giúp mình vớiiii

\(1.VP\)

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

\(=a^2+b^2=VT\left(DPCM\right)\)

1/  (a + b)² - 2ab = a² + 2ab + b² - 2ab = a² + b² + (2ab - 2ab) = a² + b²

2/  (a² + b²)² - 2a²b² = a⁴ + 2a²b² + b⁴ - 2a²b² = a⁴ + b⁴ + (2a²b² - 2a²b²) = a⁴ + b⁴

Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) 

Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)

CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)

\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)

Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu = xảy ra khi a=b=c=3

Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)

\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)

\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)

Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)

 \(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)

\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)

Vậy...

Điều kiện a;b;c dương, đặt vế trái là P

\(P=\sum\frac{a}{a^2+c^2+2\left(a^2+b^2\right)}\le\sum\frac{a}{2ac+4ab}=\frac{1}{2}\sum\frac{1}{c+2b}=\frac{1}{2}\sum\frac{1}{b+b+c}\)

\(\Rightarrow P\le\frac{1}{18}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)=\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Dấu "=" xảy ra khi \(a=b=c\)

với x+y+z=0 thì \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0< =>\)x³ +y³ +z³ =3xyz

nếu đặt x=a²; y=b² ;z=c² thì ta cần có a² +b² +c² =0 thì sẽ có a⁶ +b⁶ +c⁶ =3a²b²c²

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0< =>\frac{ab+bc+ca}{abc}=0< =>ab+bc+ca=0.\)

a+b+c=0 <=> (a+b+c)² =0 <=> \(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0< =>a^2+b^2+c^2=0.\)(chứng minh xong)

\(\dfrac{a}{6}=\dfrac{b}{9}\)

\(\Leftrightarrow9a=6b\)

\(\Rightarrow3a=2b\)(chia cả 2 vế cho 3)

\(\Rightarrow3a-2b=0\Rightarrow\dfrac{3a-2b}{3a+2b}=0\)

Chúc bn học tốt

Ta có: `a/6 = b/9` `-> 9a = 6b`

`-> 3a = 2b`

Vì `3a = 2b` nên `3a - 2b = 0`.

`-> A = (3a - 2b)/(3a + 2b) = 0/(3a + 2b) = 0` 

Vậy giá trị biểu thức `A` là `0`.