Ta có \(a+b+c+d=0\Leftrightarrow a+c=-\left(b+d\right)\Leftrightarrow\left(a+c\right)^3=\left[-\left(b+d\right)\right]^3\Leftrightarrow a^3+3a^2c+3ac^2+c^3=-b^3-3b^2d-3bd^2-d^3\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2c-3ac^2-3b^2d-3bd^2\Leftrightarrow a^3+b^3+c^3+d^3=-3ac\left(a+c\right)-3bd\left(b+d\right)\Leftrightarrow a^3+b^3+c^3+d^3=3ac\left(b+d\right)-3bd\left(b+d\right)\Leftrightarrow a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)Vậy \(a+b+c+d=0\) thì \(a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)

@Nguyễn Việt Lâm

https://hoc24.vn/id/2782086

Ta có đánh giá \(\frac{b+2}{\left(b+1\right)\left(b+5\right)}\ge\frac{3}{4\left(b+2\right)}\)

Thật vậy, BĐT trên tương đương:

\(4\left(b+2\right)^2\ge3\left(b+1\right)\left(b+5\right)\)

\(\Leftrightarrow b^2-2b+1\ge0\Leftrightarrow\left(b-1\right)^2\ge0\) (luôn đúng)

\(\Rightarrow\frac{\left(a+1\right)\left(b+2\right)}{\left(b+1\right)\left(b+5\right)}\ge\frac{3\left(a+1\right)}{4\left(b+2\right)}\)

Tương tự và cộng lại: \(P\ge\frac{3}{4}\left(\frac{a+1}{b+2}+\frac{b+1}{c+2}+\frac{c+1}{a+2}\right)\)

\(P\ge\frac{3}{4}\left(\frac{\left(a+1\right)^2}{ab+2a+b+2}+\frac{\left(b+1\right)^2}{bc+2b+c+2}+\frac{\left(c+1\right)^2}{ca+2c+a+2}\right)\)

\(P\ge\frac{3}{4}.\frac{\left(a+b+c+3\right)^2}{ab+bc+ca+3a+3b+3c+6}\)

\(P\ge\frac{3}{4}.\frac{a^2+b^2+c^2+2ab+2bc+2ca+6a+6b+6c+9}{ab+bc+ca+3a+3b+3c+6}\)

\(P\ge\frac{3}{4}.\frac{2ab+2bc+2ca+6a+6b+6c+12}{ab+bc+ca+3a+3b+3c+6}=\frac{3}{4}.2=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

>=8 nha

Tại sao lại bằng 8

 \(A=\frac{a^3+b^3-\left(a^2+b^2\right)}{\left(a-1\right)\left(b-1\right)}=\frac{a^2\left(a-1\right)+b^2\left(b-1\right)}{\left(a-1\right)\left(b-1\right)}=\frac{a^2}{b-1}+\frac{b^2}{a-1}\)

(chơi 3 cách luôn cho máu :3)

Cách 1, Áp dụng Svacxơ  đc

\(A=\frac{a^2}{b-1}+\frac{b^2}{a-1}\ge\frac{\left(a+b\right)^2}{a+b-2}=\frac{t^2}{t-2}\left(t=a+b>2\right)\)

Ta luôn có \(\frac{t^2}{t-2}\ge8\left(1\right)\)thật vậy

\(\left(1\right)\Leftrightarrow t^2\ge8t-16\Leftrightarrow t^2-8t+16\ge0\Leftrightarrow\left(t-4\right)^2\ge0\left(True\right)\)

=> Đpcm

Cách 2, \(A=\frac{a^2}{b-1}+\frac{b^2}{a-1}\ge2\sqrt{\frac{a^2.b^2}{\left(b-1\right)\left(a-1\right)}}=2.\frac{a}{\sqrt{a-1}}.\frac{b}{\sqrt{b-1}}\)

Ta đi c/m \(\frac{a}{\sqrt{a-1}}\ge2\left(#\right)\)thật vậy

\(\left(#\right)\Leftrightarrow a\ge2\sqrt{a-1}\Leftrightarrow a^2\ge4a-4\Leftrightarrow a^2-4a+4\ge0\Leftrightarrow\left(a-2\right)^2\ge0\left(true\right)\)

=> (#) đúng 

tương tự\(\frac{b}{\sqrt{b-1}}\ge2\)

\(\Rightarrow A\ge2.2.2=8\)(Đpcm)

Cách 3 , \(A=\frac{a^2}{b-1}+\frac{b^2}{a-1}=\frac{\left(a-1+1\right)^2}{b-1}+\frac{\left(b-1+1\right)^2}{a-1}\)

                 \(=\frac{\left(a-1\right)^2+2\left(a-1\right)+1}{b-1}+\frac{\left(b-1\right)^2+2\left(b-1\right)+1}{a-1}\)

               \(=\frac{\left(a-1\right)^2}{b-1}+\frac{2\left(a-1\right)}{b-1}+\frac{1}{b-1}+\frac{\left(b-1\right)^2}{a-1}+\frac{2\left(b-1\right)}{a-1}+\frac{1}{a-1}\)

                 \(=\left[\frac{\left(a-1\right)^2}{b-1}+\frac{\left(b-1\right)^2}{a-1}\right]+2\left(\frac{a-1}{b-1}+\frac{b-1}{a-1}\right)+\left(\frac{1}{b-1}+\frac{1}{a-1}\right)\)

                 \(\ge2\sqrt{\frac{\left(a-1\right)^2.\left(b-1\right)^2}{\left(b-1\right)\left(a-1\right)}}+2.2\sqrt{\frac{a-1}{b-1}.\frac{b-1}{a-1}}+\frac{2}{\sqrt{\left(a-1\right)\left(b-1\right)}}\)

                    \(=2\sqrt{\left(a-1\right)\left(b-1\right)}+\frac{2}{\sqrt{\left(a-1\right)\left(b-1\right)}}+4\)

                     \(\ge2\sqrt{2\sqrt{\left(a-1\right)\left(b-1\right)}.\frac{2}{\sqrt{\left(a-1\right)\left(b-1\right)}}}+4\)

                      \(=2.2+4=8\)

Dấu "=" xảy ra tại a = b = 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 )

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

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

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

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 :)

\(\hept{\begin{cases}a^2+b^2\ge2\sqrt{a^2b^2}=2|ab|\\b^2+c^2\ge2\sqrt{b^2c^2}=2|bc|\\c^2+a^2\ge\sqrt{c^2a^2}=2|ca|\end{cases}}\Rightarrow\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)\ge8|\left(abc\right)^2|=8a^2b^2c^2\)

(vì a²+b²; b²+c²; c²+a²;|ab|;|bc|;|ca| đều \(\ge0\))

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