Question

giúp mik vs

 

Answer 1

Bài 1:

a: 2(3x-1)-3x=10

=>6x-2-3x=10

=>3x-2=10

=>3x=12

=>\(x=\frac{12}{3}=4\)

b: ĐKXĐ: x<>0; x<>-1

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

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

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

=>\(3x^2-x+1=x^2+2x+1+x^2+x=2x^2+3x+1\)

=>\(x^2-4x=0\)

=>x(x-4)=0

=>x=0(loại) hoặc x=4(nhận)

c: \(\frac{2x+1}{3}-\frac{3x-2}{2}>\frac16\)

=>\(\frac{2\left(2x+1\right)-3\left(3x-2\right)}{6}>\frac16\)

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

=>4x+2-9x+6>1

=>-5x+8>1

=>-5x>-7

=>x<7/5

Bài 2:

a: \(A=\left(\frac{x^2-3}{x^2-9}+\frac{1}{x-3}\right):\frac{x}{x+3}\)

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

\(=\frac{x^2-3+x+3}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x}=\frac{x^2+x}{x\left(x-3\right)}\)

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

b: |A|=3

=>A=3 hoặc A=-3

TH1: A=3

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

=>3(x-3)=x+1

=>3x-9=x+1

=>2x=10

=>x=5(nhận)

TH2: A=-3

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

=>-3(x-3)=x+1

=>-3x+9=x+1

=>-4x=-8

=>x=2(nhận)

Bài 2:

1:

a: 7x+35=0

=>7x=-35

=>x=-5

b: \(\frac{8-x}{x-7}-8=\frac{1}{x-7}\) (ĐKXĐ: x<>7)

=>\(\frac{8-x-1}{x-7}=8\)

=>\(\frac{7-x}{x-7}=8\)

=>-1=8(vô lý)

2: \(18-3x\left(1-x\right)\le3x^2+3x\)

=>\(18-3x+3x^2\le3x^2+3x\)

=>-3x+18<=3x

=>-6x<=-18

=>x>=3