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a)
\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}=\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}\)
\(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\)
MTC: \(2\left(x-1\right)\left(x+1\right)\left(x-5\right)\)
\(\dfrac{3x-6}{x^2-6x+5}=\dfrac{3x-6}{x^2-x-5x+5}=\dfrac{3x-6}{x\left(x-1\right)-5\left(x-1\right)}\\ =\dfrac{3x-6}{\left(x-1\right)\left(x-5\right)}=\dfrac{2\left(x+1\right)\left(3x-6\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)
\(\dfrac{5x-5}{2x^2-2}=\dfrac{5x-5}{2\left(x^2-1\right)}=\dfrac{5x-5}{2\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x-5\right)\left(5x-5\right)}{2\left(x-1\right)\left(x+1\right)\left(x-5\right)}\)
Bài giải
a) \(\dfrac{1}{x+2}=\dfrac{x.\left(x-2\right)}{\left(x+2\right)\left(x-2\right).x}=\dfrac{x^2-2x}{x\left(x+2\right)\left(x-2\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=-\dfrac{8}{x\left(x-2\right)}=-\dfrac{8.\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
b) \(x^2+1=\dfrac{x^2+1}{1}=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x^2-1}=\dfrac{x^4-1}{x^2-1}\)
\(\dfrac{x^4}{x^2-1}\) giữ nguyên.
c) \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3.y}{\left(x-y\right)^3.y}=\dfrac{x^3y}{y\left(x-y\right)^3}\)
\(\dfrac{x}{y^2-xy}=\dfrac{x}{y.\left(y-x\right)}=-\dfrac{x}{y.\left(x-y\right)}=-\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right).\left(x-y\right)^2}=\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right)^3}\)
bài 1)
a) \(\dfrac{2ax-2x-3y+3ay}{4ax+6x+9y+6ay}\)
= \(\dfrac{\left(2ax-2x\right)+\left(3ay-3y\right)}{\left(4ax+6x\right)+\left(6ay+9y\right)}\)
= \(\dfrac{2x\left(a-1\right)+3y\left(a-1\right)}{2x\left(2a+3\right)+3y\left(2a+3\right)}\)
= \(\dfrac{\left(2x+3y\right)\left(a-1\right)}{\left(2x+3y\right)\left(2a+3\right)}\)
= \(\dfrac{a-1}{2a+3}\)
Vậy biểu thức \(\dfrac{2ax-2x-3y+3ay}{4ax+6x+9y+6ay}\) ko phụ thuộc vào biến x,y mà phụ thuộc vào biến a
a: \(=\dfrac{1-2x+3+2y+2y-4}{6x^3y}=\dfrac{-2x+4y}{6x^3y}=\dfrac{-2\left(x-2y\right)}{6x^3y}=\dfrac{-x+2y}{3x^3y}\)
b: \(=\dfrac{x^2-2+2-x}{x\left(x-1\right)^2}=\dfrac{x\left(x-1\right)}{x\left(x-1\right)^2}=\dfrac{1}{x-1}\)
c: \(=\dfrac{3x+1+x^6-3x}{x^2-3x+1}\)
\(=\dfrac{x^6+1}{x^2-3x+1}\)
d: \(=\dfrac{x^2+38x+4+3x^2-4x-2}{2x^2+17x+1}\)
\(=\dfrac{4x^2+34x+2}{2x^2+17x+1}=2\)
a: \(\dfrac{5}{2x+6}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}\)
3/x^2-9=6/2(x+3)(x-3)
b: \(\dfrac{2x}{x^2-8x+16}=\dfrac{2x}{\left(x-4\right)^2}=\dfrac{6x^2}{3x\left(x-4\right)^2}\)
\(\dfrac{x}{3x^2-12x}=\dfrac{x}{3x\left(x-4\right)}=\dfrac{x\left(x-4\right)}{3x\left(x-4\right)^2}\)
c: \(\dfrac{x+y}{x}=\dfrac{\left(x+y\right)\cdot\left(x-y\right)}{x\left(x-y\right)}\)
x/x-y=x^2/x(x-y)
e: \(\dfrac{1}{x+2}=\dfrac{2x-x^2}{x\left(x+2\right)\left(2-x\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8\left(x+2\right)}{x\left(2-x\right)\left(2+x\right)}\)
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
1,
\(x^2-2ax+a^2=\left(x-a\right)^2\)
\(x^2-ax=x\left(x-a\right)\)
Vậy MSC: \(\left(x-a\right)^2x\)
2,
\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(x^2-x=x\left(x-1\right)\)
\(x^2+x+1\)
vậy MSC là: \(x\left(x-1\right)\left(x^2+x+1\right)\)
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