\(a+b=10\) và \(ab=4\)

1. Có: \(A=a^2+b^2=\left(a+b\right)^2-2ab=10^2-2.4=92\)

2. \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=10^3-3.4.10=880\)

3. \(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=92^2-2.4^2=8432\)

4. \(a^5+b^5=\left(a^2+b^2\right)\left(a^3+b^3\right)-a^2b^2\left(a+b\right)=92.880-4^2.10=80800\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

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

\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow a^2b+ab^2+abc+abc+b^2c+bc^2+a^2c+abc+ac^2-abc=0\)

\(\Leftrightarrow ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc\right)+ac\left(a+c\right)=0\)

\(\Leftrightarrow\left(a+c\right)\left(ab+b^2+bc+ac\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

\(\circledast Với:a=-b\) , ta có :

\(P=\left(-b+b\right)\left(b^3+c^3\right)\left(c^5+a^5\right)=0\)

\(\circledast Với:b=-c\) , ta có :

\(P=\left(a+b\right)\left(b^3-b^3\right)\left(c^5+a^5\right)=0\)

\(\circledast Với:c=-a\) , ta có :

\(P=\left(a+b\right)\left(b^3+c^3\right)\left(-a^5+a^5\right)=0\)

KL..............

    \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Rightarrow\frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}\)

\(\Rightarrow\left(ab+ac+bc\right)\left(a+b+c\right)=abc\)

\(\Rightarrow a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+3abc=âbc\)

\(\Rightarrow\left(a^2b+ab^2\right)+\left(ac^2+bc^2\right)+\left(a^2c+2abc+b^2c\right)=0\)

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

\(\Rightarrow ab\left(a+b\right)+c^2\left(a+b\right)+\left(ac+bc\right)\left(a+b\right)=0\)

\(\Rightarrow\left(a+b\right)\left(ab+c^2+ac+bc\right)=0\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a=-b\\\frac{b=-c}{a=-c}\end{cases}}\)

Từ đó: P = 0.

Mình giải hơi tắt. Mong bạn hiểu bài.

Chúc bạn học tốt.

Bài 5.

1. Chứng minh

$\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}$

Ta có:

$\dfrac{2}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}$

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

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

$=\dfrac{(a-b)^2+b^2}{ab(a+b)}\ge0$

Vậy: $\boxed{\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}}$

2. Chứng minh

$\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}$

Vì $a,b,c>0$ nên:

$\dfrac1a+\dfrac1b+\dfrac1c>\dfrac1a$

Mà: $\dfrac1a>\dfrac{a}{a+b+c}$

Suy ra: $\boxed{\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}}$

1.

$a^3+b^4-ab(a+b)$

$=a^3+b^4-a^2b-ab^2$

$=a^2(a-b)+b^2(b-a)$

$=(a-b)(a^2-b^2)$

$=(a-b)^2(a+b)\ge0$

Suy ra: $\boxed{a^3+b^4\ge ab(a+b)}$

2.

$a^4+b^4-ab(a^2+b^2)$

$=a^4+b^4-a^3b-ab^3$

$=a^3(a-b)+b^3(b-a)$

$=(a-b)(a^3-b^3)$

$=(a-b)^2(a^2+ab+b^2)\ge0$

Vậy: $\boxed{a^4+b^4\ge ab(a^2+b^2)}$

3.

$a^5+b^5-ab(a^3+b^3)$

$=a^5+b^5-a^4b-ab^4$

$=a^4(a-b)+b^4(b-a)$

$=(a-b)(a^4-b^4)$

$=(a-b)^2(a+b)(a^2+b^2)\ge0$

Vậy: $\boxed{a^5+b^5\ge ab(a^3+b^3)}$

\(2\left(x-2\right)\left(x+3\right)-x^2+4=0\)

\(2\left(x^2+3x-2x-6\right)-x^2+4=0\)

\(2x^2+6x-4x-12-x^2+4=0\)

\(x^2+2x-8=0\)

\(x^2+4x-2x-8=0\)

\(x\left(x+4\right)-2\left(x+4\right)=0\)

\(\left(x+4\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+4=0\rightarrow x=\left(-4\right)\\x-2=0\rightarrow x=2\end{cases}}\)

3/ 

a/ \(2\left(x+1\right)^2-3\left(x-1\right)^2+\left(x+2\right)\left(5-x\right)\)

\(=2\left(x^2+2x+1\right)-3\left(x^2-2x+1\right)+\left(5x-x^2+10-2x\right)\)

\(=2x^2+4x+2-3x^2+6x-3+5x-x^2+10-2x\)

\(=-2x^2+13x+9\)

b/ \(\left(3x-1\right)^3+\left(3x-1\right)^3-6x^2+9\)

\(=2\left(3x-1\right)^3-6x^2+9\)

\(=2\left(\left(3x\right)^3-3\left(3x\right)^2\cdot1+3\cdot3x\cdot1-1\right)-6x^2+9\)

\(=2\left(27x^3-27x^2+9x-1\right)-6x^2+9\)

\(=54x^3-54x^2+18x-2-6x^2+9\)

\(=54x^3-60x^2+18x+7\)

Số hơi dài, nên dễ tính sai -,- tính mik hay cẩu thả có j sai ibbb ạ

Các bạn giúp mình nhanh nhanh sáng mai kiểm tra rồi !!!!!!!!

a) \(a^3-b^3\)

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

Ta có:

\(a-b=1\)

\(\Rightarrow\left(a-b\right)^2=1\)

\(\Rightarrow a^2-2ab+b^2=1\)

\(\Rightarrow a^2+b^2=1+2ab\left(1\right)\)

Ta lại có: \(ab=6\left(2\right)\)

Thay (1) và (2) vào (*) ta được

\(=1.\left(1+2ab+ab\right)\)

\(=1+3ab\)

\(=1+3.6\)

\(=19\)

b) \(a^3-b^3\)

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

Ta có:

\(a+b=1\)

\(\Rightarrow\left(a+b\right)^2=1\)

\(\Rightarrow a^2+2ab+b^2=1\)

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

Ta lại có: \(ab=-1\left(2\right)\)

Thay (1) và (2) vào (*) ta được

\(=1\left(1-2ab+ab\right)\)

\(=1-ab\)

\(=1-\left(-1\right)\)

\(=2\)

c) \(2\left(a^3+b^3\right)-3\left(a^2+b^2\right)\)

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

Ta có:

\(a+b=1\)

\(\Rightarrow\left(a+b\right)^2=1\)

\(\Rightarrow a^2+2ab+b^2=1\)

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

Ta lại có: \(ab=-2\left(2\right)\)

Thay (1) và (2) vào (*) ta được

\(=2.1\left(1-2ab-ab\right)-3\left(1-2ab\right)\)

\(=2\left(1-3ab\right)-3\left(1-2ab\right)\)

\(=2\left[1-3.\left(-2\right)\right]-3\left[1-2.\left(-2\right)\right]\)

\(=2.7-3.5\)

\(=29\)

d) \(x^3+y^3+3xy\)

\(=x^3+y^3+3xy\left(x+y\right)\) ( Vì x + y = 1 nên GTBT không đổi )

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

\(=1\)

e) \(x^3-y^3-3xy\)

\(=x^3-y^3-3xy\left(x-y\right)\) ( Vì x - y = 1 nên GTBT không đổi )

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

\(=1\)