tích đúng mình làm cho

bạn giải giùm với ạk

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. 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) 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)

Ta có :\(\left(a-b\right)^2+4ab\)

       \(=a^2-2ab+b^2+4ab\)

         \(=a^2+2ab+b^2hay\left(a+b\right)^2\)

Vậy:\(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

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

(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

ban su dung hang dang thuc la ra

\(\left(a+3b\right)\left(b+3a\right)\le\left(\frac{4a+4b}{2}\right)^2=\left(2a+2b\right)^2\)

=>\(\frac{1}{2}\sqrt{\left(a+3b\right)\left(b+3a\right)}\le\frac{1}{2}\left(2a+2b\right)=a+b\)

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.

eos bieets