\(\dfrac{x}{x-2}+\dfrac{x+2}{x}>2\)
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22 tháng 6 2021
đk: x khác 2;4
\(\dfrac{2}{\left(x-4\right)\left(2-x\right)}-\dfrac{x-1}{x-2}-\dfrac{x+3}{x-4}=0\)
<=> \(\dfrac{-2-\left(x-1\right)\left(x-4\right)-\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=0\)
<=> \(-2-x^2+5x-4-x^2-x+6=0\)
<=> -2x2 +4x = 0
<=> 2x(x-2) = 0
Mà x khác 2
<=> 2x = 0
<=> x = 0
19 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)
19 tháng 10 2021
b: Ta có: \(3^x+3^{x+2}=20\)
\(\Leftrightarrow3^x\cdot10=20\)
\(\Leftrightarrow3^x=2\left(loại\right)\)
22 tháng 6 2021
`1/(x-1)-(3x^2)/(x^3-1)=(2x)/(x^2+x+1)`
ĐK:`x ne 1`
`pt<=>(x^2+x+1)/(x^3-1)-(3x^2)/(x^3-1)=(2x(x-1))/(x^3-1)`
`<=>x^2+x+1-3x^2=2x^2-2x`
`<=>4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+x-1=0`
`<=>(x-1)(4x+1)=0`
`x ne 1=>x-1 ne 0`
`<=>4x+1=0`
`<=>x=-1/4`
Vậy `S={-1/4}`
PH
1
22 tháng 6 2021
đk: x khác -1; 3
\(\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{x}{2\left(3-x\right)}\)
<=> \(\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}+\dfrac{x}{2\left(x-3\right)}=0\)
<=> \(\dfrac{x\left(x-3\right)-4x+x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}=0\)
<=> \(x^2-3x-4x+x^2+x=0\)
<=> 2x2 - 6x = 0
<=> 2x(x-3) = 0
Mà x khác 3
<=> 2x = 0
<=> x = 0
PH
2
30 tháng 6 2021
\(\dfrac{-3}{x+2}< \dfrac{2}{3-x}\) Dk ( x ≠ -2 ; x ≠ 3)
MTC : (x + 2) (3 - x)
\(\Leftrightarrow\dfrac{-3\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}< \dfrac{2\left(x+2\right)}{\left(x+2\right)\left(3-x\right)}\)
\(\Leftrightarrow\) -3(3 - x) < 2(x + 2)
\(\Leftrightarrow\) -9 + 3x < 2x + 4
\(\Leftrightarrow\) 3x - 2x < 4 + 9
\(\Leftrightarrow\) x < 13
Vay x < 13
Chuc ban hoc tot
30 tháng 6 2021
Ta có: \(\dfrac{-3}{x+2}< \dfrac{2}{3-x}\)
\(\Leftrightarrow\dfrac{-3}{x+2}-\dfrac{2}{3-x}< 0\)
\(\Leftrightarrow\dfrac{-3}{x+2}+\dfrac{2}{x-3}< 0\)
\(\Leftrightarrow\dfrac{-3\left(x-3\right)+2\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}< 0\)
\(\Leftrightarrow\dfrac{-3x+9+2x+4}{\left(x+2\right)\left(x-3\right)}< 0\)
\(\Leftrightarrow\dfrac{-x+13}{\left(x+2\right)\left(x-3\right)}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}x>13\\-2< x< 3\end{matrix}\right.\)
29 tháng 7 2021
a) Ta có: \(\left(x-\dfrac{3}{4}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{3}{4}=0\)
hay \(x=\dfrac{3}{4}\)
b) Ta có: \(\left(x+\dfrac{4}{9}\right)^2=\dfrac{49}{144}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{9}=\dfrac{7}{12}\\x+\dfrac{4}{9}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{36}\\x=\dfrac{-37}{36}\end{matrix}\right.\)
29 tháng 10 2021
a: \(x\cdot\dfrac{3}{4}+x=\dfrac{7}{8}\)
\(\Leftrightarrow x\cdot\dfrac{7}{4}=\dfrac{7}{8}\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
6 tháng 2 2022
ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
\(A=\left(\dfrac{3\sqrt{x}}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{x+6\sqrt{x}+9}\)
\(=\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{\left(\sqrt{x}+3\right)^2}\)
\(=\dfrac{3\sqrt{x}+9-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)
\(=\dfrac{\sqrt{x}+13}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
Vậy...
LT
7 tháng 6 2021
\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)\(\left(ĐK:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
Chúc bạn học tốt ^.^
AT
7 tháng 6 2021
\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11}{9-x}\left(x\ge0,x\ne9\right)\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
\(\dfrac{x}{x-2}\) =\(\dfrac{x+2}{x}\) > 2
MTC : x(x - 2)
\(\dfrac{x^2}{x\left(x-2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}=\dfrac{2x\left(x-2\right)}{x\left(x-2\right)}\)
\(\Leftrightarrow\) x2 + (x - 2) (x + 2) > 2x(x - 2)
\(\Leftrightarrow\) x2 + x2 - 4 > 2x2 - 4x
\(\Leftrightarrow\) x2 + x2 + 4 > 2x2 - 4x
\(\Leftrightarrow\) x2 + x2 - 2x2 + 4x > -4
\(\Leftrightarrow\) 4x > -4
\(\Leftrightarrow\) x > \(\dfrac{-4}{4}=-1\)
Vay x > - 1
Minh xin loi ban nhe , ban sua lai giup minh :
\(\dfrac{x^2}{x\left(x-2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}>\dfrac{2x\left(x-2\right)}{x\left(x-2\right)}\)
Chuc ban hoc tot