Question

Câu 1: Giải các phương trình sau:
a) \(\frac{x^2-1}{15}\)+\(\frac{x^2-3}{13}\)=\(\frac{x^2-5}{11}\)+\(\frac{x^2-7}{9}\)
b) \(\frac{1}{x^2+5x+6}\)+\(\frac{1}{x^2+7x+12}\)+\(\frac{1}{x^2+9x+20}\)+\(\frac{1}{x^2+11x+30}\)=2

Mong các bạn giúp mình mai mình cần gấp

Answer 1

Câu 1: Giải các phương trình sau:
a) $\frac{x^2-1}{15}$+$\frac{x^2-3}{13}$=$\frac{x^2-5}{11}$+$\frac{x^2-7}{9}$

<=> $\frac{x^2-1}{15}$-1+ $\frac{x^2-3}{13}$-1=$\frac{x^2-5}{11}$-1+$\frac{x^2-7}{9}$-1

<=> $\frac{x^2-16}{15}$
+$\frac{x^2-16}{13}$-$\frac{x^2-16}{11}$-\(\frac{x^2-16}{9}\)=0
<=>(\(x^2-16\))(\(\frac{1}{15}\)+\(\frac{1}{13}\)-\(\frac{1}{11}\)-\(\frac{1}{9}\))=0

<=>\(x^2\)-16=0

<=>\(x^2\)=16

<=>\(x^2\)= \(4^2\)

<=>\(x^2\)= \(4^2\)

<=>\(x\)= 4 hoặc x=-4

Vậy pt có tập nghiệm S=\(\left\{4;-4\right\}\)

Answer 2

b) $\frac{1}{x^2+5x+6}$+$\frac{1}{x^2+7x+12}$+$\frac{1}{x^2+9x+20}$+$\frac{1}{x^2+11x+30}$=2

<=>\(\frac{1}{\left(x+2\right)\left(x+3\right)}\)+\(\frac{1}{\left(x+3\right)\left(x+4\right)}\)+\(\frac{1}{\left(x+4\right)\left(x+5\right)}\)+\(\frac{1}{\left(x+5\right)\left(x+6\right)}\)=2

<=>\(\frac{1}{x+2}\)-\(\frac{1}{x+3}\)+\(\frac{1}{x+3}\)-\(\frac{1}{x+4}\)+\(\frac{1}{x+4}\)-\(\frac{1}{x+5}\)+\(\frac{1}{x+5}\)-\(\frac{1}{x+6}\)=2

<=>\(\frac{1}{x+2}\)-\(\frac{1}{x+6}\)=2

<=>\(\frac{4}{\left(x+2\right)\left(x+6\right)}\)=\(\frac{2x^2+16x+24}{\left(x+2\right)\left(x+6\right)}\)

<=>4=2x\(^2\)+16x+24

<=>2x\(^2\)+16x+20=0

...

Answer 3

<=>2(x\(^2\)+8x+10)=0

<=>x\(^2\)+8x+10=0

<=>x\(^2\)+8x+16=26

<=>(x+4)\(^2\)=26

<=>x+4= \(\sqrt{26}\) hoặc -\(\sqrt{26}\)

<=>x=\(\sqrt{26}\)- 4 hoặc -\(\sqrt{26}\)-4

Vậy pt có tập nghiệm S={\(\sqrt{26}\)- 4 ;-\(\sqrt{26}\)- 4}