Vì a=b=c nên:

A=ab^2c.(-1/2bc^2)+(3/2abc).(-bc)^2

A=a^4.(-1/2a^3)+(3/2a^3).a^4

A=a^4.(-1/2a^3+3/2abc)

A=a^4.a^3=a^7

Thay a=1 vào A ta có: A=(-1)^7=-1

Ta có: \(A=ab^2c\cdot\left(-\dfrac{1}{2}bc^2\right)+\dfrac{3}{2}abc\cdot\left(-bc\right)^2\)

\(=\dfrac{-1}{2}ab^3c^3+\dfrac{3}{2}abc\cdot b^2c^2\)

\(=\dfrac{-1}{2}ab^3c^3+\dfrac{3}{2}ab^3c^3\)

\(=ab^3c^3\)

Thay a=-1; b=-1; c=-1 vào A, ta được:

\(A=-1\cdot\left(-1\right)^3\cdot\left(-1\right)^3=-1\)

Vì a=b=c nên:

A=ab^2c.(-1/2bc^2)+(3/2abc).(-bc)^2

A=a^4.(-1/2a^3)+(3/2a^3).a^4

A=a^4.(-1/2a^3+3/2abc)

A=a^4.a^3=a^7

Thay a=1 vào A ta có: A=(-1)^7=-1

** Bạn lưu ý lần sau viết đề bằng công thức toán để được hỗ trợ tốt hơn.

Lời giải:

$\frac{a+b}{c}+\frac{a+c}{b}+\frac{b+c}{a}=-2$

$\Leftrightarrow \frac{a+b}{c}+1+\frac{a+c}{b}+1+\frac{b+c}{a}=0$

$\Leftrightarrow (a+b+c)(\frac{1}{c}+\frac{1}{b})+\frac{b+c}{a}=0$

$\Leftrightarrow \frac{(a+b+c)(b+c)}{bc}+\frac{b+c}{a}=0$

$\Leftrightarrow (b+c)(\frac{a+b+c}{bc}+\frac{1}{a})=0$

$\Leftrightarrow (b+c).\frac{a(a+b+c)+bc}{abc}=0$

$\Leftrightarrow \frac{(b+c)(a+b)(a+c)}{abc}=0$

$\Rightarrow (a+b)(b+c)(c+a)=0$

$\Rightarrow a+b=0$ hoặc $b+c=0$ hoặc $c+a=0$

Không mất tổng quát giả sử $a+b=0\Rightarrow a=-b$

$1=a^3+b^3+c^3=(-b)^3+b^3+c^3=c^3\Rightarrow c=1$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{-1}{b}+\frac{1}{b}+\frac{1}{1}=1$

Vậy..........

\(A^2+B^2=\left(A+B\right)^2-2AB=5\)

\(A^3+B^3=\left(A+B\right)^3-3AB\left(A+B\right)=9\)

\(A^5+B^5=\left(A^2+B^2\right)\left(A^3+B^3\right)-\left(AB\right)^2\left(A+B\right)=5.9-2^2.3=...\)

B.

\(A^2+B^2=\left(A+B\right)^2-2AB=2\)

\(A^6+B^6=\left(A^2\right)^3+\left(B^2\right)^3=\left(A^2+B^2\right)^3-3\left(AB\right)^2\left(A^2+B^2\right)=2^3-3.1^2.2=...\)

Ta có: \(A^2+B^2=\left(A+B\right)^2-2AB=3^2-2.2=5\)

\(A^5+B^5=\left(A^3+B^3\right)\left(A^2+B^2\right)-A^2B^2\left(A+B\right)=\left(A+B\right)\left(A^2-AB+B^2\right)\left(A^2+B^2\right)-A^2B^2\left(A+B\right)=3\left(5-2\right).5-2^2.3=33\)

Bài 2:

C=A-B

\(=2x^2-6xy+4y^2+5x^2-4xy-7y^2\)

\(=7x^2-10xy-3y^2\)

\(=7\cdot1^2-10\cdot1\cdot\dfrac{1}{2}-3\cdot\dfrac{1}{4}=7-5-\dfrac{3}{4}=2-\dfrac{3}{4}=\dfrac{5}{4}\)

a: (x+1)^3-x(x-2)^2+x-1=0

=>x^3+3x^2+3x+1-x(x^2-4x+4)+x-1=0

=>x^3+3x^2+4x-x^3+4x^2-4x=0

=>7x^2=0

=>x=0

b: =>x^3-3x^2+3x-1-x^3-27+3x^2-12=2

=>3x=2+1+27+12=39+3=42

=>x=14

b: \(A=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{58}\right)⋮13\)

\(a,\Leftrightarrow2A=8+2^3+2^4+...+2^{21}\\ \Leftrightarrow2A-A=8+2^3+2^4+...+2^{21}-4-2^2-2^3-...-2^{20}\\ \Leftrightarrow A=2^{21}+8-4-2^2=2^{21}\left(đpcm\right)\\ b,A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+3^4+...+3^{58}\right)\\ A=13\left(3+3^4+...+3^{58}\right)⋮13\)

a,Chứng minh rằng A là một lũy thừa của 2

A=4+2^2+2^3+2^4+......+2^20

b,Chứng tỏ A=3^1+3^2+3^3+.....+3^60 chia hết cho 13

= \(\dfrac{-3}{20}\)

Chọn D

D. \(\dfrac{-3}{20}\)

b. 6x(x - 5) - x(6x + 3)

= x(6x - 30) - x(6x + 3)

= x(6x - 30 - 6x - 3)

= x(-33)

= -33x

\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)