Bài 4:
a: \(\frac{\left(a+b\right)^2-c^2}{a+b+c}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{a+b+c}\)
=a+b-c
b: \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+c+b\right)}=\frac{a+b-c}{a-b+c}\)
c: \(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
\(=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)
\(=\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}=\frac{2x^2-x-15}{3x^2-10x+3}\)
\(=\frac{2x^2-6x+5x-15}{3x^2-9x-x+3}=\frac{\left(x-3\right)\left(2x+5\right)}{\left(x-3\right)\left(3x-1\right)}=\frac{2x+5}{3x-1}\)
Bài 3:
a: \(A=\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)
\(=\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x^2-4\right)\left(x+1\right)}=\frac{2\left(x-2\right)^2}{x^2-4}\)
\(=\frac{2\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x-2\right)}{x+2}\)
THay x=1/2 vào A, ta được;
\(A=\frac{2\left(\frac12-2\right)}{\frac12+2}=\frac{2\cdot\frac{-3}{2}}{\frac52}=-3:\frac52=-3\cdot\frac25=-\frac65\)
b: \(B=\frac{x^3-x^2y+xy^2}{x^3+y^3}\)
\(=\frac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\cdot\left(x^2-xy+y^2\right)}=\frac{x}{x+y}\)
Thay x=-5; y=10 vào B, ta được:
\(B=\frac{-5}{-5+10}=\frac{-5}{5}=-1\)
