5x + 8xy + 5y với x + y \(\dfrac{2}{5}\) ; x . y = \(\dfrac{3}{4}\)
2xy + 7xyz - 2xz với x = \(\dfrac{3}{7}\); y - z = \(\dfrac{5}{2}\) ; y . z = -1
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NN
3 tháng 12 2017
a) \(\dfrac{3x-2}{2xy}+\dfrac{7x+2}{2xy}\)
\(=\dfrac{\left(3x-2\right)+\left(7x+2\right)}{2xy}\)
\(=\dfrac{3x-2+7x+2}{2xy}\)
\(=\dfrac{10x}{2xy}\)
\(=\dfrac{5}{y}\)
b) \(\dfrac{5x+y^2}{x^2y}+\dfrac{x^2-5y}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}+\dfrac{x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)+x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3+x^3-5xy}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
c) \(\dfrac{3x-2}{2xy}-\dfrac{7x-y}{2xy}\)
\(=\dfrac{\left(3x-2\right)-\left(7x-y\right)}{2xy}\)
\(=\dfrac{3x-2-7x+y}{2xy}\)
\(=\dfrac{-2-4x+y}{2xy}\)
d) \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
e) \(\dfrac{16xy}{3x-1}.\dfrac{3-9x}{12xy^3}\)
\(=\dfrac{16xy\left(3-9x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{4\left(3-9x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4\left(9x-3\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4.3\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-12}{3y^2}\)
\(=\dfrac{-4}{y^2}\)
f) \(\dfrac{8xy}{3x-1}:\dfrac{12xy^3}{5-15x}\)
\(=\dfrac{8xy}{3x-1}.\dfrac{5-15x}{12xy^3}\)
\(=\dfrac{8xy\left(5-15x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{2\left(5-15x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2\left(15x-5\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2.5\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-10}{3y^2}\)
30 tháng 6 2019
\(a,A=5x+8xy+5y=(5x+5y)+8xy\)
\(=5(x+y)+8xy\)
\(=5\cdot\frac{2}{5}+8\cdot(-1)=2+(-8)=-6\)
\(b,B=2xy+7xyz-2xz\)
\(=2\cdot\frac{3}{7}y+7\cdot\frac{3}{7}yz-2\cdot\frac{3}{7}z\)
\(=\frac{6}{7}y+3yz-\frac{6}{7}z\)
\(=\frac{6}{7}y+3\cdot(-1)-\frac{6}{7}z\)
\(=\frac{6}{7}y+(-3)-\frac{6}{7}z\)
Làm nốt :v
30 tháng 6 2019
a)
A=\(5\left(x+y\right)+8xy\)
\(=5.\frac{2}{5}+8.\left(-1\right)\)
\(=2-8\)
\(=-6\)
Vậy.......
hc tốt
19 tháng 12 2021
b: \(=\dfrac{x^3+6x^2-25}{x\left(x+5\right)\left(x-2\right)}+\dfrac{x+5}{x\left(x-2\right)}\)
\(=\dfrac{x^3+6x^2-25+x^2+10x+25}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x^3+7x^2+10x}{x\left(x+5\right)\left(x-2\right)}=\dfrac{x+2}{x-2}\)
B
2
TA
30 tháng 8 2016
ta có:
A= 5(x+y)+8xy
Thay xy=3/4 và x+y=2/5 vào biểu thức A ta có:
A=5*2/5+8*3/4=8
14 tháng 6 2018
A= ( 5x + 5y ) + 8xy
A = 5(x+y)+8xy
A= 5.2/5+8.3/4
A= 2 + 6
A = 8.
4 tháng 3 2018
1. \(x^2+y^2+z^2+3=2\left(x+y+z\right)< =>x^2-2x+1+y^2-2y+1+z^2-2z+1=0< =>\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\)
=>x-1=0<=>x=1
y-1=0<=>y=1
z-1=0<=>z=1
vậy....
2. \(\dfrac{2-x}{2008}-1=\dfrac{1-x}{2009}-\dfrac{x}{2010}\)
<=>\(\dfrac{2-x}{2008}+1=\dfrac{1-x}{2009}+1-\dfrac{x}{2010}+1\)
<=>\(\dfrac{2010-x}{2008}=\dfrac{2010-x}{2009}+\dfrac{2010-x}{2010}\)
<=>(2010-x)(1/2008-1/2009-1/2010)=0
vì 1/2008-1/2009-1/2010 khác 0 nên 2010-x=0<=>x=2010
RD
4 tháng 3 2018
1)\(x^2+y^2+z^2+3=2\left(x+y+z\right)\)
\(\Leftrightarrow x^2-2x+1+y^2-2y+1+z^2-2z+1=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\)
\(\Leftrightarrow x=y=z=1\)
2)\(\dfrac{2-x}{2008}-1=\dfrac{1-x}{2009}-\dfrac{x}{2010}\)
\(\Leftrightarrow\dfrac{2-x}{2008}+1=\dfrac{1-x}{2009}+1-\dfrac{x}{2010}+1\)
\(\Leftrightarrow\dfrac{2010-x}{2008}=\dfrac{2010-x}{2009}+\dfrac{2010-x}{2010}\)
\(\Leftrightarrow\left(2010-x\right)\left(\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{1}{2010}\right)=0\)
\(\Leftrightarrow x=2010\)(vì \(\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{1}{2010}\ne0\))
DT
1
