Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = x2(x - 1) + 6(1 - x)
A = x3 - x2 + 6 - 6x
A = (x3 - 6x) - (x2 - 6)
A = x.(x2 - 6) - (x2 - 6)
A = (x - 1)(x2 - 6)
C = x2 + 2xy + y2 - yz - xz
C = (x + y)2 - z.(x + y)
C = (x + y - z).(x + y)
B = \(\frac{8xy-6x^2}{3y\left(3x-4y\right)}=\frac{2x\left(4y-3x\right)}{-3y\left(4y-3x\right)}=-\frac{2x}{3y}\)
C = \(\frac{2x^3-18x}{x^4-81}=\frac{2x\left(x^2-9\right)}{\left(x^2-9\right)\left(x^2+9\right)}=\frac{2x}{x^2+9}\)
1) 3-81x3
=3-3.(3x)3
=3[13-(3x)3)]
=3(1-3x)(1+3x+9x2)
2) 250x3y3-2
=2(125x3y3-1)
=2[(5xy)3-13]
=2(5xy-1)(25x2y2+5xy+1)
3) a4+4b4
=[(a2)2+4a2b2+(2b2)]-4a2b2
=(a2+2b2)-(2ab)2
=(a2+2b2-2ab)(a2+2b2+2ab)
4) x5+x+1
=x5-x2+x2+x+1
=x2(x3-13)+(x2+x+1)
=x2(x-1)(x2+x+1)+(x2+x+1)
=(x2+x+1)[x2(x-1)+1]
=(x2+x+1)(x3-x2+1)
Chúc bn học giỏi nhoa!!!
x4+x=x(x3+1)=x(x+1)(x2-x+1)
x4+64=x4+16x2+64-16x2=(x2+8)2-(4x)2=(x2+8+4x)(x2+8-4x)
4x4+81=4x4+36x2+81-36x2=(2x2+9)2-(6x)2=(2x2+9+6x)(2x2+9-6x)
64x4+y4=64x4+16(xy)2+y4-16(xy)2=(8x2+y2)-(4xy)2=(8x2+y2-4xy)(8x2+y2=4xy)
x4+4y4=x4+4(xy)2+4y4-4(xy)2=(x2+2y2-2xy)(x2+2y2+2xy)
x4+x2+1=(x4+2x2+1)-x2=(x2+1-x)(x2+1+x)
Mình làm có vài đoạn hơi tắt nha.
1, \(x^3+8x^2+17x+10=\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)
\(=x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)\(=\left(x+1\right)\left(x^2+7x+10\right)=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
2. \(2x^3-3x^2+3x-1=\left(2x^3-x^2\right)-\left(2x^2-x\right)+\left(2x-1\right)\)
\(=x^2\left(2x-1\right)-x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(2x-1\right)\left(x^2-x+1\right)\)
3. \(x^4+x^2+1=\left(x^4+1\right)+x^2=\left(x^2+1\right)^2-2x^2+x^2\)\(=\left(x^2+1\right)^2-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
4. \(81x^4+4=\left(9x^2\right)^2+2^2=\left(9x^2+2\right)^2-2.9x^2.2=\left(9x^2+2\right)^2-\left(6x\right)^2\)
\(=\left(9x^2+6x+2\right)\left(9x^2-6x+2\right)\)
\(2x^4-9x^3-11x^2+81x=63\)
\(\Rightarrow2x^4-9x^3-11x^2+81x-63=0\)
\(\Rightarrow2x^4-2x^3-7x^3+7x^2-18x^2+18x+63x-63=0\)
\(\Rightarrow2x^3\left(x-1\right)-7x^2\left(x-1\right)-18x\left(x-1\right)+63\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x^3-7x^2-18x+63\right)=0\)
\(\Rightarrow\left(x-1\right)\left[x^2\left(2x-7\right)-9\left(2x-7\right)\right]=0\)
\(\Rightarrow\left(x-1\right)\left(2x-7\right)\left(x^2-9\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x-7\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-7=0\\x-3=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=3\\x=-3\end{matrix}\right.\)
Vậy.................
Chúc bạn học tốt!!!
a) \(\left(x^2+4\right)^2-4x\left(x^2+4\right)=0\)
\(=\left(x^2+4\right)\left(x^2+4-4x\right)=0\)
\(=\left(x^2+4\right)\left(x+2\right)^2=0\)
Mà \(x^2\ge0\Rightarrow x^2+4>0\)
\(\Rightarrow x+2=0\)
\(\Rightarrow x=-2\)
b) \(x^5-18x^3+81x=0\)
\(=\left(x^5-9x^3\right)-\left(9x^3-81x\right)=0\)
\(=x^3\left(x^2-9\right)-9x\left(x^2-9\right)=0\)
\(=\left(x^3-9x\right)\left(x^2-9\right)=0\)
\(=x\left(x^2-9\right)\left(x^2-9\right)=0\)
\(=x\left(x^2-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x\in\left\{-3;3\right\}\end{cases}}\)
a) (x2 + 4)2 - 4x(x2 + 4) = 0
(x2 + 4)(x2 + 4 - 4x) = 0
(x2 + 4)(x - 2)2 = 0
\(\Rightarrow\) x2 + 4 = 0 hoặc (x - 2)2 = 0
\(\Rightarrow\) x2 = - 4 hoặc x - 2 = 0
\(\Rightarrow\) x \(\in\) tập hợp rỗng hoặc x = 2
Vậy x = 2
b) x5 - 18x3 + 81x = 0
x(x4 - 18x2 + 81) = 0
x(x2 - 9) = 0
x(x - 3)(x + 3) = 0
\(\Rightarrow\) x = 0 hoặc x - 3 = 0 hoặc x + 3 = 0
\(\Rightarrow\) x = 0 hoặc x = 3 hoặc x = - 3
Vậy \(x\in\left\{0;3;-3\right\}\)
a) \(x^4+x^2+1\)
\(=x^2\left(x^2+1\right)+\left(x^2+1\right)-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
b) \(\left(1+x\right)^2-4x\left(1-x^2\right)\)
\(=\left(1+x\right)^2-4x\left(1-x\right)\left(1+x\right)\)
\(=\left(1+x\right)\left[1+x-4x+4x^2\right]\)
\(=\left(1+x\right)\left[1-3x+4x^2\right]\)
81x4+4=(9x2)2+22+2.9x2-36x2=(9x2+2)2-(6x)2
=(9x2-6x+2)(9x2+6x+2)
chuc hk tot.