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+ Chứng minh (a + b)2 = (a – b)2 + 4ab
Ta có:
VP = (a – b)2 + 4ab = a2 – 2ab + b2 + 4ab
= a2 + (4ab – 2ab) + b2
= a2 + 2ab + b2
= (a + b)2 = VT (đpcm)
+ Chứng minh (a – b)2 = (a + b)2 – 4ab
Ta có:
VP = (a + b)2 – 4ab = a2 + 2ab + b2 – 4ab
= a2 + (2ab – 4ab) + b2
= a2 – 2ab + b2
= (a – b)2 = VT (đpcm)
+ Áp dụng, tính:
a) (a – b)2 = (a + b)2 – 4ab = 72 – 4.12 = 49 – 48 = 1
b) (a + b)2 = (a – b)2 + 4ab = 202 + 4.3 = 400 + 12 = 412.
Chứng minh rằng
a/ (a+b)^2=(a-b)^2+4ab
b/ (a-b)^2=(a+b)^2-4ab
c/ (a^2+b^2)(x^2+y^2)=(ax-by)^2+(ay+bx)^2
a) \(\left(a+b\right)^2=a^2+2ab+b^2\left(1\right)\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2-2ab+4ab+b^2=a^2+2ab+b^2\left(2\right)\)
Từ (1) và (2) => đpcm
b) \(\left(a-b\right)^2=a^2-2ab+b^2\left(3\right)\)
\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2+2ab-4ab+b^2=a^2-2ab+b^2\left(4\right)\)
Từ (3) và (4) =>đpcm
c) \(\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right)\)
\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\left(5\right)\)
\(\left(ax-by\right)^2+\left(ay+bx\right)^2=a^2x^2-2axby+b^2y^2+a^2y^2+2aybx+b^2x^2\)
\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\left(6\right)\)
Từ (5) và (6) =>đpcm
a) VP=(a-b)2+4ab
=a2-2ab+b2+4ab
=a2+b2+2ab
=(a+b)2=VT
Vậy (a+b)^2 = (a-b)^2 +4ab
b) VP=(a+b)2-4ab
=a2+2ab+b2-4ab
=a2-2ab+b2
=(a-b)2=VT
Vậy (a-b)^2 = (a+b)^2 - 4ab
c)
VP=(ax-by)2+(ay+bx)2
=a2x2-2axby+b2y2+a2y2+2axby+b2x2
=a2x2+b2y2+a2y2+b2x2
=(a2x2+b2x2)+(b2y2+a2y2)
=x2.(a2+b2)+y2.(a2+b2)
=(a2+b2)(a2+y2)=VT
Vậy ( a^2 + b^2 ).(x^2 +y^2) = (ax - by)^2 +(ay+bx)^2
\(\left(a+b\right)^2=a^2+2ab+b^2\)(1)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2\)(2)
từ (1) và (2) => đpcm
\(\left(a-b\right)^2=a^2-2ab+b^2\)(3)
\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2\)(4)
từ (1) và (2) => đpcm
1) biến đổi vế trái:
= a2+2ab+b2 -a2 +2ab -b2
=4ab = vế phải ( đpcm)
3;5 tuong tu
1) (a + b)2 - (a - b)2 = a2 + 2ab + b2 - a2 + 2ab - b2 = 4ab
3) (a + b)2 - 4ab = a2 + 2ab + b2 - 4ab = a2 - 2ab + b2 = (a - b)2
5) a3 + b3 = a3 + 3a2b + 3ab2 + b3 - 3a2b - 3ab2 = (a + b)3 - 3ab(a + b)
a,
Ta có : \(VP=\left(A-B\right)^2+4AB=A^2-2AB+B^2+4AB=A^2+2AB+B^2=\left(A+B\right)^2\)=> \(\left(A+B\right)^2=\left(A-B\right)^2+4AB\) ( đpcm )
Vậy \(\left(A+B\right)^2=\left(A-B\right)^2+4AB\).
b,
Ta có : \(VP=\left(A+B\right)^2-4AB=A^2+2AB+B^2-4AB=A^2-2AB+B^2=\left(A-B\right)^2\)
=> \(\left(A-B\right)^2=\left(A+B\right)^2-4AB\)
Vậy \(\left(A-B\right)^2=\left(A+B\right)^2-4AB\).
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)
a) \(\left(a+b\right)^2=\left(a-b^2\right)+4ab\)
VP = \(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2\)
VT = \(\left(a+b\right)^2=a^2+2ab+b^2\)
=> VT = VP
b) \(\left(a-b\right)^2=a^2-2ab+b^2\)
\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2\)
Mình làm theo ý hiểu của mik thôi chứ đề bài bn viết khó hiểu lắm