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, \(VP=\left(a+b\right)\left(a+b\right)-4ab\)

\(=a^2+2ab+b^2-4ab\)

\(=a^2+b^2-2ab=\left(a-b\right)^2=VT\left(đpcm\right)\)

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

Ta có : VP = \(x^4-y^4\)

\(=\left(x^2\right)^2-\left(y^2\right)^2\)

\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)

\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)

Vp\(=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\) = VT

Vậy  \(x^4-y^4\) \(=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\) (đpcm)

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

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

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

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 )

ban su dung hang dang thuc la ra

\(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)

\(=x\left(x^3+x^2y+xy^2+y^3\right)-y\left(x^3+x^2y+xy^2+y^3\right)\)

\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)

\(=\left(x^4-y^4\right)+\left(x^3y-x^3y\right)+\left(x^2y^2-x^2y^2\right)+\left(xy^3-xy^3\right)\)

\(=x^4-y^4=VP\)

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

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

\(=a^2+2ab+b^2-a^2+2ab-b^2\)

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

\(=4ab=VP\)

Câu a :

\(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)

Nhân 2 vế lại ta được \(x^4-y^4=VP\)

\(\Rightarrowđpcm\)

Câu b :

\(VT=\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b-a+b\right)\left(a+b+a-b\right)=2b.2a=4ab=VP\)

\(\Rightarrowđpcm\)