b,\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

=>\(\dfrac{bc}{abc}+\dfrac{ac}{bac}+\dfrac{ab}{abc}=0\)

=>\(\dfrac{ab+ac+bc}{abc}=0\)

=>ab+ac+bc=0

=>ab=-ac-bc

ac=-ab-bc

bc=-ab-ac

N=\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ca}+\dfrac{1}{c^2+2ab}\)

N=\(\dfrac{1}{a^2+bc+bc}+\dfrac{1}{b^2+ca+ca}+\dfrac{1}{c^2+ab+ab}\)

N=\(\dfrac{1}{a^2-ab-ac+bc}+\dfrac{1}{b^2-ab-bc+ca}+\dfrac{1}{c^2-ac-bc+ab}\)

N=\(\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-a\right)-c\left(b-a\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)

N=\(\dfrac{1}{\left(a-c\right)\left(a-b\right)}+\dfrac{1}{\left(b-c\right)\left(b-a\right)}+\dfrac{1}{\left(c-b\right)\left(c-a\right)}\)

N=\(\dfrac{b-c}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}-\dfrac{a-c}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\dfrac{a-b}{\left(b-c\right)\left(a-c\right)\left(a-b\right)}\)

N=\(\dfrac{b-c-a+c+a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)=0

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

\(=\left(b-c\right)\left(c-a\right)\left(c-b\right)\left(c+b+a\right)\)

nguồn câu hỏi tương tự

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

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

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

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

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

\(=\left(a-c\right)\left[b^2+ac-b\left(c+a\right)\right]\)

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

\(=\left(a-c\right)\left[b\left(b-c\right)+a\left(c-b\right)\right]\)

\(=\left(a-c\right)\left[b\left(b-c\right)-a\left(b-c\right)\right]\)

\(=\left(a-c\right)\left(b-c\right)\left(b-a\right)\)

Cách khác:

Ta có:
\(a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)\)

\(=a(b^2-c^2)-b[(b^2-c^2)+(a^2-b^2)]+c(a^2-b^2)\)

\(=a(b^2-c^2)-b(b^2-c^2)-b(a^2-b^2)+c(a^2-b^2)\)

\(=(a-b)(b^2-c^2)-(b-c)(a^2-b^2)\)

\(=(a-b)(b-c)(b+c)-(b-c)(a-b)(a+b)\)

\(=(a-b)(b-c)[(b+c)-(a+b)]=(a-b)(b-c)(c-a)\)

a) \(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)

\(=\left(x^4-2x^3+5x^2-4x+4\right)+\left(x^2-4x+4\right)\)

\(=x^4-2x^3+6x^2-8x+8\)

\(=\left(x^4-2x^3+2x^2\right)+\left(4x^2-8x+8\right)\)

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

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

\(x^4-9x^3+28x^2-36x+16\)

\(=x^4-x^3-8x^3+8x^2+20x^2-20x-16x+16\)

\(=\left(x^4-x^3\right)-\left(8x^3-8x^2\right)+\left(20x^2-20x\right)-\left(16x-16\right)\)

\(=x^3\left(x-1\right)-8x^2\left(x-1\right)+20x\left(x-1\right)-16\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3-8x^2+20x-16\right)\)

\(=\left(x-1\right)\left(x^3-2x^2-6x^2+12x+8x-16\right)\)

\(=\left(x-1\right)[x^2\left(x-2\right)-6x\left(x-2\right)+8\left(x-2\right)]\)

\(=\left(x-1\right)\left(x-2\right)\left(x^2-6x+8\right)\)

\(=\left(x-1\right)\left(x-2\right)\left(x^2-4x-2x+8\right)\)

\(=\left(x-1\right)\left(x-2\right)[x\left(x-4\right)-2\left(x-4\right)]\)

\(=\left(x-1\right)\left(x-2\right)\left(x-2\right)\left(x-4\right)\)

\(=\left(x-1\right)\left(x-2\right)^2\left(x-4\right)\)

Để tui cân hết choa 

a, Lời giải 

\(=\left[x^2-\left(x-2\right)\right]^2+\left(x-2\right)^2\)

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

\(=x^4-2x^3+4x^2+2x^2-8x+8\)

\(=x^4-2x^3+6x^2+8x+8\)

\(=\left(x^4-2x^3+2x^2\right)+\left(4x^2-8x+8\right)\)

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

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

Quy đồng đi, ta sẽ được  \(A=0\)

\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-a\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)

\(A=\frac{-b+c}{-\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{-c+a}{-\left(a-b\right)\left(a-c\right)\left(b-a\right)}+\frac{-a+b}{-\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(A=\frac{-b+c-c+a-a+b}{-\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(A=\frac{0}{-\left(a-b\right)\left(a-c\right)\left(b-a\right)}\)

A = 0

đặt \(\hept{\begin{cases}a+b=x\\b+c=y\\c+a=z\end{cases}}\)

cậu tính A theo x,y,x rồi chứng minh 

\(B=\frac{x}{z-y}.\frac{y}{x-z}+\frac{y}{x-z}.\frac{z}{y-x}+\frac{z}{y-x}.\frac{x}{z-y}=-1\)

thì ta có A+2B>=0   -->A>=-2B=2

\(\frac{\left(a+b\right)^2}{a-b}+\frac{\left(b+c\right)^2}{\left(b-c\right)}+\frac{\left(c+a\right)^2}{\left(c-a\right)}\ge2\)

Subtract 2 from both sides:

\(\frac{\left(a+b\right)^2}{a-b}+\frac{\left(b+c\right)^2}{b-c}+\frac{\left(c+a\right)^2}{c-a}-2\ge2-2\)

Refine:

\(\frac{\left(a+b\right)^2}{a-b}+\frac{\left(b+c\right)^2}{b-c}+\frac{\left(c+a\right)^2}{c-a}\ge0\)

Simplyfy : \(\frac{\left(a+b\right)^2}{\left(a-b\right)}+\frac{\left(b+c\right)^2}{b-c}+\frac{\left(c+a\right)^2}{c-a}:\)     \(\frac{4a^2bc-4a^2c^2-4a^2b^2+2a^2b-2a^2c+4ab^2c+4abc^2+2ac^2-2ab^2-4b^2c^2+2b^2c-2bc^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(\frac{\left(a+b\right)^2}{\left(a-b\right)}+\frac{\left(b+c\right)^2}{\left(b-c\right)}+\frac{\left(c+a\right)^2}{\left(c-a\right)}-2\)

Convert element to fraction: \(2=\frac{2}{1}\)

\(=\frac{\left(a+b\right)^2}{\left(a-b\right)}+\frac{\left(b+c\right)^2}{\left(b-c\right)}+\frac{\left(c+a^2\right)}{\left(c-a\right)}-\frac{2}{1}\)

Find LCD for: \(\frac{\left(a+b\right)^2}{\left(a-b\right)}+\frac{\left(b+c\right)^2}{\left(b-c\right)}+\frac{\left(c+a\right)^2}{c-a}-\frac{2}{1}\):

Find the least common denominator 1   (a  - b) (b - c) (c- a) = (a  - b) (b - c) (c- a)(a  - b) (b - c) (c- a)

Sau đó vào đây để xem bài giải tiếp theo nhá! Lười đánh máy tiếp lắm!   Có gì mai mốt sử dụng phần mềm đó giải khỏi phải lên đây hỏi.

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