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DN
0
LQ
0
15 tháng 11 2021
\(a,=\dfrac{x^4\left(x-2\right)+2x^2\left(x-2\right)-3\left(x-2\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4+2x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4-x^2+3x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x-1\right)\left(x^2+3\right)}{x+4}\)
\(b,=\dfrac{x^4-3x^2-x^2+3}{x^4-x^2+7x^2-7}=\dfrac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\dfrac{x^2-3}{x^2+7}\\ c,=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}=\dfrac{x^2-1}{x^2+1}\)
H9
HT.Phong (9A5)
CTVHS
7 tháng 8 2023
a) \(x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left|x-2y\right|\)
TH1: \(x-2y--\left(x-2y\right)\)
\(=x-2y+x-2y\)
\(=2x-4y\)
TH2: \(x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
b) \(x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+\left|x^2-4\right|\)
TH1:
\(x^2+-\left(x^2-4\right)\)
\(=x^2-x^2+4\)
\(=4\)
TH2:
\(x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
c) \(2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\) (x>5)
\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}\)
\(=2x-1-\sqrt{x-5}\)
d) \(\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}\) (\(x>\sqrt{2}\))
\(=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}\)
\(=\sqrt{x^2-2}\)
e) \(\sqrt{\left(x^2-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}\)
\(=\left|x^2-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}\)
\(=\left|x^2-4\right|+\sqrt{\dfrac{\left(x-4\right)^2}{\left(x-4\right)^2}}\)
\(=\left|x^2-4\right|+1\)
TH1:
\(x^2-4+1\)
\(=x^2-3\)
TH2:
\(-\left(x^2-4\right)+1\)
\(=-x^2+4+1\)
\(=-x^2+5\)
7 tháng 8 2023
a: \(A=x-2y-\sqrt{x^2-4xy+4y^2}\)
=x-2y-|x-2y|
Khi x>=2y thì A=x-2y-x+2y=0
Khi x<2y thì A=x-2y+x-2y=2x-4y
b: \(B=x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\left|x^2-4\right|\)
TH1: x>=2 hoặc x<=-2
B=x^2+x^2-4=2x^2-4
TH2: -2<=x<=2
B=x^2+4-x^2=4
c: \(C=2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\)
\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}=2x-1-\sqrt{x-5}\)
d: \(D=\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}=\sqrt{x^2-2}\)
2 tháng 12 2020
a, \(2\left(x+3\right)\left(x-4\right)=\left(2x-1\right)\left(x+2\right)-27\)
\(\Leftrightarrow2\left(x^2-4x+3x-12\right)=2x^2+4x-x-2-27\)
\(\Leftrightarrow2x^2-2x-24=2x^2+3x-29\Leftrightarrow-5x+5=0\Leftrightarrow x=1\)
b, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
\(\Leftrightarrow x^3-8-x\left(x^2-9\right)=26\Leftrightarrow-8+9x=26\)
\(\Leftrightarrow9x=18\Leftrightarrow x=2\)
1 tháng 8 2023
1: =>x^2-x=3-x
=>x^2=3
=>x=căn 3 hoặc x=-căn 3
2: =>x^2-4x+3=x^2-4x+4 và x>=2
=>3=4(vô lý)
3: =>2|x-1|=6
=>|x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=-2 hoặc x=4
4: =>|2x-3|=|x-2|
=>2x-3=x-2 hoặc 2x-3=-x+2
=>x=1 hoặc x=5/3
5: =>\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)
=>x+2=0
=>x=-2
N
Rút gọn:
\(M=\dfrac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\dfrac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
2
23 tháng 10 2018
điều kiện \(\left[{}\begin{matrix}x\ge2\\x< -2\end{matrix}\right.\)
ta có : \(M=\dfrac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\dfrac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
\(=\dfrac{\left(x+2+\sqrt{x^2-4}\right)^2+\left(x+2-\sqrt{x^2-4}\right)^2}{\left(x+2-\sqrt{x^2-4}\right)\left(x+2+\sqrt{x^2-4}\right)}\)
\(=\dfrac{4x^2+8x}{4x+8}=x\)
N
Rút gọn:
\(M=\dfrac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\dfrac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
1
AL
22 tháng 10 2018
M= \(\dfrac{x+2+\sqrt{x^2-4}}{x+2-\sqrt{x^2-4}}+\dfrac{x+2-\sqrt{x^2-4}}{x+2+\sqrt{x^2-4}}\)
M=\(\dfrac{(x+2+\sqrt{x^2-4})^2}{(x+2)^2+\sqrt{x^2-4}^2}+\dfrac{(x+2-\sqrt{x^2-4})^2}{(x+2)^2+\sqrt{x^2-4}^2}\)
M=\(\dfrac{\left(x+2\right)^2+2.\left(x+2\right).\sqrt{x^2-4}+x^2-4}{x^2+4x+4-x^2+4}+\dfrac{\left(x+2\right)^2-2.\left(x+2\right).\sqrt{x^2-4}+x^2-4}{x^2+4x+4-x^2+4}\)
M=\(\dfrac{\left(x+2\right)^2+x^2-4+\left(x+2\right)^2+x^2-4}{x^2+4x+4-x^2+4}\)
M=\(\dfrac{x^2+4x+4+x^2-4+x^2+4x+4+x^2-4}{4x+8}\)
M= \(\dfrac{4x^2+8}{4x+8}\)
M= \(\dfrac{x.\left(4x+8\right)}{4x+8}\)
M= x
OK nha bn !!
6 tháng 2 2022
a: \(=\dfrac{\left|x+2\right|}{x-1}\)
b: \(=x-2y-\left|x-2y\right|\)\(=\left[{}\begin{matrix}x-2y-x+2y=0\\x-2y+x-2y=2x-4y\end{matrix}\right.\)
c: \(=\dfrac{\left|x+2\right|}{\left(x+2\right)\left(x-2\right)}=\pm\dfrac{1}{x-2}\)
thiếu đề kìa