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\(\dfrac{2a\cdot x^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)
\(=\dfrac{-2a\left(x-1\right)^2}{5b\left(x-1\right)\left(x+1\right)}=\dfrac{-2a\left(x-1\right)}{5b\left(x+1\right)}\)
\(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)
\(=\dfrac{4x\cdot x-4x\cdot y}{5x^2\cdot x-5x^2\cdot y}\)
\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)
\(=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}\)
=x+y-z
\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3+y^3\right)\left(x^3-y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)
\(A=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-b\right)}=\dfrac{2a\left(x-1\right)^2}{5b\left(1-b\right)}\)
\(B=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)
1) \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{-5b\left(x^2-1\right)}\)
\(=\dfrac{2a\left(x-1\right)^2}{-5b\left(x-1\right)\left(x+1\right)}=\dfrac{2a\left(x-1\right)}{-5b\left(x+1\right)}\)
2) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+x+3x+3}{2\left(x+3\right)}=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{2\left(x+3\right)}\)\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
3)\(\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
4) \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)Học tốt nha you<3
p/s: tớ ko bk đã rút gọn hết chưa:(
1, \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)
\(=\dfrac{2a\left(x-1\right)^2}{5b\left(1-x\right)\left(1+x\right)}\)
\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)
2, \(\dfrac{x^2+4x+3}{2x+6}\)
\(=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}\)
\(=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
3, \(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)
\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
4, \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)
5, \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)
a) \(\dfrac{x^2-16}{4x-x^2}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(4-x\right)}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left[-\left(x-4\right)\right]}\)
\(=\dfrac{\left(-1\right)\left(x+4\right)}{x}=\dfrac{-x-4}{x}\)
b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
c) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)
d) \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}=\dfrac{2a\left(x-1\right)^2}{5b\left[-\left(x-1\right)\right]\left(1+x\right)}\)
\(=\dfrac{2a\left[-\left(x-1\right)\right]}{5b\left(1+x\right)}=-\dfrac{2ax-2a}{50\left(1+x\right)}=-\dfrac{-\left(2a-2ax\right)}{5b\left(1+x\right)}-\dfrac{2a-2ax}{5b-5bx}\)
e) \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=z+y-z\)
f) \(\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4x}{5x^2}=\dfrac{4}{5x}\)
a, \(\dfrac{x^2-16}{4x-x^2}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(4-x\right)}=\dfrac{-x-4}{x}=-1-\dfrac{4}{x}\)
b, \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+x+3x+3}{2\left(x+3\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{2\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{x+1}{2}\)
c,\(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{4}{5}\)
d, \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}=\dfrac{a\left(2x^2-4x+2\right)}{-5b\left(x^2-1\right)}\)
\(=\dfrac{2a\left(x-1\right)^2}{-5b\left(x-1\right)\left(x+1\right)}=\dfrac{2a\left(x-1\right)}{-5b\left(x+1\right)}\)
e, \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)
f, \(\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
Chúc bạn học tốt!!!
B=\(\frac{5\left(x-y\right)-3\left(x-y\right)}{10\left(x-y\right)}\)
B=\(\frac{\left(x-y\right)\left(5-3\right)}{10\left(x-y\right)}\)
B= \(\frac{\left(x-y\right)2}{10\left(x-y\right)}\)
B= 5
vậy B=5
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)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
\(\frac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=-\frac{2a\left(x-1\right)^2}{5b\left(x+1\right)\left(x-1\right)}\)
\(=-\frac{2a\left(x-1\right)}{5b\left(x+1\right)}\)
\(a,\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)
\(=\dfrac{2a\left(x-1^2\right)}{5b\left(x-1\right)\left(1+x\right)}\)
\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)
\(b,\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)
\(=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)