Question

[Giang Phạm]

Bt: Giải các pt sau:
a, \(\dfrac{x+3}{x+1}\) + \(\dfrac{x-2}{x}\)=\(\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)
b, \(\dfrac{2}{x-2}-\dfrac{x}{x+3}=\dfrac{5x}{\left(x-2\right)\left(x+3\right)}-1\)
c, \(\dfrac{1}{x^2+x+1}-\dfrac{1}{x^2-x+1}=\dfrac{1-2x}{x^4+x^2+1}\)
d, \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
e, \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

Answer 1

[Ánh Lê]

a) Đk : \(x\ne0;\ne1\)

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

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

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

\(\Rightarrow\dfrac{0}{x-1}=0\)

=> Phương trình có vô số nghiệm x

b) Đk : \(x\ne2;x\ne3\)

\(\dfrac{2}{x-2}-\dfrac{x}{x+3}=\dfrac{5x}{\left(x-2\right)\left(x+3\right)}-1\)

\(\Rightarrow\dfrac{2x+6}{\left(x-2\right)\left(x+3\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+3\right)}-\dfrac{5x}{\left(x-2\right)\left(x+3\right)}+\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}\)

=0

\(\Rightarrow\dfrac{2x+6-x^2+2x-5x+x^2+x+6}{\left(x-2\right)\left(x+3\right)}=0\)

\(\Rightarrow\dfrac{12}{\left(x-2\right)\left(x+3\right)}=0\)

=> Phương trình vô nghiệm

c)

\(\Leftrightarrow\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+x+1}{x^4+x^2+1}-\dfrac{1-2x}{x^4+x^2+1}=0\)

\(\Rightarrow\dfrac{x^2-x+1-x^2-x-1-1+2x}{x^4+x^2+1}=0\)

\(\Rightarrow\dfrac{-1}{x^4+x^2+1}=0\)

=> PTVN

d) Thôi tự làm đi, câu này dễ :Vvv

e)

\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)\)=40

\(\Rightarrow\left[\left(x+1\right)\left(x+5\right)\right]\cdot\left[\left(x+2\right)\left(x+4\right)\right]=40\)

\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)

Đặt

\(x^2+6x+7=t\)

Phương trình tương đương

\(\left(t-1\right)\left(t+1\right)=40\)

\(t^2=41\)

\(\)\(t=\pm\sqrt{41}\)

Thay vào tìm x.

Answer 2

[Giang Phạm]

Thanks ;)