Toán 9

N

nguyenbahiep1

2/ Giải phương trình $x^4-30x^2+31x-30=0$


[laTEX](x-5)(x+6)(x^2-x+1) = 0 \Rightarrow x = 5 , x = - 6 [/laTEX]
 
N

nguyenbahiep1

3/ Cho $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$
CMR: $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0$



[laTEX]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1 \Rightarrow \frac{a^2}{b+c}+\frac{ba}{c+a}+\frac{ca}{a+b}=a \\ \\ \Rightarrow \frac{a^2}{b+c} = a - \frac{ba}{c+a}-\frac{ca}{a+b} \\ \\ \frac{b^2}{c+a} = b - \frac{ba}{c+b}-\frac{cb}{a+b}\\ \\ \frac{c^2}{a+b} = c - \frac{ca}{c+b}-\frac{cb}{a+c} \\ \\ \Rightarrow \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} =a+b+c - (\frac{ab+bc}{c+a}+ \frac{ac+bc}{a+b}+\frac{ab+ca}{b+c}) \\ \\ \Rightarrow \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} = a+b+c - (a+b+c) = 0 [/laTEX]
 
V

vipboycodon

2. $x^4-30x^2+31x-30 = 0$
<=> $x^4+x-30x^2+30x-30 = 0$
<=> $x(x+1)(x^2-x+1)-30(x^2-x+1) = 0$
<=> $(x-5)(x+6)(x^2-x+1) = 0$
=> $x = 5 , x = -6$.
 
Last edited by a moderator:
Top Bottom