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a(b^3-c^3) +b(c^3-a^3)+c(a^3-b^3)
=> a(b-c)(b^2+bc+c^2)+bc^3-ba^3+ca^3-cb^3
=>a(b-c)(b^2+bc+c^2)-(cb^3-bc^3)-(ba^3-ca^3)
=>a(b-c)(b^2+bc+c^2)-bc(b-c)(b+c)-a^3(b-c)
=>(b-c)(ab^2+abc+ac^2-cb^2-bc^2-a^3)
=>(b-c)(
1) a3+b3+c3-3abc = (a+b)3-3ab(a+b)+c3-3abc
= (a+b+c)(a2+2ab+b2-ab-ac+c2) -3ab(a+b+c)
= (a+b+c)( a2+b2+c2-ab-bc-ca)
1
(x2-8)2+36
=x4-16x2+64+36
=x4+20x2+100-36x2
=(x2+10)2-(6x)2
HĐT số 3
Câu 1:
\(=x^4-16x^2+64+36\)
\(=x^4-16x^2+100\)
\(=x^4+20x^2+100-36x^2\)
\(=\left(x^2+10\right)^2-\left(6x\right)^2\)
\(=\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)
Câu 2: \(=x^4+2x^2+1-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
a,x3-27+3x(x-3)
=(x-3)(x2+3x+9)+3x(x-3)
=(x-3)(x2+6x+9)
=(x-3)(x+3)2
b,5x3-7x2+10x-14
= x2(5x-7)+2(5x-7)
= (5x-7)(x2+2)
a,x3-27+3x(x-3)
=(x-3)(x2+3x+9)+3x(x-3)
=(x-3)(x2+3x+9+3x)
=(x-3)(x2+6x+9)
=(x-3)(x+3)2
b,5x3-7x2+10x-14
=(5x3+10x)-(7x2+14)
=5x(x2+2)-7(x2+2)
=(x2+2)(5x-7)
Câu 1:
Áp dụng BĐT Cô si cho 4 số dương, ta có:
\(a^4+b^4+c^4+d^4\ge4.^4\sqrt{\left(abcd\right)^4}=4abcd\)
Dấu "=" \(\Leftrightarrow a=b=c=d\)
Câu 2:
Gọi quãng đường AB là x km (x>0)
\(V_{tb}=\dfrac{S}{t}=\dfrac{x}{\dfrac{x}{\dfrac{2}{20}}+\dfrac{x}{\dfrac{2}{30}}}=\dfrac{x}{\dfrac{x}{40}+\dfrac{x}{60}}=\dfrac{x}{\dfrac{5x}{120}}=\dfrac{120x}{5x}=\dfrac{120}{5}=24\left(\text{km/h}\right)\)
Vậy ...
\(x^4+2x^2+5x^3+10x-2x^2-4\)
\(x^2\left[x^2+2\right]+5x\left(x^2+2\right)-2\left(x^2+2\right)\)
\(\left(x^2+5x-2\right)\left(x^2+2\right)\)
b,a4(b-c) +b4(c-a) +c4(a-b)
=(a-b)*c^4+(b^4-a^4)*c-a*b^4+a^4*b
=-(b-a)*(c-a)*(c-b)*(c^2+b*c+a*c+b^2+a*b+a^2)