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Bài 1:
Ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow \left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow \frac{a+b}{ab}+\frac{a+b+c-c}{c(a+b+c)}=0\)
\(\Leftrightarrow (a+b)\left(\frac{1}{ab}+\frac{1}{c(a+b+c)}\right)=0\)
\(\Leftrightarrow (a+b).\frac{c(a+b+c)+ab}{abc(a+b+c)}=0\Leftrightarrow (a+b).\frac{c(c+a)+b(a+c)}{abc(a+b+c)}=0\)
\(\Leftrightarrow \frac{(a+b)(b+c)(c+a)}{abc(a+b+c)}=0\Rightarrow (a+b)(b+c)(c+a)=0\)
\(\Rightarrow \left[\begin{matrix} a+b=0\\ b+c=0\\ c+a=0\end{matrix}\right.\)
Ta xét TH $a+b=0\Rightarrow a=-b$, các TH khác làm tương tự:
Khi đó: \(\frac{1}{a^{2017}+b^{2017}+c^{2017}}=\frac{1}{(-b)^{2017}+b^{2017}+c^{2017}}=\frac{1}{c^{2017}}\)
Và: \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{(-b)^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{c^{2017}}\)
Do đó: \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2017}+b^{2017}+c^{2017}}\)
Ta có đpcm.
Bài 2:
Ta có:
Áp dụng công thức quen thuộc (suy ra trực tiếp từ hằng đẳng thức đáng nhớ): \(x^3+y^3=(x+y)^3-3xy(x+y)\) ta có:
\(a^3+b^3=2c^3\)
\(\Leftrightarrow a^3+b^3+c^3=3c^3\)
\(\Leftrightarrow (a+b)^3-3ab(a+b)+c^3=3c^3\)
\(\Leftrightarrow (a+b)^3+c^3-3ab(a+b)=3c^3\)
\(\Leftrightarrow (a+b+c)^3-3(a+b).c(a+b+c)-3ab(a+b)=3c^3\)
\(\Leftrightarrow (a+b+c)^3=3c^3+3ab(a+b)+3(a+b)c(a+b+c)\vdots 3\)
Mà $3\in\mathbb{P}$ nên \(\Rightarrow a+b+c\vdots 3\)
Ta có đpcm.
1. x^3-19x-30
=x^3-25x+6x-30
=x(x^2-25)+6(x-5)
=x(x+5)(x-5)+6(x-5)
=(x-5)(x^2+5x+6)
=(x-5)(x^2+2x+3x+6)
=(x-5)[x(x+2)+3(x+2)]
=(x-5)(x+2)(x+3)
2.
a + b + c = 0
<=> (a + b + c)² = 0
<=> a² + b² + c² + 2(ab + bc + ca) = 0
<=> a² + b² + c² = -2(ab + bc + ca) ------------(1)
CẦn chứng minh:
2(a^4 + b^4 + c^4) = (a² + b² + c²)²
<=> 2(a^4 + b^4 + c^4) = a^4 + b^4 + c^4 + 2(a²b² + b²c² + c²a²)
<=> a^4 + b^4 + c^4 = 2(a²b² + b²c² + c²a²)
<=> (a² + b² + c²)² = 4(a²b² + b²c² + c²a²) ---(cộng 2 vế cho 2(a²b² + b²c² + c²a²) )
<=> [-2(ab + bc + ca)]² = 4(a²b² + b²c² + c²a²) ----(do (1))
<=> 4.(a²b² + b²c² + c²a²) + 8.(ab²c + bc²a + a²bc) = 4(a²b² + b²c² + c²a²)
<=> 8.(ab²c + bc²a + a²bc) = 0
<=> 8abc.(a + b + c) = 0
<=> 0 = 0 (đúng), Vì a + b + c = 0
=> Đpcm
Ta có a-b+b-c+c-a=0 nên (a−b)^3+(b−c)^3+(c−a)^3=3(a−b)(b−c)(c−a)
Do đó 3(a−b)(b−c)(c−a)=210⇔(a−b)(b−c)(c−a)=70
mà a;b;cϵZ→a−b;b−c;c−aϵZ
→a−b;b−c;c−a là ước của 70
Mặt khác 70=(−2)(−5)^7 (do tổng 3 số này bằng 0)
Do đó A=2+5+7=14
a, 4b2c2 - (b2+c2-a2)2
= (2bc)²-(b²+c²-a²)²
=(2bc+b²+c²-a²)(2bc-b²-c²+a²)
=[(b+c)²-a²][a²-(b-c)²]
=(b+c+a)(b+c-a)(a+b-c)(a-b+c).
