\(\dfrac{x+9}{x+2}\)
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27 tháng 8 2021
a: Ta có: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{x-4}\)
\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+22}{x-4}\)
d: Ta có: \(D=\dfrac{1}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-12}{x-9}\)
\(=\dfrac{\sqrt{x}-3+x+3\sqrt{x}+2\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+6\sqrt{x}-15}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
4 tháng 7 2021
Ta có: \(\left(\dfrac{x^2-3x}{x^2-9}-1\right):\left(\dfrac{9-x^2}{x^2+x-6}-\dfrac{x-3}{2-x}+\dfrac{x-2}{x+3}\right)\)
\(=\left(\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-1\right):\left(\dfrac{9-x^2+x^2-9+\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\right)\)
\(=\left(\dfrac{x}{x+3}-1\right):\dfrac{x-2}{x+3}\)
\(=\dfrac{x-x-3}{x+3}\cdot\dfrac{x+3}{x-2}\)
\(=\dfrac{-3}{x-2}\)
H
4 tháng 7 2021
Điều kiện : x ≠ 2 ; x ≠ 3 ; x ≠ - 3
\(\left(\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-1\right):\left(\dfrac{\left(3-x\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}-\dfrac{x-3}{2-x}+\dfrac{x-2}{x+3}\right)\)
\(=\left(\dfrac{x}{x+3}-1\right):\left(\dfrac{9-x^2+\left(x-3\right)\left(x+3\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\right)\)
\(=\dfrac{x-x-3}{x+3}:\dfrac{9-x^2+x^2-9+\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{-3}{x+3}:\dfrac{x-2}{\left(x+3\right)}\)
\(=\dfrac{-3}{x-2}\)
28 tháng 7 2021
a) Ta có: \(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(=\dfrac{x^2}{x\left(x-3\right)}-\dfrac{6\left(x-3\right)}{x\left(x-3\right)}-\dfrac{9}{x\left(x-3\right)}\)
\(=\dfrac{x^2-6x+18-9}{x\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
b) Ta có: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(=\dfrac{7\left(x+6\right)-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-7x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-13x+6x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left[x\left(x-13\right)+6\left(x-13\right)\right]}{x\left(x+6\right)}\)
\(=\dfrac{13-x}{x}\)
c) Ta có: \(\dfrac{6}{x-3}-\dfrac{2x-6}{x^2-9}-\dfrac{4}{x+3}\)
\(=\dfrac{6\left(x+3\right)-2x+6-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{6x+18-2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
30 tháng 12 2021
\(\text{a)}\dfrac{-3}{5}-x=\dfrac{21}{10}\)
\(x=\dfrac{-3}{5}-\dfrac{21}{10}=\dfrac{-27}{10}\)
\(\text{b)}x:\dfrac{2}{9}=\dfrac{9}{2}\)
\(x\) \(=\dfrac{9}{2}.\dfrac{2}{9}=1\)
\(\text{c) }\dfrac{x}{9}=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{9.5}{3}=15\)
\(\text{d)}x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)
\(x:\dfrac{8}{125}=\dfrac{125}{8}\)
\(x\) \(=\dfrac{125}{8}.\dfrac{8}{125}=1\)
20 tháng 12 2022
a: \(=\dfrac{x^2-x+x+1+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\)
b: \(=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
c: \(=\dfrac{2x^2-3x-9-x^2+3x+x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2+6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x}{x-3}\)
F
26 tháng 12 2021
a)\(\dfrac{x^2}{x-1}+\dfrac{1-2x}{x-1}\)
=\(\dfrac{x^2+1-2x}{x-1}\)
=\(\dfrac{x^2-2x+1}{x-1}\)
=\(\dfrac{\left(x-1\right)^2}{x-1}\)
= x - 1
F
26 tháng 12 2021
b) \(\dfrac{x}{x-3}\) + \(\dfrac{-9}{x^2-3x}\)
=\(\dfrac{x}{x-3}\)+ \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x.x}{x\left(x-3\right)}\) + \(\dfrac{-9}{x\left(x-3\right)}\)
=\(\dfrac{x^2+3^2}{x\left(x-3\right)}\)
=\(\dfrac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}\)
=\(\dfrac{x+3}{x}\)
#Fiona
19 tháng 12 2020
a) Ta có: \(B=\left(\dfrac{x}{3x-9}+\dfrac{2x-3}{3x-x^2}\right)\cdot\dfrac{3x^2-9x}{x^2+6x+9}\)
\(=\left(\dfrac{x}{3\left(x-3\right)}-\dfrac{2x-3}{x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\left(\dfrac{x^2}{3x\left(x-3\right)}-\dfrac{3\left(2x-3\right)}{3x\left(x-3\right)}\right)\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x^2-6x+9}{3x\left(x-3\right)}\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x^2-6x+9}{x^2+6x+9}\)
b) Ta có: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{1}{x+2}\)
\(=\left(\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{1}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{1}\)
\(=\dfrac{-6}{x-2}\)
11 tháng 2 2022
1 x 7 x 6 x 9 1 x 7 x 3 x 2 x 3 x 3
----------------- = --------------------------- = \(\dfrac{1}{5}\)
7 x 9 x 6 x 5 7 x 3 x 3 x 3 x 2 x 5
=> (x+2+7)/x+2
nên 7 chia hết x+2 nên
x thuộc 5
giúp tui điiii