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a) mk chỉnh đề:
Chứng minh: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\) (1)
hoặc \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\) (2)
BÀI LÀM
TH1:
\(VP=\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2=VP\) (đpcm)
TH2:
\(VP=\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2=VT\) (đpcm)
b) \(a+b=9\)\(\Rightarrow\)\(a=9-b\)
Ta có: \(ab=20\)\(\Rightarrow\)\(\left(9-b\right).b=20\)
\(\Leftrightarrow\)\(b^2-9b+20=0\)
\(\Leftrightarrow\)\(\left(b-4\right)\left(b-5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}b=4\\b=5\end{cases}}\)
Nếu \(b=4\)thì: \(a=5\)\(\Rightarrow\)\(\left(a-b\right)^{2011}=\left(5-4\right)^{2011}=1\)
Nếu \(b=5\)thì \(a=4\)\(\Rightarrow\)\(\left(a-b\right)^{2011}=\left(4-5\right)^{2011}=-1\)
a, sửa đề CM: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
\(VP=\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2=VT\left(đpcm\right)\)
b, \(a+b=9\Leftrightarrow\left(a+b\right)^2=81\Leftrightarrow\left(a-b\right)^2+4ab=81\Leftrightarrow\left(a-b\right)^2=81-4.20=1\Leftrightarrow a-b=\pm1\)
Với \(a-b=1\Rightarrow\left(a-b\right)^{2011}=1\)
Với \(a-b=-1\Rightarrow\left(a-b\right)^{2011}=-1\)
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ta có: a+b = 9
=> (a+b)2 = 81
a2 + 2ab + b2 = 81
=> a2 - 2ab + b2 + 4ab = 81
(a-b)2 + 4ab = 81
(a-b)2 + 80= 81
(a-b)2 = 1 = 12 = (-1)2
=> a-b = 1 hoặc a-b = -1
=> (a-b)2015 = 12015 = 1
(a-b)2015 = (-1)2015 = -1
KL:...
a + b = 9 => ( a + b )2 = 81
=> a2 + 2ab + b2 = 81
=> a2 + 2.20 + b2 = 81
=> a2 + b2 + 40 = 81
=> a2 + b2 = 41
Xét ( a - b )2 = a2 - 2ab + b2 = ( a2 + b2 ) - 2 . 20 = 41 - 40 = 1
=> ( a - b )2 = 1
=> a - b = { 1; -1 }
mà a > b => a - b = 1
=> ( a - b )2015 = 12015 = 1
Vậy,......
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Bài 2:
\(A=\dfrac{x\left(x^3+1\right)}{x^2-x+1}-\dfrac{x\left(x^3-1\right)}{x^2+x+1}\)
\(=x\left(x+1\right)-x\left(x-1\right)\)
=x^2+x-x^2+x
=2x
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\(3,\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\left[\left(\frac{1}{x}\right)^2-2.\frac{1}{x}.\frac{1}{y}+\left(\frac{1}{y}\right)^2\right]-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\left[\frac{1}{x^2}-\frac{2}{xy}+\frac{1}{y^2}\right]-\frac{x^2+y^2}{x^2-2xy+y^2}\)
\(=\frac{2}{xy}:\left[\frac{y^2-2.xy+x^2}{x^2y^2}\right]-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}.\frac{x^2y^2}{x^2-2xy+y^2}-\frac{x^2+y^2}{x^2-2xy+y^2}\)
\(=\frac{2xy}{x^2-2xy+y^2}+\frac{-x^2-y^2}{x^2-2xy-y^2}\)
\(=\frac{2xy-x^2-y^2}{x^2-2xy+y^2}=\frac{-\left(x^2-2xy+y^2\right)}{x^2-2xy+y^2}=-1\)
\(\frac{2011^3+11^3}{2011^3+2000^3}\)
\(=\frac{\left(2011+11\right)\left(2011^2-2011.11+11^2\right)}{\left(2011+2000\right)\left(2011^2-2011.2000+2000^2\right)}\)
\(=\frac{\left(2011+11\right)\left[2011^2-11\left(2011-11\right)\right]}{\left(2011+2000\right)\left[2011^2-2000\left(2011-2000\right)\right]}\)
\(=\frac{\left(2011+11\right)\left(2011^2-11.2000\right)}{\left(2011+2000\right)\left(2011^2-2000.11\right)}\)
\(=\frac{2011+11}{2011+2000}\left(2011^2-11.2000\ne0\right)\)
đpcm
ta có a+b=9
=>(a+b)^2=81
=>(â-b)^2+4ab=81
=>(a-b)^2=80-4.20
=>(a-b)^2=80-81
=>(a-b)^2=(-1)
mà a<b nên a-b<0
=> a-b = -1
vậy (a-b)^2011 =(-1) ^ 2011=(-1)
Ta có : \(a+b=9\Leftrightarrow a^2+b^2+2ab=81\Rightarrow a^2+b^2+40=81\)
\(\Rightarrow a^2+b^2=41\Rightarrow a^2+b^2-2ab=41-40=1\)
\(\Leftrightarrow\left(a-b\right)^2=1\Rightarrow a-b=-1\left(a< b\right)\)
\(\Rightarrow\left(a-b\right)^{2011}=-1^{2011}=-1\)