Question

Giải các phương trình sau:

a) 2x(x+2)-3(x+2)=0

b) 5/(x-3)+4(x+3)=x-5/x^2-0

Answer 1

A> <X+2><2X-3>=0

X+2=0 HOẶC 2X-3=0

X=-2 HOẶC X=\(\frac{3}{2}\)

B> NHÌN KÌ KÌ SAO ẤY BẠN ƠI

Answer 2

a)  (x+2)(2x - 3)=0   <=>  \(\orbr{\begin{cases}x+2=0\\2x-3=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-2\\x=\frac{3}{2}\end{cases}}\)

Answer 3

ý b mình đánh nhầm 

5/(x-3)+4(x+3)=x-5/x^2-9

Answer 4

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

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

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}\)

           Vậy \(x=\frac{3}{2};-2\)

b)\(\frac{5}{x-3}+4\left(x+3\right)=\frac{x-5}{x^2}\)

\(\Rightarrow\frac{5x^2}{\left(x-3\right)x^2}+\frac{4x^2\left(x+3\right)\left(x-3\right)}{x^2\left(x-3\right)}=\frac{\left(x-5\right)\left(x-3\right)}{x^2\left(x-3\right)}\)

\(\Rightarrow5x^2+4x^2\left(x^2-9\right)=x^2-8x+15\)

\(\Rightarrow4x^3-4x^2-x^2+8x-15=0\)

\(\Rightarrow4x^3-5x^2+8x-15=0\)

                Đến đây mk chịu

Answer 5

a) (x+2)(2x-3)=0 <=> x=-2 hoặc x=3/2

Answer 6

B>\(\frac{5}{X-3}\)+\(\frac{4}{X+3}\)=\(\frac{5}{X^2-9}\)

\(\frac{5< X+3>}{X^2-9}+\frac{4< X-3>}{X^2-9}=\frac{5}{X^2-9}\)

5X+15+4X-12=5

9X=2

X=\(\frac{2}{9}\)