Chứng minh đẳng thức:
\(\frac{x-2}{-x}=\frac{2^3-x^3}{x\left(x^2+2x+4\right)}\)
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22 tháng 3 2020
\(\left(\frac{2x+2y-z}{3}\right)^2+\left(\frac{2y+2z-x}{3}\right)^2+\left(\frac{2z+2x-y}{3}\right)^2\\ =\frac{4x^2+4y^2+z^2+8xy-4xz-4yz}{9}+\frac{4y^2+4z^2+x^2+8yz-4xy-4xz}{9}+\frac{4z^2+4x^2+y^2+8xz-4yz-4xy}{9}\\ =\frac{9x^2+9y^2+9z^2}{9}=x^2+y^2+z^2\)
22 tháng 3 2020
- Ta có : \(\left(\frac{2x+2y-z}{3}\right)^2+\left(\frac{2y+2z-x}{3}\right)^2+\left(\frac{2x+2z-y}{3}\right)^2\)
\(=\frac{\left(2x+2y-z\right)^2}{9}+\frac{\left(2y+2z-x\right)^2}{9}+\frac{\left(2x+2z-y\right)^2}{9}\)
\(=\frac{\left(2x+2y-z\right)^2+\left(2y+2z-x\right)^2+\left(2x+2z-y\right)^2}{9}\)
\(=\frac{4x^2+4y^2+z^2+8xy-4yz-4xz+4y^2+4z^2+x^2+8yz-4xy-4xz+4x^2+4z^2+y^2+8xz-4xy-4yz}{9}\)
\(=\frac{9x^2+9y^2+9z^2}{9}=\frac{9\left(x^2+y^2+z^2\right)}{9}=x^2+y^2+z^2\)
TT
21 tháng 2 2020
Mấy câu này bạn nhân chéo là được, sử dụng biến đổi tương đương nhé ! Mình làm mẫu câu a)
Cách 1 :\(\frac{3y}{4}=\frac{6xy}{8x}\) \(\Leftrightarrow3y\cdot8x=6xy\cdot4\)
\(\Leftrightarrow24xy=24xy\) ( đúng )
Do đó : \(\frac{3y}{4}=\frac{6xy}{8x}\)
Cách 2 : Rút gọn 1 biểu thức : Ta có : \(\frac{6xy}{8x}=\frac{6y}{8}=\frac{3y}{4}=VT\)
KN
24 tháng 11 2019
b) \(\left[\frac{2}{3x}-\frac{2}{x+1}.\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{x+1}.\left(\frac{x+1}{3x}-\left(x+1\right)\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{x+1}.\left(x+1\right)\left(\frac{1}{3x}-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-2\left(\frac{1}{3x}-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{3x}+2\right]:\frac{x-1}{x}\)
\(=2.\frac{x}{x-1}=\frac{2x}{x-1}\left(đpcm\right)\)
KN
24 tháng 11 2019
a) \(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
\(=\left(\frac{9}{x\left(x^2-9\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\left(\frac{9}{x\left(x+3\right)\left(x-3\right)}+\frac{x^2-3x}{x\left(x+3\right)\left(x-3\right)}\right)\)
\(:\left(\frac{3x-9}{3x\left(x+3\right)}-\frac{x^2}{3x\left(x+3\right)}\right)\)
\(=\frac{x^2-3x+9}{x\left(x+3\right)\left(x-3\right)}:\frac{-x^2+3x-9}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x+3\right)\left(x-3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}\)
\(=\frac{x^2-3x+9}{x-3}.\frac{3}{x^2+3x-9}\)
\(=\frac{x^2-3x+9}{3-x}.\frac{3}{x^2-3x+9}\)
\(=\frac{3}{3-x}\left(đpcm\right)\)
28 tháng 2 2022
a: \(VT=\dfrac{a^2\left(a-4\right)-\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}\)
\(=\dfrac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-1\right)\left(a^2-5a+4\right)}\)
\(=\dfrac{\left(a-4\right)\left(a+1\right)}{\left(a-4\right)\left(a-1\right)}=\dfrac{a+1}{a-1}=VP\)
b: \(VT=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2}{x^2+1}=VP\)
2 tháng 7 2019
\(\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2\left(x+1\right)}{\left(x+1\right).3x}-\frac{2\left(-x-1\right)}{x+1}\right]:\frac{x-1}{x}\)
\(=\)\(\left[\frac{2}{3x}-\frac{2\left(x+1\right)}{\left(x+1\right).3x}+\frac{2\left(x+1\right)}{x+1}\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{3x}+2\right]:\frac{x-1}{x}\)
\(=2.\frac{x}{x-1}=\frac{2x}{x-1}\)\(\left(đpcm\right)\)
6 tháng 11 2018
hùi nãy mem nào k sai cho t T_T t buồn
\(VT\ge6\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-2\left(xy+yz+zx\right)+2.\frac{9}{4\left(x+y+z\right)}\)
\(=6\left(x+y+z\right)^2-2.\frac{\left(x+y+z\right)^2}{3}+\frac{9}{2\left(x+y+z\right)}=6.\left(\frac{3}{4}\right)^2-2.\frac{\left(\frac{3}{4}\right)^2}{3}+\frac{9}{2.\frac{3}{4}}\)
\(=\frac{27}{8}-\frac{3}{8}+6=9\)
\(\Rightarrow\)\(VT\ge9\) ( đpcm )
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{4}\)
Chúc bạn học tốt ~
Ta có: \(\frac{2^3-x^3}{x\left(x^2+2x+4\right)}=\frac{\left(2-x\right)\left(4+2x+x^2\right)}{x\left(4+2x+x^2\right)}=\frac{2-x}{x}\)\(=-\frac{2-x}{-x}=\frac{-\left(2-x\right)}{-x}=\frac{-2+x}{-x}=\frac{x-2}{-x}\)(đpcm)
VP: \(\frac{2^3-x^3}{x\left(x^2+2x+4\right)}\) = \(\frac{\left(2-x\right)\left(4+2x+x^2\right)}{x\left(x^2+2x+4\right)}\) = \(\frac{2-x}{x}\) = \(\frac{-\left(2-x\right)}{-x}\) = \(\frac{x-2}{-x}\) (VT)