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Ta có : ( 2a2 - a - 7 ) / ( a-2) = \(\frac{2a^2-a-7}{a-2}\)
= \(\frac{\left(2a+3\right)\left(a-2\right)}{\left(a-2\right)}+\frac{\left(-1\right)}{\left(a-2\right)}\)
= 2a + 3 + \(\frac{\left(-1\right)}{ \left(a-2\right)}\)
Để biểu thức trên chia hết cho ( a - 2 ) thì ( -1) phải chia hết cho ( a-2)
=> ( a - 2 ) thuộc Ư(-1) = \(\left\{-1;1\right\}\)
- a - 2 = -1 => a = 1
- a - 2 = 1 => a = 3
Vậy a=1 hoặc a=3 thì 2a2 - a - 7 chia hết cho a-2
hihi
Ta có: 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)
Mặt khác:
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.
ab2+ac2+abc+a2b+bc2+abc+a2c+b2c+abc
=ab2+ac2+a2b+bc2+a2c+b2c+3abc
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
a.) \\(\\left(a+b+c\\right)^3-a^3-b^3-c^3\\)
\\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc-a^3-b^3-c^3\\)\\(=3\\left(3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\right)\\)
\\(=3\\left(abc+a^2b+a^2c+ac^2+b^2c+ab^2+abc+bc^2\\right)\\)
\\(=3\\left[ab\\left(a+c\\right)+ac\\left(a+c\\right)+b^2\\left(a+c\\right)+bc\\left(a+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(ab+ac+bc+b^2\\right)\\)
\\(=3\\left(a+c\\right)\\left[a\\left(b+c\\right)+b\\left(b+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(a+b\\right)\\left(b+c\\right)\\)
b) 4a2b2-(a2 +b2-c2)2
=（2ab+a2+b2-c2)(2ab-a2-b2+c2）
=[(a+b)2-c2][c2-(a-b)2]
=(a+b+c)(a+b-c)(c+a-b)(c-a+b)
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc-a^3-b^3-c^3\)
\(=3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)
\(=3\left(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\right)\)
\(=3\left(ab\left(a+b\right)+b^2c+abc+bc^2+c^2a+ca^2+abc\right)\)
\(=3\left(ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\right)\)
\(=3\left(a+b\right)\left(ab+bc+c^2+ac\right)\)
\(=3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(2\left(a+b+c\right)=a^2+b^2+c^2+3\)
\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}a-1=0\\b-1=0\\c-1=0\end{array}\right.\)\(\Leftrightarrow a=b=c=1\)
sai đề r`