Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1.\)
\(a,\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2\)
\(\Rightarrow\left(a+b\right)^2=\left(a-b\right)^2+4ab\left(đpcm\right)\)
a) \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)(luôn dương)
b) \(x^2-x+\frac{1}{2}=x^2-x+\frac{1}{4}+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\frac{1}{4}>0\)(luôn dương)
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+b)^2-(a-b)^2=4ab
a^2+2ab+b^2-a^2+2ab-b^2=4ab
a^2+2ab+b^2-a^2+2ab-b^2-4ab=0
a^2-a^2+2ab+2ab-4ab+b^2-b^2=0
0=0
=>dpcm
Biến đổi vế trái ta có:
\(\left(a+b\right)^2-\left(a-b\right)^2=a^2+2ab+b^2-a^2+2ab-b^2=4ab=VP\)
=>đpcm
Bài giải:
a) (a + b)2 = (a – b)2 + 4ab
- Biến đổi vế trái:
(a + b)2 = a2 +2ab + b2 = a2 – 2ab + b2 + 4ab
= (a – b)2 + 4ab
Vậy (a + b)2 = (a – b)2 + 4ab
- Hoặc biến đổi vế phải:
(a – b)2 + 4ab = a2 – 2ab + b2 + 4ab = a2 + 2ab + b2
= (a + b)2
Vậy (a + b)2 = (a – b)2 + 4ab
b) (a – b)2 = (a + b)2 – 4ab
Biến đổi vế phải:
(a + b)2 – 4ab = a2 +2ab + b2 – 4ab
= a2 – 2ab + b2 = (a – b)2
Vậy (a – b)2 = (a + b)2 – 4ab
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
CMR: (a + b)2 = (a - b)2 + 4ab
(a - b)2 = (a + b)2 - 4ab
Ta có: (a + b)2 = a2 + 2ab + b2
= a2 +2ab + b2 - 2ab +2ab
= a2 - 2ab + b2 + 2ab +2ab
= (a - b)2 +4ab
Ta có: (a - b)2 = a2 - 2ab + b2
= a2 - 2ab + b2 + 2ab - 2ab
= a2 + 2ab + b2 - 2ab - 2ab
= (a + b)2 - 4ab
Áp dụng:
a) Tính (a - b)2 , biết a + b = 7 và a.b = 12
Ta có: (a - b)2 = (a + b)2 - 4ab
= 72 - 4.12
= 49 - 48
Vậy (a - b)2 = 1
b) Tính (a + b)2 , biết a - b = 7 và a.b = 3
Ta có: (a + b)2 = (a - b)2 + 4ab
= 72 + 4.3
= 49 + 12
Vậy ( a + b)2 = 61
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)
a) mk chỉnh đề:
Chứng minh: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\) (1)
hoặc \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\) (2)
BÀI LÀM
TH1:
\(VP=\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2=VP\) (đpcm)
TH2:
\(VP=\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2=VT\) (đpcm)
b) \(a+b=9\)\(\Rightarrow\)\(a=9-b\)
Ta có: \(ab=20\)\(\Rightarrow\)\(\left(9-b\right).b=20\)
\(\Leftrightarrow\)\(b^2-9b+20=0\)
\(\Leftrightarrow\)\(\left(b-4\right)\left(b-5\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}b=4\\b=5\end{cases}}\)
Nếu \(b=4\)thì: \(a=5\)\(\Rightarrow\)\(\left(a-b\right)^{2011}=\left(5-4\right)^{2011}=1\)
Nếu \(b=5\)thì \(a=4\)\(\Rightarrow\)\(\left(a-b\right)^{2011}=\left(4-5\right)^{2011}=-1\)
a, sửa đề CM: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
\(VP=\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2=VT\left(đpcm\right)\)
b, \(a+b=9\Leftrightarrow\left(a+b\right)^2=81\Leftrightarrow\left(a-b\right)^2+4ab=81\Leftrightarrow\left(a-b\right)^2=81-4.20=1\Leftrightarrow a-b=\pm1\)
Với \(a-b=1\Rightarrow\left(a-b\right)^{2011}=1\)
Với \(a-b=-1\Rightarrow\left(a-b\right)^{2011}=-1\)