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14 tháng 12 2018
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
2 tháng 7 2022
a: \(A=x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)
Dấu '=' xảy ra khi x=3/2
c: \(x^2-x+2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\)
=>\(\dfrac{3}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}< =3:\dfrac{7}{4}=\dfrac{12}{7}\)
=>C>=-12/7
Dấu '=' xảy ra khi x=1/2
12 tháng 7 2017
a, \(\dfrac{4x^2-8xy}{10y-5x}=\dfrac{4x\left(x-2y\right)}{5\left(2y-x\right)}=\dfrac{-4x}{5}\)
b, \(\dfrac{\left(x-2\right)^2-1}{x^2-6x+9}=\dfrac{\left(x-2-1\right)\left(x-2+1\right)}{\left(x-3\right)^2}\)
\(=\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)^2}=\dfrac{x-1}{x-3}\)
c, \(\dfrac{x^2+8x+16}{x^2-16}=\dfrac{\left(x+4\right)^2}{\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\)
29 tháng 5 2022
a: \(=\dfrac{x}{y\left(x-y\right)}+\dfrac{2x-y}{y\left(x-y\right)}=\dfrac{x+2x-y}{y\left(x-y\right)}=\dfrac{3x-y}{y\left(x-y\right)}\)
b: \(=\dfrac{x\left(x+3\right)}{\left(x+3\right)^2}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x+3}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-3x+3x+9-6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
c: \(=\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x-3}\)
d: \(=\dfrac{x^2-1-x^2+4}{x+1}=\dfrac{3}{x+1}\)
NN
27 tháng 11 2017
1) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}-\dfrac{1}{1-x}+1\)
\(=\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{x-1}+1\)
\(=\dfrac{x^2}{x+1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x-1}+1\) MTC: \(\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2\left(x-1\right)+2x+\left(x+1\right)+\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^3-x^2+2x+x+1+x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)}\)
b) \(\dfrac{1}{x^3-x}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{x^2-1}\)
\(=\dfrac{1}{x\left(x^2-1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\) MTC: \(x\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1-\left(x+1\right)+2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1-x-1+2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
14 tháng 6 2022
a: \(=\dfrac{3x}{5\left(x+y\right)}-\dfrac{x}{10\left(x-y\right)}\)
\(=\dfrac{6x\left(x-y\right)-x\left(x+y\right)}{10\left(x-y\right)\cdot\left(x+y\right)}\)
\(=\dfrac{6x^2-6xy-x^2-xy}{10\left(x-y\right)\left(x+y\right)}=\dfrac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}\)
b: \(=\dfrac{7}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{1}{x\left(2x+3\right)}-\dfrac{1}{2\left(2x-3\right)}\)
\(=\dfrac{7x+2\left(2x-3\right)-x\left(2x+3\right)}{2x\left(2x+3\right)\left(2x-3\right)}\)
\(=\dfrac{7x+4x-6-2x^2-3x}{2x\left(2x+3\right)\left(2x-3\right)}\)
\(=\dfrac{-2x^2-6}{2x\left(2x+3\right)\left(2x-3\right)}=\dfrac{-x^2-3}{x\left(2x+3\right)\left(2x-3\right)}\)
c: \(=\dfrac{5}{x+1}+\dfrac{10}{x^2-x+1}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x}{x^2-x+1}\)
TQ
17 tháng 7 2017
a,\(\dfrac{3}{x-3}\) - \(\dfrac{6x}{9-x^2}\) + \(\dfrac{x}{x+3}\) (*)
đkxđ: x khác 3, x khác -3
(*) \(\dfrac{3(x+3)}{\left(x-3\right).\left(x+3\right)}\)- \(\dfrac{6x}{\left(x-3\right).\left(x+3\right)}\) + \(\dfrac{x\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}\)
=>3x+9 -6x + x2+3x
<=>x2 + 3x-6x+3x + 9
<=>x2 +9
<=>(x-3).(x+3)
23 tháng 7 2022
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
1 tháng 1 2023
a: \(=\dfrac{2x^2}{x^2-1}+\dfrac{6}{x-3}-\dfrac{2x-6}{\left(x-3\right)\left(x^2-1\right)}\)
\(=\dfrac{2x^3-6x^2+6x^2-6-2x+6}{\left(x-3\right)\left(x^2-1\right)}\)
\(=\dfrac{2x\left(x-1\right)}{\left(x-3\right)\left(x-1\right)\left(x+1\right)}=\dfrac{2x}{\left(x-3\right)\left(x+1\right)}\)
b: \(=\dfrac{x+3}{x\left(x-6\right)}-\dfrac{x+9}{\left(x-6\right)\left(x+4\right)}+1\)
\(=\dfrac{\left(x+3\right)\left(x+4\right)-x\left(x+9\right)+x\left(x-6\right)\left(x+4\right)}{x\left(x-6\right)\left(x+4\right)}\)
\(=\dfrac{x^2+7x+12-x^2-9x+x\left(x^2-2x-24\right)}{x\left(x-6\right)\left(x+4\right)}\)
\(=\dfrac{-2x+12+x^3-2x^2-24x}{x\left(x-6\right)\left(x+4\right)}\)
\(=\dfrac{x^3-2x^2-26x+12}{x\left(x-6\right)\left(x+4\right)}\)
\(=\dfrac{x^3-6x^2+4x^2-24x-2x+12}{x\left(x-6\right)\left(x+4\right)}\)
\(=\dfrac{\left(x-6\right)\left(x^2+4x-2\right)}{x\left(x-6\right)\left(x+4\right)}=\dfrac{x^2+4x-2}{x^2+4x}\)
\(\dfrac{6x}{15y^3}\cdot\dfrac{10y^3}{x^2}=\dfrac{6x}{x^2}\cdot\dfrac{10y^3}{15x^3}=\dfrac{6}{x}\cdot\dfrac{2}{3}=\dfrac{4}{x}\)
\(\dfrac{6x}{15y^3}\cdot\dfrac{10y^3}{x^2}=\dfrac{6x\cdot10y^3}{15y^3\cdot x^2}=\dfrac{2\cdot3\cdot x\cdot5y^3\cdot2}{5y^3\cdot3\cdot x\cdot x}=\dfrac{4}{x}\)