Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ \(\dfrac{x^2-x-6}{x^2+7x+10}=\dfrac{x^2-3x+2x-6}{x^2+2x+5x+10}=\dfrac{x\left(x-3\right)+2\left(x-3\right)}{x\left(x+2\right)+5\left(x+2\right)}=\dfrac{\left(x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}=\dfrac{x-3}{x+5}\)
b/ \(\dfrac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}=\dfrac{x^2+xy+2xy+2y^2}{x\left(x^2-y^2\right)+2y\left(x^2-y^2\right)}=\dfrac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x+2y\right)\left(x^2-y^2\right)}=\dfrac{\left(x+2y\right)\left(x+y\right)}{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}=\dfrac{1}{x-y}\)
a) \(\dfrac{x^2-x-6}{x^2+7x+10}=\dfrac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x+5\right)}=\dfrac{x-3}{x+5}\)
b) \(\dfrac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}=\dfrac{\left(x+y\right)\left(x+2y\right)}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}=\dfrac{\left(x+y\right)\left(x+2y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}=\dfrac{x+y}{x^2-y^2}=\dfrac{x+y}{\left(x+y\right)\left(x-y\right)}=\dfrac{1}{x-y}\)
2)
a) \(\dfrac{1}{x}.\dfrac{6x}{y}\)
\(=\dfrac{6x}{xy}\)
\(=\dfrac{6}{y}\)
b) \(\dfrac{2x^2}{y}.3xy^2\)
\(=\dfrac{2x^2.3xy^2}{y}\)
\(=\dfrac{6x^3y^2}{y}\)
\(=6x^3y\)
c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\)
\(=\dfrac{15x.2y^2}{7y^3.x^2}\)
\(=\dfrac{30xy^2}{7x^2y^3}\)
\(=\dfrac{30}{7xy}\)
d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)
\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)
\(=\dfrac{2y}{5x\left(x-y\right)}\)
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
Câu 1:
\(\text{a) }\dfrac{x^2-xy}{3xy-3y^2}=\dfrac{x\left(x-y\right)}{3y\left(x-y\right)}=\dfrac{x}{3y}\)
\(\text{b) }\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\\ =\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\\ =\dfrac{2a\left(x-1\right)^2}{5b\left(1-x\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)^2}{5b\left(x-1\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\\ =-\dfrac{2ax-2a}{5bx+5b}\)
\(\text{c) }\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
\(\text{d) }\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)
\(\text{e) }\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x+y\right)^3}\\ =\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\\ =\dfrac{x^3+y^3}{x^4-xy^3}\)
Câu 3:
\(\text{ a) }\dfrac{\left(a+b\right)^2-c^2}{a+b+c}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
\(\text{b) }\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\\ =\dfrac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}\\ =\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\\ =\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}\\ =\dfrac{a+b-c}{a-b+c}\)
\(\text{c) }\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\\ =\dfrac{2x^3-x^2-6x^2+3x-15x+45}{3x^3-10x^2-9x^2+3x+30x-9}\\ =\dfrac{\left(2x^3-x^2-15x\right)-\left(6x^2-3x-45\right)}{\left(3x^3-10x^2+3x\right)-\left(9x^2-30x+9\right)}\\ =\dfrac{x\left(2x^2-x-15\right)-3\left(2x^2-x-15\right)}{x\left(3x^2-10x+3\right)-3\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-6x+5x-15\right)}{\left(x-3\right)\left(3x^2-9x-x+3\right)}\\ =\dfrac{\left(x-3\right)\left[\left(2x^2-6x\right)+\left(5x-15\right)\right]}{\left(x-3\right)\left[\left(3x^2-9x\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left[x\left(x-3\right)+5\left(x-3\right)\right]}{\left(x-3\right)\left[3x\left(x-3\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left(x-3\right)\left(x+5\right)}{\left(x-3\right)\left(x-3\right)\left(3x-1\right)}\\ =\dfrac{x+5}{3x-1}\)
\(a,\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)
\(=\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\frac{2\left(x-2\right)}{x+2}\)
Với \(x=\frac{1}{2}\)
\(\Rightarrow\frac{2\left(x-2\right)}{x+2}=\frac{2\left(\frac{1}{2}-2\right)}{\frac{1}{2}+2}=\frac{2.-\frac{3}{2}}{\frac{5}{2}}=-3.\frac{2}{5}=\frac{-6}{5}\)
b,Do x = -5; y = 10=> y = -2x
Thay y = -2x vào biểu thức ta được
\(\frac{x^3-x^2\left(-2x\right)+x\left(-2x\right)^2}{x^3+\left(-2x\right)^3}\)
\(=\frac{x^3+2x^3+2x^2}{x^3-8x^3}\)
\(=\frac{3x^3+2x^2}{-7x^3}=\frac{3}{-7}+\frac{2}{-7x}\)
Thay x = -5 là đc
\(B=\dfrac{x^3+2x^2y-xy^2-2y^3}{x^2+3xy+2y^2}\)
\(B=\dfrac{x^2\left(x+2y\right)-y^2\left(x+2y\right)}{x^2+xy+2xy+2y^2}\)
\(B=\dfrac{\left(x+2y\right)\left(x^2-y^2\right)}{x\left(x+y\right)+2y\left(x+y\right)}\)
\(B=\dfrac{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(2y+x\right)}\)
\(B=x-y\)\(\left(\text{Đ}K:x+2y\ne0;x+y\ne0\right)\)
Tham khảo nhé~
\(B=\dfrac{x^3+2x^2y-xy^2-2y^3}{x^2+3xy+2y^2}\)
\(=\dfrac{x^2\left(x+2y\right)-y^2\left(x+2y\right)}{x^2+xy+2xy+2y^2}\)
\(=\dfrac{\left(x^2-y^2\right)\left(x+2y\right)}{x\left(x+y\right)+2y\left(x+y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x+2y\right)}{\left(x+2y\right)\left(x+y\right)}\)
\(=x-y\)
\(\frac{x^2-x-6}{x^2+7x+10}\)
\(=\frac{x^2-3x+2x-6}{x^2+5x+2x+10}=\frac{x.\left(x-3\right)+2.\left(x-3\right)}{x.\left(x+5\right)+2.\left(x+5\right)}\)
\(=\frac{\left(x+2\right).\left(x-3\right)}{\left(x+2\right).\left(x+5\right)}=\frac{x-3}{x+5}\)