Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\text{a)}x^3-6x^2+12x-8\)
\(=x^3-2x^2-4x^2+8x+4x-8\)
\(=\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+4x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)\left(x+2\right)^2\)
\(\text{b)}8x^2+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)
Bài 2:
\(\text{a) }x^7+1=\left(x^{\frac{7}{3}}\right)^3+1^3=\left(x^{\frac{7}{3}}+1\right)\left[\left(x^{\frac{7}{3}}\right)^2-x^{\frac{7}{3}}+1\right]=\left(x^{\frac{7}{3}}+1\right)\left(x^{\frac{14}{3}}-x^{\frac{7}{3}}+1\right)\)
\(\text{b) }x^{10}-1=\left(x^5\right)^2-1^2=\left(x^5-1\right)\left(x^5+1\right)\)
Bài 3:
\(\text{a) }69^2-31^2=\left(69-31\right)\left(69+31\right)=38.100=3800\)
\(\text{b) }1023^2-23^2=\left(1023-23\right)\left(1023+23\right)=1000.1046=1046000\)
\(ab-a+b-a^2=\left(ab-a^2\right)+\left(b-a\right)=a.\left(b-a\right)+\left(b-a\right)=\left(b-a\right).\left(a+1\right)\\ \)
\(8-\left(x-1\right)^3=2^3-\left(x-1\right)^3=\left(2-x+1\right).\left(2^2+2.\left(x-1\right)+\left(x-1\right)^2\right)\)
\(=\left(3-x\right).\left(4+2x-2+x^2-2x+1\right)=\left(3-x\right).\left(3+x^2\right)\)
\(3x^2-8x-16=\left(3x^2+4x\right)-\left(12x+16\right)=x.\left(3x+4\right)-4.\left(3x+4\right)=\left(3x+4\right).\left(x-4\right)\)
\(ab-a+b-a^2\)
\(=\left(ab-a^2\right)+\left(b-a\right)\)
\(=a\left(b-a\right)+\left(b-a\right)\)
\(=\left(a+1\right)\left(b-a\right)\)
Bài làm:
\(8x^3+4x^2-9x+30=8x^3+16x^2-12x^2-24x+15x+30\)
\(=8x^2\left(x+2\right)-12x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(8x^2-12x+15\right)\)
\(8x^3+4x^2-9x+30\)
\(=8x^3+16x^2-12x^2-24x+15x+30\)
\(=8x^2\cdot\left(x+2\right)-12x\cdot\left(x+2\right)+15x\cdot\left(x+2\right)\)
\(=\left(x+2\right)\cdot\left(8x^2-12x+15\right)\)
a,x8 +x4 +1=x6 .x2 +x3 .x+1=x6 .x2-x2 +x3 .x-x+1+x+x2=x2.(x6-1)+x.(x3-1)+1+x+x2=x2.(x3-1).(x3+1)+x.(x-1).(x2+x+1)+1+x+x2
Câu 1:
Ta có \(x^3+3x-5=x^3+2x+x-5=\left(x^2+2\right)x+x-5\)
để giá trị của đa thức \(x^3+3x-5\)chia hết cho giá trị của đa thức \(x^2+2\)
thì \(x-5⋮x^2+2\Rightarrow\left(x-5\right)\left(x+5\right)⋮x^2+2\Rightarrow x^2-25⋮x^2+2\)
\(\Leftrightarrow x^2+2-27⋮x^2+2\Rightarrow27⋮x^2+2\)
\(\Leftrightarrow x^2+2\inƯ\left(27\right)\)do \(x^2+2\inℤ,\forall x\inℤ\)
mà \(x^2+2\ge2,\forall x\inℤ\)
\(\Rightarrow x^2+2\in\left\{3;9;27\right\}\)\(\Leftrightarrow x^2\in\left\{1;7;25\right\}\)
mà \(x^2\)là số chính phương \(\forall x\inℤ\)
\(\Rightarrow x^2\in\left\{1;25\right\}\Leftrightarrow x\in\left\{\pm1;\pm5\right\}\)
**bạn nhớ thử lại nhé
\(KL...\)
\(x^4+13x^2+36=x^4+4x^2+9x^2+36\)
\(=x^2\left(x^2+4\right)+9\left(x^2+4\right)=\left(x^2+9\right)\left(x^2+4\right)\)
??????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
a) \(x^2-y^2-x-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
b) \(x^2-y^2+2yz-z^2\)
\(=x^2-\left(y^2-2yz+z^2\right)\)
\(=x^2-\left(y-z\right)^2\)
\(=\left(x-y+z\right)\left(x+y-z\right)\)
=a^2 + a^3 -b^2 +b^3 -a^2b^2(a+b)
=(a^2-b^2) + (a^3+b^3) -a^2b^2(a+b)
=(a-b)(a+b) + (a+b)(a^2-ab+b^2) - a^2b^2(a+b)
=(a+b)(a-b+a^2-ab+b^2-a^2b^2)
=(a+b) ( (a-ab) -(b-b^2) +a^2(1-b^2) )
=(a+b) ( a(1-b) - b(1-b) + a^2(1-b)(1+b) )
=(a+b) (1-b)(a-b+a^2+a^2b)
a) \(x^{m+2}-2x^m=x^m\left(x^2-2\right)\)
b) \(x^{k+1}-x^{k+2}=x^{k+1}\left(1-x\right)\)
a) xm+2 - 2xm = xm.x2 + 2.xm = xm( x2 - 2 ) = xm( x - √2 )( x + √2 )
b) xk+1 - xk+2 = xk+1 - xk+1.x = xk+1( 1 - x )
= x2 + 8x + 16 - 6 = (x2 +2.x.4 + 42) - (\(\sqrt{6}\))2 = (x+4)2 - (\(\sqrt{6}\))2 = (x+4 - \(\sqrt{6}\)). (x+4+ \(\sqrt{6}\))