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a: \(\dfrac{x^2-4x+4}{x^2-2x}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}=\dfrac{\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)}\)
\(\dfrac{x+1}{x^2-1}=\dfrac{1}{x-1}=\dfrac{x}{x\left(x-1\right)}\)
b: \(\dfrac{x^3-2^3}{x^2-4}=\dfrac{x^2+2x+4}{x+2}\)
3/x+2=3/x+2
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
b) 
d) 
f) 
h) 
k) 
n) 
bạn đăng tách cho mn cùng giúp nhé
Bài 1 :
a, \(\Leftrightarrow11-x=12-8x\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
b, \(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\Leftrightarrow x=-2\)
c, \(\Leftrightarrow3-2x=-x-4\Leftrightarrow x=7\)
d, \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)
\(\Leftrightarrow3x^2+12x-9=3x^2+3x+1\Leftrightarrow x=\dfrac{10}{9}\)
e, \(\Leftrightarrow2x^2-x-3=2x^2+9x-5\Leftrightarrow x=5\)
f, \(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-22\)
\(\Leftrightarrow-5x^2+2x-1=-5x^2-x-22\Leftrightarrow3x=-21\Leftrightarrow x=-7\)
\(a.\) Ta có:
\(MTC:\) \(\left(x+1\right)\left(x+2\right)\)
Do đó
\(\frac{3x}{x+1}=\frac{3x\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}\)
\(\frac{x+4}{x+2}=\frac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x+2\right)}\)
\(b.\) Ta có:
\(x^2+x=x\left(x+1\right)\)
\(x^2-1=\left(x-1\right)\left(x+1\right)\)
nên \(MTC:\) \(x\left(x-1\right)\left(x+1\right)\)
Do đó:
\(\frac{5}{x^2+x}=\frac{5}{x\left(x+1\right)}=\frac{5\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(\frac{6}{x^2-1}=\frac{6}{\left(x-1\right)\left(x+1\right)}=\frac{6x}{x\left(x-1\right)\left(x+1\right)}\)
\(c.\) Ta có:
\(x^2-5x+4=x^2-x-4x+4=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
\(2x^2-8x=2x\left(x-4\right)\)
nên \(MTC:\) \(2x\left(x-1\right)\left(x-4\right)\)
Do đó:
\(\frac{4}{x^2-5x+4}=\frac{4}{\left(x-1\right)\left(x-4\right)}=\frac{8x}{2x\left(x-1\right)\left(x-4\right)}\)
\(\frac{x+1}{2x^2-8x}=\frac{x+1}{2x\left(x-4\right)}=\frac{\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x-4\right)}\)
Làm nốt d :P
\(\frac{x+3}{2x^2-15x-8};\frac{3}{x^2-8x}\)
Ta có : \(2x^2-15x-8=\left(2x+1\right)\left(x-8\right)\)
\(x^2-8x=x\left(x-8\right)\)
MTC : \(x\left(x-8\right)\left(2x+1\right)\)
\(\frac{x+3}{2x^2-15x-8}=\frac{x+3}{\left(2x+1\right)\left(x-8\right)}=\frac{x^2+3x}{x\left(x-8\right)\left(2x+1\right)}\)
\(\frac{3}{x^2-8x}=\frac{3}{x\left(x-8\right)}=\frac{6x+3}{x\left(x-8\right)\left(2x+1\right)}\)
a: \(\dfrac{x}{2x^2+7x-15}=\dfrac{x}{\left(x+5\right)\left(2x-3\right)}=\dfrac{x^2-2x}{\left(x+5\right)\left(x-2\right)\left(2x-3\right)}\)
\(\dfrac{x+2}{x^2+3x-10}=\dfrac{x+2}{\left(x+5\right)\left(x-2\right)}=\dfrac{\left(x+2\right)\left(2x-3\right)}{\left(2x-3\right)\left(x+5\right)\left(x-2\right)}\)
\(\dfrac{1}{x+5}=\dfrac{\left(2x-3\right)\left(x-2\right)}{\left(2x-3\right)\left(x-2\right)\left(x+5\right)}\)
b: \(\dfrac{1}{-x^2+3x-2}=\dfrac{-1}{\left(x-1\right)\left(x-2\right)}=\dfrac{-\left(x+6\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x+6\right)\left(x-3\right)}\)
\(\dfrac{1}{x^2+5x-6}=\dfrac{1}{\left(x+6\right)\left(x-1\right)}=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+6\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\dfrac{1}{-x^2+4x-3}=\dfrac{-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{-\left(x-2\right)\left(x+6\right)}{\left(x-1\right)\left(x-3\right)\left(x+6\right)\left(x-2\right)}\)
c: \(\dfrac{3}{x^3-1}=\dfrac{3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{x}{x-1}=\dfrac{x\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
EM MỚI LỚP 3 LÊN EM KO BIẾT GÌ HẾT
CHẮC CHỊ HOẶC ANH NÊN TRA GOOGLE
\(a,\dfrac{x^3-2^3}{x^2-4}=\dfrac{\left(x-2\right)\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-2x+4}{x+2}\\ b,\dfrac{2}{2x-4}=\dfrac{2}{2\left(x-2\right)}=\dfrac{1}{x-2}\\ \dfrac{3}{3x-6}=\dfrac{3}{3\left(x-2\right)}=\dfrac{1}{x-2}\)