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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: (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

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

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

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

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

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

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

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

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

\(A=1+2^2+2^3+2^4+...+2^{2015}\)

\(2A=2+2^3+2^4+2^5+...+2^{2016}\)

\(2A-A=\left(2+2^3+2^4+2^5+...+2^{2016}\right)-\left(1+2^2+2^3+2^4+...+2^{2015}\right)\)

\(A=2^{2016}-1\)

Đề sai rồi

A=1+2^2+2^3+2^4+.....+2^2015

=>2A=2+2^3+2^4+2^5+.....+2^2016

=>2A-A=2^2016-1=>A=2^2016-1

Vậy A+1=2^x=2^2016. Hay x=2016

\(\text{Ta có:}\)

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

\(\left(a-1\right)^3+\left(b-2\right)^3+\left(c-3\right)^3-3\left(a-1\right)\left(b-2\right)\left(c-3\right)+3\left(a-1\right)\left(b-2\right)\left(c-3\right)=0\)

\(\Leftrightarrow\left(a+b+c-6\right)\left(....\right)+3\left(a-1\right)\left(b-2\right)\left(c-3\right)=0\)

\(\Leftrightarrow a=1\text{ hoặc }b=2\text{ hoặc }c=3\)

còn lại ko tính đc bạn ktra lại đề

mk nhầm , chiều mk lm tiếp

Ta có \(\left(a-1\right)+\left(b-2\right)+\left(c-3\right)=6-6=0\)

\(\Rightarrow\left(a-1\right)^3+\left(b-2\right)^3+\left(c-3\right)^3=3\left(a-1\right)\left(b-2\right)\left(c-3\right)=0\)

<=> a=1 hoặc b=2 hoặc c=3

Xét a=1 => b+c=5

Ta có : \(\left(a-1\right)^{2015}+\left(b-2\right)^{2015}+\left(c-3\right)^{2015}=0+\left(b+c-5\right).A=0\)

Tương tự với b=2,c=3 ta cũng được \(\left(a-1\right)^{2015}+\left(b-2\right)^{2015}+\left(c-3\right)^{2015}=0\)

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