1/Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=81\)

\(\Rightarrow M=ab+bc+ca=\frac{\left(81-141\right)}{2}\)

a,\(a+b+c=9\)

\(\Rightarrow\left(a+b+c\right)^2=81\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=81\)

Vì \(a^2+b^2+c^2=141\)

\(\Rightarrow2ab+2bc+2ca=-60\)

\(\Rightarrow2\left(ab+bc+ca\right)=-60\)

\(\Rightarrow ab+bc+ca=-30\)

Vậy ...

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

1/ \(a+b+c=11\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=121\)

\(\Leftrightarrow ab+bc+ca=\frac{121-\left(a^2+b^2+c^2\right)}{2}=\frac{121-87}{2}=17\)

2/ \(a^3+b^3+a^2c+b^2c-abc\)

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

\(=\left(a^2-ab+b^2\right)\left(a+b+c\right)=0\)

3/ \(x^4+3x^3y+3xy^3+y^4\)

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

\(=\left(9^2-2.4\right)^2-2.4^2+3.4.\left(9^2-2.4\right)=6173\)

bạn alibaba nguyễn có thể làm lại giúp mình được không ?

1)

$2x=a+b+c$

$\Rightarrow x-a=\dfrac{b+c-a}{2},\quad x-b=\dfrac{c+a-b}{2},\quad x-c=\dfrac{a+b-c}{2}$

$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)$

$=\dfrac{(b+c-a)(c+a-b)+(c+a-b)(a+b-c)+(a+b-c)(b+c-a)}{4}$

$=\dfrac{2ab+2bc-a^2-b^2-c^2+2bc+2ca-a^2-b^2-c^2+2ca+2ab-a^2-b^2-c^2}{4}$

$=\dfrac{4ab+4bc+4ca-3(a^2+b^2+c^2)}{4}$

$=\dfrac{(a+b+c)^2-2(a^2+b^2+c^2)}{4}$

$=\dfrac{(2x)^2-2(a^2+b^2+c^2)}{4}$

$=ab+bc+ca-x^2$

$\therefore\ (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=ab+ac+bc-x^2.$

2)

$ab+bc+ca=abc,\quad a+b+c=1$

$(a-1)(b-1)(c-1)$$=abc-ab-bc-ca+a+b+c-1$$=abc-ab-bc-ca+1-1$$=abc-(ab+bc+ca)$$=abc-abc$$=0$

$\therefore\ (a-1)(b-1)(c-1)=0.$

\(x^5+y^5\)

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

Tách ra hằng đẳng thức tiếp rồi thay vào

a/

$x^2+y^2+z^2=3$

$F=\dfrac{x^2+1}{z+2}+\dfrac{y^2+1}{x+2}+\dfrac{z^2+1}{y+2}$

$\ge \dfrac{(x+y+z)^2}{(x+y+z)+6}\qquad (\text{Titu})$

Đặt $t=x+y+z$.

Ta có $t^2\le 3(x^2+y^2+z^2)=9$

$\Rightarrow t\le 3$.

Xét $f(t)=\dfrac{t^2}{t+6}$ thì $f'(t)=\dfrac{t(t+12)}{(t+6)^2}>0$ nên $f(t)$ tăng trên $(0,+\infty)$.

Mặt khác $t\ge \sqrt{x^2+y^2+z^2}=\sqrt3$.

Suy ra $F\ge f(\sqrt3)=\dfrac{3}{6+\sqrt3}$ $=\dfrac{6-\sqrt3}{11}$.

Dấu bằng khi $x=y=z=1$.

$\boxed{\min F=\dfrac{6-\sqrt3}{11}}$.

b/

Đặt $S=\sqrt{\dfrac{a}{a+3}}+\sqrt{\dfrac{b}{b+3}}+\sqrt{\dfrac{c}{c+3}}.$

Theo bất đẳng thức Cauchy-Schwarz, $S^2\le (a+b+c)\left(\dfrac1{a+3}+\dfrac1{b+3}+\dfrac1{c+3}\right)$.

Lại có $(a+b+c)^2\ge 3(ab+bc+ca)=9$

$\Rightarrow a+b+c\ge 3$.

Theo bất đẳng thức Nesbitt dạng Engel,

$\dfrac1{a+3}+\dfrac1{b+3}+\dfrac1{c+3}\le \dfrac1{6}(3)=\dfrac12.$

Do đó $S^2\le \dfrac32$

$\Rightarrow S\le \sqrt{\dfrac32}<\dfrac32$.

Suy ra $\sqrt{\dfrac{a}{a+3}}+\sqrt{\dfrac{b}{b+3}}+\sqrt{\dfrac{c}{c+3}}\le \dfrac32.$

a/ Biến đổi tương đương:

\(\Leftrightarrow a^2c+ab^2+bc^2\ge b^2c+ac^2+a^2b\)

\(\Leftrightarrow a^2c-a^2b+ab^2-ac^2+bc^2-b^2c\ge0\)

\(\Leftrightarrow a^2\left(c-b\right)-\left(ab+ac\right)\left(c-b\right)+bc\left(c-b\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(a^2+bc-ab-ac\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(a\left(a-b\right)-c\left(a-b\right)\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(a-c\right)\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(c-b\right)\left(c-a\right)\left(b-a\right)\ge0\) luôn đúng do \(a\le b\le c\)

Vậy BĐT ban đầu đúng

Câu 2: Đề sai, cho \(a=b=c=1\Rightarrow3\ge6\) (sai)

Đề đúng phải là \(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

\(VT=\frac{a^2}{abc}+\frac{b^2}{abc}+\frac{c^2}{abc}=\frac{a^2+b^2+c^2}{abc}\ge\frac{ab+ac+bc}{abc}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Câu 3: Không phải với mọi x; y với mọi \(x;y\) dương

Biến đổi tương đương do mẫu số vế phải dương nên ta được quyền nhân chéo:

\(\Leftrightarrow3x^3\ge\left(2x-y\right)\left(x^2+xy+y^2\right)\)

\(\Leftrightarrow3x^3\ge2x^3+x^2y+xy^2-y^3\)

\(\Leftrightarrow x^3+y^3-x^2y-xy^2\ge0\)

\(\Leftrightarrow x^2\left(x-y\right)-y^2\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2-y^2\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\) (luôn đúng)