3xy+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow...\\y^2+3xy+2=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3xy=-y^2-2\Rightarrow x=\frac{-y^2-2}{3y}\)

\(\Rightarrow\left(\frac{y^2+2}{3y}\right)^2-1=3\left(1-y^2\right)\)

\(\Leftrightarrow\left(\frac{y^2-3y+2}{3y}\right)\left(\frac{y^2+3y+2}{3y}\right)=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y-1\right)\left(y-2\right)\left(y+1\right)\left(y+2\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y^2-1\right)\left(y^2-4\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\\frac{y^2-4}{9y^2}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\28y^2=4\end{matrix}\right.\)

\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{4x\left(3x-1\right)+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-3x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{\left(12x^2-3x-1\right)^2}{16x^2}=3x+1\)

\(\Leftrightarrow\left(12x^2-3x-1\right)^2=16x^2\left(3x+1\right)\)

\(\Leftrightarrow144x^4-120x^3-31x^2+6x+1=0\)

\(\Leftrightarrow144x^4-144x^3+24x^3-24x^2-7x^2+7x-x+1=0\)

\(\Leftrightarrow144x^3\left(x-1\right)+24x^2\left(x-1\right)+7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(144x^3+24x^2+7x-1\right)=0\)

Tìm được mỗi nghiệm thôi à :v

a: \(\Leftrightarrow x+2-3xm-m=5\)

\(\Leftrightarrow x\left(1-3m\right)=5+m-2=m+3\)

Để đây là pt bậc nhất thì -3m+1<>0

hay m<>1/3

b: Khi m=-1 thì pt sẽ là \(x\left(1+3\right)=-1+3=2\)

=>x=1/2

Giải pt :

\(\left(x+2\right)\left(3x+1\right)+x^2=4\)

\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)+x^2-4=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)+\left(x^2-4\right)=0\)

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

\(\Leftrightarrow\left(x+2\right)\left[\left(3x+1\right)+\left(x-2\right)\right]=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\4x-1=0\end{matrix}\right.\)

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

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

Tập nghiệm của pt là : \(S=\left\{-2;\dfrac{1}{4}\right\}\)

a)               \(3x^2+x-4=0\)

\(\Leftrightarrow\)\(3x^2-3x+4x-4=0\)

\(\Leftrightarrow\)\(3x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=-\frac{4}{3}\end{cases}}\)

Vậy..

b)    \(2x^2-x-28=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=4\\x=-3.5\end{cases}}\)

Vậy...

c)     \(6x^2-x-7=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(6x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=\frac{7}{6}\end{cases}}\)

Vậy....

d)  \(3x^2-5=0\)

\(\Leftrightarrow\)\(3x^2=5\)

\(\Leftrightarrow\)\(x^2=\frac{5}{3}\)

\(\Leftrightarrow\)\(x=\pm\sqrt{\frac{5}{3}}\)

Vậy...

=>\(\dfrac{-1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}=2\)

=>\(\dfrac{1}{x-4}-\dfrac{1}{x-1}=2\)

=>\(\dfrac{x-1-x+4}{x^2-5x+4}=2\)

=>2x^2-10x+8=3

=>2x^2-10x+5=0

=>\(x=\dfrac{5\pm\sqrt{15}}{2}\)