\(\left(a+b\right)^2=\left(a-b\right)^2+4ab\) a) tính ( a+b) ^2 biết a-b = 2 , ab = 1 các bn chỉ cách làm cho mình luôn nha phần nào khó hiểu phân tích hộ mình luôn nha
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)\(\left(a-b\right)^2+4ab=\left(a+b\right)^2\)
1. Ta có: \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b+a-b\right)\left(a+b-a+b\right)\)
\(=2a.2b=4ab\)
=> đpcm
2. Ta có: \(\left(a+b\right)^2+\left(a-b\right)^2=a^2+2ab+b^2+a^2-2ab+b^2\)
\(=2a^2+2b^2=2\left(a^2+b^2\right)\)
=> đpcm
3. Ta có:\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab\)
\(=a^2-2ab+b^2=\left(a-b\right)^2\)
=> đpcm
4. Ta có: \(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2\)
\(a,\left(a+b\right)^2-\left(a-b\right)^2=4ab\)
\(\Leftrightarrow\left(a^2+b^2+2ab\right)-\left(a^2+b^2-2ab\right)=4ab\)
\(\Leftrightarrow a^2+b^2-a^2-b^2+2ab+2ab=4ab\)
\(\Leftrightarrow4ab=4ab\Leftrightarrow4ab-4ab=0\Leftrightarrow0=0\)(đpcm)
\(b,\left(a+b\right)^2+\left(a-b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow\left(a^2+b^2+2ab\right)+\left(a^2+b^2-2ab\right)=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+b^2+a^2+b^2+\left(2ab-2ab\right)=2\left(a^2+b^2\right)\)
\(\Leftrightarrow2\left(a^2+b^2\right)=2\left(a^2+b^2\right)\Leftrightarrow2\left(a^2+b^2\right)-2\left(a^2+b^2\right)=0\Leftrightarrow0=0\)(đpcm)
\(c,\left(a+b\right)^2-4ab=\left(a-b\right)^2\)
\(\Leftrightarrow\left(a^2+b^2+2ab\right)-4ab=a^2+b^2-2ab\)
\(\Leftrightarrow a^2+b^2-2ab=a^2+b^2-2ab\)
\(\Leftrightarrow\left(a-b\right)^2=\left(a-b\right)^2\Leftrightarrow\left(a-b\right)^2-\left(a-b\right)^2=0\Leftrightarrow0=0\)(đpcm)
\(d,\left(a-b\right)^2+4ab=\left(a+b\right)^2\)
\(\Leftrightarrow\left(a^2+b^2-2ab\right)+4ab=\left(a+b\right)^2\)
\(\Leftrightarrow a^2+b^2-2ab+4ab=\left(a+b\right)^2\)
\(\Leftrightarrow a^2+b^2+2ab=\left(a+b\right)^2\Leftrightarrow\left(a+b\right)^2=\left(a+b\right)^2\)
\(\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)^2=0\Leftrightarrow0=0\)(đpcm)
1) \(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=2b.2a=4ab\)( đpcm )
2) \(\left(a+b\right)^2+\left(a-b\right)^2=a^2+2ab+b^2+a^2-2ab+b^2\)
\(=2\left(a^2+b^2\right)\)( đpcm )
3) \(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab\)
\(=a^2-2ab+b^2=\left(a-b\right)^2\)( đpcm )
4) \(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2\)( đpcm )
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)
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 \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\left(đpcm\right)\)
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
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
chứng minh rằng
\(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
\(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
\(a\left(b-c\right)^2+b\left(a-c\right)^2+c\left(a-b\right)^2-a^3-b^3-c^3+4ab\) nếu ko thấy thì là +4ab
\(a\left(b^3-c3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
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 : \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
\(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
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
Chứng minh rằng
a) ( a + b ) = \(\left(a-b\right)^2\)+ 4ab
b) \(\left(a-b\right)^2\)= \(\left(a+b\right)^2\)- 4ab
Ta có: \(VP=\left(a-b\right)\left(a-b\right)+4ab\)
\(=a^2-2ab-b^2+4ab\)
\(=a^2-b^2+2ab=\left(a+b\right)^2=VT\left(đpcm\right)\)
b, \(VP=\left(a+b\right)\left(a+b\right)-4ab\)
\(=a^2+2ab+b^2-4ab\)
\(=a^2+b^2-2ab=\left(a-b\right)^2=VT\left(đpcm\right)\)
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
Chứng minh bất đẳng thức: \(4ab\left(a+b\right)\left(a+1\right)\left(a+b+1\right)+b^2\ge0\)