a) Chứng minh \(\left(a+b\right)^2=\left(a+b\right)^2+4ab\)
b) Cho \(a+b=9\), \(ab=20\)
Tính\(\left(a-b\right)^{2011}\)
Cho a, b >0 thỏa mãn a + b = 1. Chứng minh:\(2\sqrt{ab}+\frac{\left(a+1\right)\left(b+1\right)\left(a+b\right)}{4ab}\ge\frac{9}{4}+\left(a+b\right)^2\)
P/s: Có ai như em không, ra đề xong quên mất hướng giải:)))
chứng minh rằng
\(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
\(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
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, \(VP=\left(a+b\right)\left(a+b\right)-4ab\)
\(=a^2+2ab+b^2-4ab\)
\(=a^2+b^2-2ab=\left(a-b\right)^2=VT\left(đpcm\right)\)
1. CHỨNG MINH RẰNG
a) \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)
b) \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
c) \(\left(a^2+b^2\right).\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(ay+bx\right)^2\)
2. CHỨNG MINH RẰNG : a = b = c KHI
\(\left(a+b+c\right)^2=3\left(ab+ac+bc\right)\)
3. CHO a + b + c = 0 VÀ \(a^2+b^2+c^2=1\)
Tính \(M=a^4+b^4+c^4\)
4. CHỨNG MINH RẰNG GIÁ TRỊ CÁC BIỂU THỨC SAU LUÔN LUÔN DƯƠNG
a) \(x^2+x+1\)
b) \(x^2-x+\frac{1}{2}\)
\(1.\)
\(a,\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2\)
\(\Rightarrow\left(a+b\right)^2=\left(a-b\right)^2+4ab\left(đpcm\right)\)
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)
cho các số thực dương a,b thỏa mãn \(\sqrt{a}+\sqrt{b}=1\)
Chứng minh rằng \(3\left(a+b\right)^2-\left(a+b\right)+4ab\ge\frac{1}{2}\sqrt{\left(a+3b\right)\left(b+3a\right)}\)
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 )
Chứng minh :
\(\left(a+b\right)^2-\left(a-b\right)^2=4ab\)
(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
Chứng minh rằng:
a)\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
b)\(\left(a-b\right)^3+3ab\left(a-b\right)=a^3-b^3\)
c)\(\left(a+b\right)^2-\left(a-b\right)^2=4ab\)
cho các số thực dương thỏa mãn \(\sqrt{a}+\sqrt{b}=1\)
Chứng minh rằng \(3\left(a+b\right)^2-\left(a+b\right)+4ab\ge\frac{1}{2}\sqrt{\left(a+3b\right)\left(b+3a\right)}\)
\(\left(a+3b\right)\left(b+3a\right)\le\left(\frac{4a+4b}{2}\right)^2=\left(2a+2b\right)^2\)
=>\(\frac{1}{2}\sqrt{\left(a+3b\right)\left(b+3a\right)}\le\frac{1}{2}\left(2a+2b\right)=a+b\)
Mình làm phần dễ nhất rồi, còn lại của bạn đó ^^
Đặt . Do đó
. Cần chứng minh:
Or
Bình phương 2 vế và xét hiệu, ta cần chứng minh:
Đó là điều hiển nhiên vì:
Done.