Question

Bài 1: Giải pt:

a) (x² - 5x)² + 10(x²-5x) +24=0

b)( x²+ x +1)( x²+x+2)=12

c) x(x+1)(x² +x +1)=42

Bài 2: Giải pt:

1/x² +5x +6 + 1/x²+7x+12 + 1/ x²+9x+20 = 3/40

giúp mình với ạ,mình cần gấp

Answer 1

Bài 2

Ta có :

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

\(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

\(x^2+9x+20=\left(x+4\right)\left(x+5\right)\)

Khi đó:

\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}=\dfrac{3}{40}\)

=> \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}=\dfrac{3}{40}\)

=> \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{3}{40}\)

=> \(\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{3}{40}\)

Giải phương trình ta được x = 3