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a) \(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
\(=\left(x^4-2x^3+5x^2-4x+4\right)+\left(x^2-4x+4\right)\)
\(=x^4-2x^3+6x^2-8x+8\)
\(=\left(x^4-2x^3+2x^2\right)+\left(4x^2-8x+8\right)\)
\(=x^2\left(x^2-2x+2\right)+4\left(x^2-2x+2\right)\)
\(=\left(x^2+4\right)\left(x^2-2x+2\right)\)
\(x^4-9x^3+28x^2-36x+16\)
\(=x^4-x^3-8x^3+8x^2+20x^2-20x-16x+16\)
\(=\left(x^4-x^3\right)-\left(8x^3-8x^2\right)+\left(20x^2-20x\right)-\left(16x-16\right)\)
\(=x^3\left(x-1\right)-8x^2\left(x-1\right)+20x\left(x-1\right)-16\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3-8x^2+20x-16\right)\)
\(=\left(x-1\right)\left(x^3-2x^2-6x^2+12x+8x-16\right)\)
\(=\left(x-1\right)[x^2\left(x-2\right)-6x\left(x-2\right)+8\left(x-2\right)]\)
\(=\left(x-1\right)\left(x-2\right)\left(x^2-6x+8\right)\)
\(=\left(x-1\right)\left(x-2\right)\left(x^2-4x-2x+8\right)\)
\(=\left(x-1\right)\left(x-2\right)[x\left(x-4\right)-2\left(x-4\right)]\)
\(=\left(x-1\right)\left(x-2\right)\left(x-2\right)\left(x-4\right)\)
\(=\left(x-1\right)\left(x-2\right)^2\left(x-4\right)\)
\(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=ab^2-ac^2+ca^2-cb^2+b\left(c^2-a^2\right)\)
\(=\left(ab^2-cb^2\right)-\left(ac^2-ca^2\right)+b\left(c-a\right)\left(c+a\right)\)
\(=b^2\left(a-c\right)-ac\left(c-a\right)+b\left(c-a\right)\left(c+a\right)\)
\(=b^2\left(a-c\right)+ac\left(a-c\right)-b\left(a-c\right)\left(c+a\right)\)
\(=\left(a-c\right)\left[b^2+ac-b\left(c+a\right)\right]\)
\(=\left(a-c\right)\left(b^2+ac-bc-ab\right)\)
\(=\left(a-c\right)\left[b\left(b-c\right)+a\left(c-b\right)\right]\)
\(=\left(a-c\right)\left[b\left(b-c\right)-a\left(b-c\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left(b-a\right)\)
Cách khác:
Ta có:
\(a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)\)
\(=a(b^2-c^2)-b[(b^2-c^2)+(a^2-b^2)]+c(a^2-b^2)\)
\(=a(b^2-c^2)-b(b^2-c^2)-b(a^2-b^2)+c(a^2-b^2)\)
\(=(a-b)(b^2-c^2)-(b-c)(a^2-b^2)\)
\(=(a-b)(b-c)(b+c)-(b-c)(a-b)(a+b)\)
\(=(a-b)(b-c)[(b+c)-(a+b)]=(a-b)(b-c)(c-a)\)
\(a\left(b+c\right)^2\left(b-c\right)+b\left(c+a\right)^2\left(c-2\right)+c\left(a+b\right)^2\left(a-b\right)\)
\(=\left(b-c\right)\left(c-a\right)\left(c-b\right)\left(c+b+a\right)\)
nguồn câu hỏi tương tự
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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, 3x^2-7x+2
=3x^2-2x-3x+2
=x(3x-2)-(3x-2)
=(x-1)(3x-2)
b, a(x^2+1)-x(a^2+1)
=(a-x)(x^2+1)
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