Ý a mình viết nhầm để có nghiệm là -2 nha các bạn

a) Thay x = -2 vào:

\(8+2\left(4m-1\right)+15-m=0\)

\(\Leftrightarrow21+7m=0\Leftrightarrow m=-3\)

b)Thay m = - 3 vào pt: \(2x^2+13x+18=0\Leftrightarrow\left(x+2\right)\left(2x+9\right)=0\)

Đến đây bí.

ko du de ai de lam

Đề có gì thiếu à bn

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Leftrightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

Vậy....

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

nên => x+10 = 0

=> x= -10 

\(3-4x\left(25-2x\right)=8x^2+x-300.\)

\(3-100x+8x^2=8x^2+x-300\)

\(3-100x=x-300\)

\(3+300=x+100x\)

\(303=101x\)

\(x=3\)

Vậy x cần tìm bằng 3

3-4x(25-2x)=8x^2 + x -300 
<=> 3-100x+8x^2=8x^2 + x -300 
<=>3-100x=x-300 
<=>101x=303 
<=>x=3 

fuck you beast

cái đệch gì ?