b, 8x3-64 = 23.x3-43 = (2x)3-43
= (2x-4)[(2x)2+2.x.4+42] = (2x-4)(4x2+8x+16)
c, 8x3-27= 23.x3-33 = (2x)3-33
= (2x-3)[(2x)2+2.x.3+32] = (2x-3)(4x2+6x+9)
a) Ta có:
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
b) Đặt a + b - c = x
b + c - a = y
c + a - b = z
=> x + y + z = a + b - c + b + c - a + c + a - b = a + b + c
Áp dụng hằng đẳng thức \(\left(x+y+z\right)^3-x^3-y^3-z^3=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\) ( Câu a )
Ta có:
\(\left(a+b+c\right)^3-\left(a+b-c\right)^3-\left(b+c-a\right)^3+\left(c+a-b\right)^3\)
\(=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\)
\(=3.2b.2c.2a=24abc\)
Bài 4:
=>x(x^2+1)=0
=>x=0
Bài 5:
=>\(3n^3+n^2+9n^2-1-4⋮3n+1\)
=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{0;-1;1\right\}\)
Bài 3:
\(a^3+b^3+c^3-3bac\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
a) A = (a - b)3 + (b - c)3 + (c - a)3
Đặt : a - b = x ; b - c = y; c - a = z thì x + y + z = 0
Do đó: \(x^3+y^3+z^3=3xyz\)
Vậy A = \(3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
b) B = (a + b - 2c)3 + (b + c - 2a)3 + (c + a - 2b)3
Đặt : a + b - 2c = x ; b + c - 2a = y ; c + a - 2b = z
Thì x + y + z = 0 do đó \(x^3+y^3+z^3=3xyz\)
Vậy B = 3(a + b - 2c)(b + c - 2a)(c + a - 2b)
a) Ta có: \(A=\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
\(=a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3\)
\(=-3\left(a^2b+ac^2-ab^2-bc^2+b^2c-a^2c\right)\)
\(=3\left[\left(a^2b-ab^2\right)+\left(ac^2-bc^2\right)-\left(a^2c-b^2c\right)\right]\)
\(=3\left[ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a^2-b^2\right)\right]\)
\(=3\left[ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)\right]\)
\(=3\left(a-b\right)\left[ab+c^2-c\left(a+b\right)\right]\)
\(=3\left(a-b\right)\left(ab+c^2-ca-cb\right)\)
\(=3\left(a-b\right)\left[\left(ab-ac\right)-\left(bc-c^2\right)\right]\)
\(=3\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]\)
\(=3\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
b) 
b)a(b-c)3+b(c-a)3+c(a-b)3
Bạn tự tách trong ngoặc ra nhá
=ab3-ac3+bc3-a3b+a3c-b3c
=b3(a-c)+ac(a2-c2)-b(a3-c3)
=b3(a-c)+ac(a-c)(a+c)-b(a-c)(a2+ac+c2)
=(a-c)[b3+ac(a+c)-b(a2+ac+c2)]
=(a-c)(b3+a2c+ac2-ba2-abc-bc2)
=(a-c)[ac(a+c)+b(b2-a2)-bc(a+c)]
=(a-c)[c(a+c)(a-b)-b(a-b)(a+b)]
=(a-c)(ca+c2-ab-b2)(a-b)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b+c-a\right)\left[\left(a+b+c\right)^2+a\left(a+b+c\right)+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ac\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)