viết dưới dạng tổng 2 bình phương
\(a^2-4ab+5b^2-4bc+4c^2\)
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\(1,\)\(4x^2-4x+y^2+2y+2\)
\(=4x^2+4x+1+y^2+2y+1\)
\(=\left[\left(2x\right)^2-2.2x+1\right]+\left(y^2+2.y.1+1^2\right)\)
\(=\left(2x-1\right)^2+\left(y+1\right)^2\)
\(2,\)\(a^2-4ab+5b^2-4bc+4c^2\)
\(=a^2-4ab+4b^2+b^2-4bc+4c^2\)
\(=\left[a^2-2.a.2b+\left(2b\right)^2\right]+\left[b^2-2.b.2c+\left(2c\right)^2\right]\)
\(=\left(a-2b\right)^2+\left(b-2c\right)^2\)
\(3,\)\(16x^2+5+8x-4y+y^2\)
\(=16x^2+8x+1+y^2-4y+4\)
\(=\left[\left(4x\right)^2+2.4x.1+1^2\right]+\left[y^2-2.y.2+2^2\right]\)
\(=\left(4x+1\right)^2+\left(y-2\right)^2\)
67x73 = (70-3)(70+3) = 702 - 32 = 4900 - 9 = 4801.
a) \(16a^2-24ab+9b^2=\left(4a-3b\right)^2.\)
b) \(a^2+4ab+4b^2=\left(a+2b\right)^2\)
TL:
67 x 73 = ( 70 - 3 ) ( 70 + 3 ) = 702 - 32 = 4900 - 9 = 4801
a) \(16a^2\)\(-24ab+9ab=\left(4a-3b\right)^2\)
b) \(a^2\)\(+4ab+4b^2\)\(=\left(a+2b\right)^2\)
~HT~
Bài 1 :
a) \(x^2+y^2\)
\(\Leftrightarrow x^2+2xy+y^2-2xy\)
\(\Leftrightarrow\left(x+y\right)^2-2xy=\left(-3\right)^2-2.\left(-28\right)=65\)
b) \(x^3+y^3\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(\Leftrightarrow\left(x+y\right)\left(x^2+2xy+y^2-3xy\right)\)
\(\Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=\left(-3\right)\left[\left(-3\right)^2-3.\left(-28\right)\right]=-279\)
c) \(x^4+y^4\)
\(\Leftrightarrow\left(x+y\right)^4-4x^3y-4xy^3-6x^2y^2=\left(-3\right)^4-4\left(-28\right).65-6\left(-28\right)^2=2657\)
Ta có: \(\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
\(=\left(a^2c^2+2abcd+b^2d^2\right)+\left(a^2d^2-2abcd+b^2c^2\right)\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
=> đpcm
\(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
( a2 + b2 )( c2 + d2 )
= a2c2 + a2d2 + b2c2 + b2d2
= ( a2c2 + 2abcd + b2d2 ) + ( a2d2 - 2abcd + b2c2 )
= ( ac + bd )2 + ( ad - bc )2
a2 - 4ab + 5b2 - 4bc + 4c2
= a2 - 4ab + 4b2 + 4c2 - 4bc + b2
= (a - 2b)2 + (2c - b)2
\(a^2-4ab+5b^2-4bc+4c^2\)
\(=a^2-4ab+4b^2+b^2-4bc+4c^2\)
\(=a^2-2.a.2b+\left(2b\right)^2+b^2-2.b.2c+\left(2c\right)^2\)
\(=\left(a-2b\right)^2+\left(b-2c\right)^2\)