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b. Sử dụng các hằng đẳng thức
\(a^3+b^3+c^2-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=3\left(a^2+b^2+c^2-ab-bc-ca\right)\)
và \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
nên \(A=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{1}{2}.\frac{\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Do (a - b) + (b - c) + (c - a) = 0 nên áp dụng hđt \(X^2+Y^2+Z^2=-2\left(XY+YZ+ZX\right)\)khi X + Y + Z = 0, ta có:
\(A=-2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right).\)
Bài 1 :
\(b,ax^2+3ax+9=a^2\)
\(\Leftrightarrow a^2x+3ax+9-a^2=0\)
\(\Leftrightarrow ax\left(a+3\right)+\left(a+3\right)\left(3-a\right)=0\)
\(\Leftrightarrow\left(a+3\right)\left(ax+3-a\right)=0\)
Vì \(a\ne3\Rightarrow\left(a+3\right)\ne0\Rightarrow ax+3-a=0\)
\(\Leftrightarrow ax=a-3\)
Vì \(a\ne0\Rightarrow x=\frac{a-3}{a}\)
a) Do đa thức chia có bậc là 3 , đa thức bị chia có bậc 2 nên thương sẽ có bậc 1
Ta có : (x3+ ax2 + 5x +3) = (x2+ 2x + 3)( x + d)
(x3+ ax2 + 5x +3) = x3 + dx2 + 2x2 + 2dx + 3x + 3d
(x3+ ax2 + 5x +3) = x3 + x2( d + 2) + x( 2d + 3) + 3d
Đồng nhất hệ số , ta có :
d + 2 = a --> a = 1 + 2 = 3
2d + 3 = 5 --> 2.1 + 3 = 5
3d = 3 --> d = 1
Vậy , a = 3 thỏa mãn điều kiện đề bài
b) Tẹo tớ gửi nha
ĐKXĐ: \(x\ne\pm1;x\ne0\)
a)\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)
\(=\left(\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)
\(=\dfrac{x^2+2x+1-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{2x}{5x-5}-\dfrac{x^2-1}{x^2+2x+1}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}.\dfrac{5\left(x-1\right)}{2x}-\dfrac{x^2-1}{x^2+2x+1}\)
\(=\dfrac{10}{x+1}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)^2}\)
\(=\dfrac{10}{x+1}-\dfrac{x-1}{x+1}\)
\(=\dfrac{11-x}{x+1}\)
b) \(A=\dfrac{11-x}{x+1}=2\)
\(\Leftrightarrow11-x=2\left(x+1\right)\)
\(\Leftrightarrow11-x=2x+2\)
\(\Leftrightarrow-x-2x=2-11\)
\(\Leftrightarrow-3x=-9\)
\(\Leftrightarrow x=3\left(nhận\right)\)
c) -Để \(A=\dfrac{11-x}{x+1}\in Z\) thì:
\(\left(11-x\right)⋮\left(x+1\right)\)
\(\Rightarrow\left(12-x-1\right)⋮\left(x+1\right)\)
\(\Rightarrow12⋮\left(x+1\right)\)
\(\Rightarrow\left(x+1\right)\inƯ\left(12\right)\)
\(\Rightarrow\left(x+1\right)\in\left\{1;2;3;4;6;12;-1;-2;-3;-4;-6;-12\right\}\)
\(\Rightarrow x\in\left\{2;3;5;11;-2;-3;-4;-5;-7;-13\right\}\)
a)\(\frac{x^2+4}{x^2}+\frac{4}{x+1}\left(\frac{1}{x}+1\right)\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x+1}.\frac{x+1}{x}\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x}\)
\(=\frac{x^2+4x+4}{x^2}\)
\(\left(\frac{x+2}{x}\right)^2\)
=>phép chia = 1 với mọi x # 0 và x#-1
b)Cm tương tự
a: \(A=\left(\dfrac{2x^2+2}{x^3-1}+\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+3}{x^3-x^2+3x-3}\right):\dfrac{1}{x-1}\)
\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{x^4+2x^2+1-x^2}-\dfrac{x^2+3}{x^2\left(x-1\right)+3\left(x-1\right)}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x^2-x+1\right)}{\left(x^2+1\right)^2-x^2}-\dfrac{x^2+3}{\left(x-1\right)\left(x^2+3\right)}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2-x+1}{\left(x^2+1+x\right)\left(x^2+1-x\right)}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)
\(=\left(\dfrac{2x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x^2+x+1}-\dfrac{1}{x-1}\right)\cdot\dfrac{x-1}{1}\)
\(=\dfrac{2x^2+3+x-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x-1}{1}\)
\(=\dfrac{x^2+1}{x^2+x+1}\)
b: Để A là số nguyên thì \(x^2+1⋮x^2+x+1\)
=>\(x^2+x+1-x⋮x^2+x+1\)
=>\(x⋮x^2+x+1\)
=>\(x^2+x⋮x^2+x+1\)
=>\(x^2+x+1-1⋮x^2+x+1\)
=>\(-1⋮x^2+x+1\)
=>\(x^2+x+1\in\left\{1;-1\right\}\)
=>\(x^2+x+1=1\)
=>x2+x=0
=>x(x+1)=0
=>\(x\in\left\{0;-1\right\}\)
a: Ta có: \(ax-x+1=a^2\)
\(\Leftrightarrow x\left(a-1\right)=a^2-1\)
hay x=a+1