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\(a,b)\)Ta có: \(\left(a\pm b\right)^2\)
\(=\left(a\pm b\right)\left(a\pm b\right)\)
\(=a^2\pm ab\pm ab+b^2\)
\(=a^2\pm ab+b^2\)
\(c)\)\(\left(a+b\right)\left(a-b\right)=a^2-ab+ab-b^2=a^2-b^2\)
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biến đổi vế trái : a. \(\left(a+b\right)^2=a^2+2ab+B^2=VP\)
b. \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3=VP\)
c. \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=VP\)
xem 7 hằng đẳng thức đáng nhớ
a)\(=\left(a+b\right)^2=\left(a+b\right)\left(a+b\right)=a^2+ab+ab+b^2\)
\(=a^2+2ab+b^2\)
b)\(\left(a-b\right)^3=\left(a-b\right)\left(a-b\right)\left(a-b\right)=\left(a^2-ab-ab+b^2\right)\left(a-b\right)\)
\(=\left(a^2-2ab+b^2\right)\left(a-b\right)\)
\(=a^3-a^2b-2a^2b+2ab^2+ab^2-b^3\)
\(=a^3-3a^2b-3ab^2-b^3\)
c)\(\left(a+b+c\right)^2=\left(a+b+c\right)\left(a+b+c\right)\)
\(=a^2+ab+ac+ab+b^2+bc+ac+cb+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac\)
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a) \(x^2+2x+1=\left(x+1\right)^2\)
b) \(9x^2+y^2+6xy=\left(3x+y\right)^2\)
c) \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)
d) \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
e) \(\left(2x+3y\right)^3+2\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
f) mk chỉnh lại đề nha:
\(2xy^2+x^2y^4+1=\left(xy^2+1\right)^2\)
g) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
h) \(x^2-10xy+25y^2=\left(x-5y\right)^2\)
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a) \(a^2+b^2=\left(a+b\right)^2-2ab\)
\(VP=\left(a+b\right)^2-2ab=a^2+2ab+b^2-2ab\)\(=a^2+b^2=VT\)
\(\Rightarrowđpcm\)
b)\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2\)
\(VP=a^4+b^4+2a^2b^2-2a^2b^2=a^4+b^4=VT\)\(\Rightarrowđpcm\)
c) \(a^6+b^6=\left(a^2+b^2\right)\left[\left(a^2+b^2\right)^2-3a^2b^2\right]\)
\(VP=\left(a^2+b^2\right)\left(a^4-a^2b^2+b^4\right)=a^6+b^6\)
\(VP=VT\Rightarrowđpcm\)
d)\(a^6-b^6=\left(a^2-b^2\right)[\left(a^2+b^2\right)^2-a^2b^2]\)
\(VP=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)=a^6-b^6=VT\)
\(VP=VT\Rightarrowđpcm\)
<=> a^2 - 2ab + b^2 - 2ab = a^2 +b^2
<=> a^2 +b^2 - a^2 - b^2 = 0
<=> 0 = 0 (luôn đúng)
= bien doi ve phai co a2 + b2 = a2 + b2 +2ab - 2ab =[ a+b]2 - 2ab
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