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\(\left(\frac{1}{2}x+y\right).\left(\frac{1}{2}x+y\right)\)
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NV
19 tháng 6 2016
a/ \(\left(\frac{1}{2}x+y\right)\left(\frac{1}{2}x+y\right)=\left(\frac{1}{2}x+y\right)^2=\frac{1}{4}x^2+xy+y^2\)
b/ \(\left(x-\frac{1}{2}y\right)\left(x-\frac{1}{2}y\right)=\left(x-\frac{1}{2}y\right)^2=x^2-xy+\frac{1}{4}y^2\)
19 tháng 6 2016
đơn giản lăm hdt đáng nhớ đó bn
a) = ( x/2 +y)2
b) = ( x - y/2)2
BV
2
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
12 tháng 9 2018
Điều kiên \(y\ne0\)
\(\frac{x}{y}=16\Rightarrow x=16y\)thế vào \(\frac{x}{y^2}=2\)
\(\Rightarrow\frac{16y}{y^2}=2\Rightarrow\frac{16}{y}=2\Rightarrow y=8\) thế vào \(\frac{x}{y}=16\Rightarrow\frac{x}{8}=16\Rightarrow x=8.16=128\)
12 tháng 9 2018
tìm x biếtxy2 =2vàxy =16(y≠0)
\(\Rightarrow x=y^2.2\)
Vì Y khác 0 nên y = 1
\(\Rightarrow x=1^2.2\)
\(\Rightarrow x=2\)
Vậy x = 2
TA
22 tháng 8 2017
\(B=\left[\left(\frac{x}{y}-\frac{y}{x}\right):\left(x-y\right)-2.\left(\frac{1}{y}-\frac{1}{x}\right)\right]:\frac{x-y}{y}\)
\(=\left[\frac{x^2-y^2}{xy}.\frac{1}{x-y}-2.\frac{x-y}{xy}\right].\frac{y}{x-y}\)
\(=\left(\frac{\left(x-y\right)\left(x+y\right)}{xy.\left(x-y\right)}-\frac{2.\left(x-y\right)}{xy}\right).\frac{y}{x-y}\)
\(=\left(\frac{x+y}{xy}-\frac{2x-2y}{xy}\right).\frac{y}{x-y}=\frac{x+y-2x+2y}{xy}.\frac{y}{x-y}=\frac{y.\left(3y-x\right)}{xy.\left(x-y\right)}=\frac{3y-x}{x.\left(x-y\right)}\)
\(C=\left(\frac{x+y}{2x-2y}-\frac{x-y}{2x+2y}-\frac{2y^2}{y-x}\right):\frac{2y}{x-y}\)
\(=\left(\frac{x+y}{2.\left(x-y\right)}-\frac{x-y}{2.\left(x+y\right)}+\frac{2y^2}{x-y}\right).\frac{x-y}{2y}\)
\(=\frac{\left(x+y\right)^2-\left(x-y\right)^2+2.2y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{\left(x+y+x-y\right)\left(x+y-x+y\right)+4y^2.\left(x+y\right)}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}\)
\(=\frac{4xy+4xy^2+4y^3}{2.\left(x-y\right)\left(x+y\right)}.\frac{x-y}{2y}=\frac{4y.\left(x+xy+y^2\right).\left(x-y\right)}{4y.\left(x-y\right)\left(x+y\right)}=\frac{x+xy+y^2}{x+y}\)
\(D=3x:\left\{\frac{x^2-y^2}{x^3+y^3}.\left[\left(x-\frac{x^2+y^2}{y}\right):\left(\frac{1}{x}-\frac{1}{y}\right)\right]\right\}\)
\(=3x:\left\{\frac{\left(x+y\right)\left(x-y\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}.\left[\frac{xy-x^2-y^2}{y}:\frac{y-x}{xy}\right]\right\}\)
\(=3x:\left[\frac{x-y}{x^2-xy+y^2}.\left(\frac{xy-x^2-y^2}{y}.\frac{xy}{y-x}\right)\right]\)
\(=3x:\left(\frac{x-y}{x^2-xy+y^2}.\frac{xy.\left(x^2-xy+y^2\right)}{y.\left(x-y\right)}\right)\)
\(=3x:\frac{xy.\left(x-y\right)\left(x^2-xy+y^2\right)}{y.\left(x-y\right)\left(x^2-xy+y^2\right)}=3x:x=3\)
\(E=\frac{2}{x.\left(x+1\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+3\right)}\)
\(=2.\left(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)\)
\(=2.\frac{\left(x+2\right)\left(x+3\right)+x.\left(x+3\right)+x.\left(x+1\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{x^2+2x+3x+6+x^2+3x+x^2+x}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=2.\frac{3x^2+9x+6}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=2.\frac{3.\left(x^2+3x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x^2+x+2x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6.\left[x.\left(x+1\right)+2.\left(x+1\right)\right]}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\frac{6.\left(x+1\right)\left(x+2\right)}{x.\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{6}{x.\left(x+3\right)}\)
PN
11 tháng 11 2015
Công thức tống quát:
\(1+\frac{1}{\left(n-1\right)\left(n+1\right)}=1+\frac{1}{n^2-1}=\frac{n^2-1+1}{n^2-1}=\frac{n^2}{n^2-1}\)
Theo đó, ta có:
\(1+\frac{1}{1.3}=1+\frac{1}{\left(2-1\right)\left(2+1\right)}=\frac{2^2}{2^2-1}\)
\(1+\frac{1}{2.4}=1+\frac{1}{\left(3-1\right)\left(3+1\right)}=\frac{3^2}{3^2-1}\)
\(1+\frac{1}{3.5}=\frac{1}{\left(4-1\right)\left(4+1\right)}=\frac{4^2}{4^2-1}\)
\(....................\)
\(1+\frac{1}{2015.2017}=1+\frac{1}{\left(2016-1\right)\left(2016+1\right)}=\frac{2016^2}{2016^2-1}\)
Nhân lần lượt các đẳng thức trên, ta được:
\(S=\frac{\left(2.3.4....2016\right)^2}{\left(2^2-1\right)\left(3^2-1\right)\left(4^2-1\right)...\left(2016^2-1\right)}=\frac{2^2.3^2.4^2...2016^2}{\left(1.3\right)\left(2.4\right)\left(3.5\right)....\left(2015.2017\right)}=\frac{2^2.3^2.4^2...2016^2}{1.2.3^2.4^2.5^2...2014^2.2015^2.2016.2017}=\frac{2.2016}{2017}\)
có ai đang onl thì tạo phòng với mình đê
\(\left(\frac{1}{2}x+y\right)\left(\frac{1}{2}x+y\right)=\left(\frac{1}{2}x+y\right)^2\)