Bài 8.CM các hằng dẳng tức sau
1) \(\left(a+b\right)^2-\left(a-b\right)^2=4ab\)
2) \(\left(a+b\right)^2+\left(a-b\right)^2=2\left(a^2+b^2\right)\)
3) \(\left(a+b\right)^2-4ab=\left(a-b\right)^2\)
4)\(\left(a-b\right)^2+4ab=\left(a+b\right)^2\)
Những câu hỏi liên quan
a) \(\left(a+b\right)^2=\left(a-b\right)+4ab
\)
b) \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
c) \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(ay+bx\right)^2\)
a) Sửa đề: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
Ta có: \(VP=\left(a-b\right)^2+4ab\)
\(=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2\)
\(=\left(a+b\right)^2=VT\)(đpcm)
b) Ta có: \(VT=\left(a-b\right)^2\)
\(=a^2-2ab+b^2\)
\(=a^2+2ab+b^2-4ab\)
\(=\left(a+b\right)^2-4ab=VP\)(đpcm)
c) Ta có: \(VP=\left(ax-by\right)^2+\left(ay+bx\right)^2\)
\(=a^2x^2-2axby+b^2y^2+a^2y^2+2aybx+b^2x^2\)
\(=a^2x^2+b^2y^2+a^2y^2+b^2x^2\)
\(=a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right)\)
\(=\left(x^2+y^2\right)\left(a^2+b^2\right)=VT\)(đpcm)
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Cho \(\left(a+b\right)^3+4ab\ge2.\)Tìm min \(A=3\left(a^4+b^4+a^2b^2\right)-2\left(a^2+b^2\right)+1\)
1. CMR:
a)\(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
b)\(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
a)VT=\(\left(a+b\right)^2=a^2+2ab+b^2\)(1)VP=\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab\)(2)
từ (1) và (2)\(\Rightarrow\)VT=VP.Vậy \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\left(đpcm\right)\)
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a) Ta có \(VP=\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2=VT\)
\(\Rightarrow\)đpcm
b) Ta có \(VP=\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab\)
\(=a^2-2ab+b^2=\left(a-b\right)^2=VT\)
\(\Rightarrow\)đpcm
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a, Ta có:
\(\left(a-b\right)^2+4ab\)
\(=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2=VT\)
=>đpcm
b, ta có:
\(Vp=\left(a+b\right)^2-4ab\)
\(=a^2+2ab+b^2-4ab\)
\(=a^2-2ab+b^2=\left(a-b\right)^2=VT\)
=>đpcm
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Xem thêm câu trả lời
\(a\left(b-c\right)^2+b\left(a-c\right)^2+c\left(a-b\right)^2-a^3-b^3-c^3+4ab\) nếu ko thấy thì là +4ab
\(a\left(b^3-c3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
chứng minh rằng
\(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
\(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
Cho a, b, là số hữu tỉ thỏa mãn: \(\left(a^2+b^2-2\right).\left(a+b\right)^2+\left(1-ab\right)^2=-4ab\). CM: \(\sqrt{1+ab}\) là số hữu tỉ
1/ CMR : \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
\(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
2/ Tính :
\(\left(a+b+c\right)^2\)
1)VP=(a-b)2+4ab=a2-2ab+b2+4ab
=a2+2ab+b2=(a+b)2=VT
Vậy (a+b)2=(a-b)2+4ab
VP = (a+b)2-4ab=a2+2ab+b2-4ab
=a2-2ab+b2=(a-b)2=VT
Vậy (a-b)2=(a+b)2-4ab
2)(a+b+c)2=[(a+b)+c]2=(a+b)2+2(a+b)c+c2=(a2+2ab+b2)+2ac+2bc+c2
=a2+b2+c2+2ab+2ac+2bc
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Chứng minh các đẳng thức:
a)\(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)=x^4-y^4\)
b)\(\left(a+b\right)^2-\left(a-b\right)^2=4ab\)
Ta có : VP = \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
Vp\(=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\) = VT
Vậy \(x^4-y^4\) \(=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\) (đpcm)
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Cho \(^{\left(a^3+b^3\right)+4ab\ge2}\)
Tìm min của \(Q=3\left(a^4+b^4+a^2b^2\right)-2\left(a^2+b^2\right)+1\)