Quy đồng rồi phân tích nhân tử bình thường đi
\(\left(x-1\right)\left(x-ab-bc-ca\right)\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\)
thách ai làm được . cho a,b,c là hằng số và đôi một khác nhau giải pt: \(\frac{\left(b-c\right)\left(1+a\right)^2}{x+a^2}+\frac{\left(c-a\right)\left(1+b\right)^2}{x+b^2}+\frac{\left(a-b\right)\left(1+c\right)^2}{x+c^2}=0\)
Giai pt sau:
\(\frac{\left(b-c\right)\left(1+a\right)^2}{x+a^2}+\frac{\left(c-a\right)\left(1+b\right)^2}{x+b^2}+\frac{\left(a-b\right)\left(1+c\right)^2}{x+c^2}=0\)
giải phương trình
\(\frac{\left(b-c\right)\left(1+a\right)^2}{x+a^2}+\frac{\left(c-a\right)\left(1+b\right)^2}{x+b^2}+\frac{\left(a-b\right)\left(1+c\right)^2}{x+c^2}=0\)
1)\(\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-a\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
2)\(\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
3)\(\frac{1}{x^2+3x+2}+\frac{2x}{x^3+4x^2+4x}+\frac{1}{x^2+5x+6}\)
Giải phương trình :
\(a,\frac{1}{a+b-x}=\frac{1}{a}+\frac{1}{b}-\frac{1}{x}\)(x là ẩn số )
\(b,\frac{\left(b-c\right)\left(1+a\right)^2}{x+a^2}+\frac{\left(c-a\right)\left(1+b\right)^2}{x+b^2}+\frac{\left(a-b\right)\left(1+c\right)^2}{x+c^2}\)
Tìm x \(\frac{\left(b-c\right)\left(1+a\right)^2}{x+a^2}+\frac{\left(c-a\right)\left(1+b\right)^2}{x+b^2}+\frac{\left(a-b\right)\left(1+c\right)^2}{x+c^2}=0\)
gpt
a) \(\frac{1}{a+b-x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}\)
b)\(\frac{\left(b-c\right)\left(1+a\right)^2}{x+a^{^2}}+\frac{\left(c-a\right)\left(1+b\right)^2}{x+b^2}+\frac{\left(c-b\right)\left(1+c\right)^2}{x+c^2}=0\)
Giải phương trình :
a, \(\frac{1}{a+b-x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}\)
b, \(\frac{\left(b-c\right)\left(1+a\right)^2}{x+a^2}+\frac{\left(c-a\right)\left(1+b\right)^2}{x+b^2}+\frac{\left(a+b\right)\left(1+c\right)^2}{x+c^2}\)
giải phương trình:
\(\frac{a^2\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{b^2\left(x-a\right)\left(x-c\right)}{\left(b-a\right)\left(b-c\right)}+\frac{c^2\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}=x^2\)
