Question

[Mình Anh]

Bài 1) giải phương trình sau a) 4x-20=0

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

C) x+2/2008 + x+3/2007 + x+4/2006 + x+2028/6 = 0

d) 2x(x+3)+5(x+3)= 0

bài2 tìm điều kiện xác định của phương trình sau: 2 /x - 1 = 1 /x + l

bài6 giải phương trình

1/x-1 + 1/x-2 = 1/ x+2 + 1/x+1.

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 1:

a) Ta có: 4x-20=0

\(\Leftrightarrow4\left(x-5\right)=0\)

mà \(4\ne0\)

nên x-5=0

hay x=5

Vậy: x=5

b) Ta có: 3-2x=3(x+1)-x-2

\(\Leftrightarrow3-2x=3x+3-x-2\)

\(\Leftrightarrow3-2x=2x+1\)

\(\Leftrightarrow3-2x-2x-1=0\)

\(\Leftrightarrow2-4x=0\)

\(\Leftrightarrow2\left(1-2x\right)=0\)

mà \(2\ne0\)

nên 1-2x=0

\(\Leftrightarrow2x=1\)

hay \(x=\frac{1}{2}\)

Vậy: Tập nghiệm \(S=\left\{\frac{1}{2}\right\}\)

c) Ta có: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)

\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)

\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)

mà \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)

nên x+2010=0

hay x=-2010

Vậy: Tập nghiệm S={-2010}

d) Ta có: 2x(x+3)+5(x+3)=0

\(\Leftrightarrow\left(x+3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{-5}{2}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{-3;\frac{-5}{2}\right\}\)

Bài 2:

ĐKXĐ: \(x\notin\left\{-1;1\right\}\)

Bài 6:

ĐKXĐ: \(x\notin\left\{1;2;-1;-2\right\}\)

Ta có: \(\frac{1}{x-1}+\frac{1}{x-2}=\frac{1}{x+2}+\frac{1}{x+1}\)

\(\Leftrightarrow\frac{x-2}{\left(x-1\right)\left(x-2\right)}+\frac{x-1}{\left(x-1\right)\left(x-2\right)}=\frac{x+1}{\left(x+1\right)\left(x+2\right)}+\frac{x+2}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{x-2+x-1}{\left(x-1\right)\left(x-2\right)}=\frac{x+1+x+2}{\left(x+1\right)\left(x+2\right)}\)

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

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

\(\Leftrightarrow2x^3+3x^2-5x-6-\left(2x^3-3x^2-5x+6\right)=0\)

\(\Leftrightarrow2x^3+3x^2-5x-6-2x^3+3x^2+5x-6=0\)

\(\Leftrightarrow6x^2-12=0\)

\(\Leftrightarrow6x^2=12\)

\(\Leftrightarrow x^2=2\)

hay \(x=\pm\sqrt{2}\)

Vậy: Tập nghiệm \(S=\left\{\sqrt{2};-\sqrt{2}\right\}\)