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8 tháng 4 2020
a/ \(\left(x-1\right)^2=x^2-2x+1\) nên chọn đáp án D
b/ \(\left(x+2\right)^2=x^2+4x+4\) nên chọn đáp án C
c/ \(\left(a-b\right)\left(b-a\right)=-\left(a-b\right)\left(a-b\right)=-\left(a-b\right)^2\) nên chọn đáp án A
d/ \(-x^2+6x-9=-\left(x^2-6x+9\right)=-\left(x-3\right)^2\) nên chọn đáp án D
8 tháng 10 2019
Bài 1
a. \(x^4+2x^3+x^2-y^2\)\(=\left(x^2+x\right)^2-y^2\)\(=\left(x^2+x-y\right)\left(x^2+x+y\right)\)
b. \(x^3+3x^2-3x-1=\left(x-1\right)\left(x^2+4x+1\right)\)
c. \(x^3-4x^2-4x+1=\left(x+1\right)\left(x^2-5x+1\right)\)
d. \(x^3-3x^3-3x+1=\left(x+1\right)\left(x^2-4x+1\right)\)
e. \(x^3+x^2-2x-8=\left(x-2\right)\left(x^2+3x+4\right)\)
21 tháng 4 2020
Sorry Ngân Chu, đoạn chia hết cho 120 thì thêm cả chia hết cho 2 nữa, nên nhân vào mới ra 120 nhé!!
21 tháng 4 2020
Bài 1:
a, (n + 3)2 - (n - 1)2
= (n + 3 - n + 1)(n + 3 + n - 1)
= 4(2n - 2)
= 8(n - 1)
Vì 8 \(⋮\) 8 nên 8(n - 1) \(⋮\) 8 với n \(\in\) Z
b, n5 - 5n3 + 4n
= n(n4 - 5n2 + 4)
= n(n4 - n2 - 4n2 + 4)
= n[n2(n2 - 1) - 4(n2 - 1)]
= n(n2 - 1)(n2 - 4)
= n(n - 1)(n + 1)(n - 2)(n + 2)
= (n - 2)(n - 1)n(n + 1)(n + 2)
Vì (n - 2)(n - 1)n(n + 1)(n + 2) là tích của 5 số nguyên liên tiếp nên chia hết cho 3, 5, 8
Mà 3 x 5 x 8 = 120
\(\Rightarrow\) (n - 2)(n - 1)n(n + 1)(n + 2) \(⋮\) 120 hay n5 - 5n3 + 4n \(⋮\) 120 với n \(\in\) Z
Bài 2:
a, 4x(x + 1) = 8(x + 1)
\(\Leftrightarrow\) 4x(x + 1) - 8(x + 1) = 0
\(\Leftrightarrow\) (x + 1)(4x - 8) = 0
\(\Leftrightarrow\) 4(x + 1)(x - 2) = 0
\(\Leftrightarrow\) (x + 1)(x - 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy S = {-1; 2}
b, x2 - 6x + 8 = 0
\(\Leftrightarrow\) x2 - 6x + 9 - 1 = 0
\(\Leftrightarrow\) (x - 3)2 - 1 = 0
\(\Leftrightarrow\) (x - 3 - 1)(x - 3 + 1) = 0
\(\Leftrightarrow\) (x - 4)(x - 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy S = {4; 2}
c, x3 + x2 + x + 1 = 0
\(\Leftrightarrow\) x2(x + 1) + (x + 1) = 0
\(\Leftrightarrow\) (x + 1)(x2 + 1) = 0
Vì x2 + 1 > 0 với mọi x
\(\Rightarrow\) x + 1 = 0
\(\Leftrightarrow\) x = -1
Vậy S = {-1}
d, x3 - 7x - 6 = 0
\(\Leftrightarrow\) x3 - x - 6x - 6 = 0
\(\Leftrightarrow\) (x3 - x) - (6x + 6) = 0
\(\Leftrightarrow\) x(x2 - 1) - 6(x + 1) = 0
\(\Leftrightarrow\) x(x - 1)(x + 1) - 6(x + 1) = 0
\(\Leftrightarrow\) (x + 1)[x(x - 1) - 6] = 0
\(\Leftrightarrow\) (x + 1)(x2 - x - 6) = 0
\(\Leftrightarrow\) (x + 1)(x2 - 3x + 2x - 6) = 0
\(\Leftrightarrow\) (x + 1)[x(x - 3) + 2(x - 3)] = 0
\(\Leftrightarrow\) (x + 1)(x - 3)(x + 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)
Vậy S = {-1; 3; -2}
Câu e hình như bạn viết nhầm 2 lần số 17x thì phải, mình sửa lại rồi!!
e, 3x3 - 7x2 + 17x - 5 = 0
\(\Leftrightarrow\) 3x3 - x2 - 6x2 + 2x + 15x - 5 = 0
\(\Leftrightarrow\) (3x3 - x2) + (-6x2 + 2x) + (15x - 5) = 0
\(\Leftrightarrow\) x2(3x - 1) - 2x(3x - 1) + 5(3x - 1) = 0
\(\Leftrightarrow\) (3x - 1)(x2 - 2x + 5) = 0
\(\Leftrightarrow\) (3x - 1)(x2 - 2x + \(\frac{1}{4}\) + \(\frac{19}{4}\)) = 0
\(\Leftrightarrow\) (3x - 1)[(x - \(\frac{1}{2}\))2 + \(\frac{19}{4}\)] = 0
Vì (x - \(\frac{1}{2}\))2 + \(\frac{19}{4}\) > 0 với mọi x nên
\(\Rightarrow\) 3x - 1 = 0
\(\Leftrightarrow\) x = \(\frac{1}{3}\)
Vậy S = {\(\frac{1}{3}\)}
Bài 3:
Hình như phần a thì 16(1 - x) mới đúng chứ!!
a, x2(x - 1) + 16(1 - x)
= x2(x - 1) - 16(x - 1)
= (x - 1)(x2 - 16)
= (x - 1)(x - 4)(x + 4)
Câu b, d, g mình chịu, hình như đề sai thì phải, mình ko nghĩ ra được!!
c, x3 - 3x2 - 3x + 1
= (x3 + 1) - (3x2 + 3x)
= (x + 1)(x2 + x + 1) - 3x(x + 1)
= (x + 1)(x2 + x + 1 - 3x)
= (x + 1)(x2 - 2x + 1)
= (x + 1)(x - 1)(x - 1)
e, x4 - 13x2 + 36
= x4 - 4x2 - 9x2 + 36
= x2(x2 - 4) - 9(x2 - 4)
= (x2 - 4)(x2 - 9)
= (x - 2)(x + 2)(x - 3)(x + 3)
f, (x2 + x)2 + 4x2 + 4x - 12
= (x2 + x)2 + 4x2 + 4x + 4 - 16
= (x2 + x)2 + 4(x2 + x) + 4 - 16
= (x2 + x + 2)2 - 16
= (x2 + x + 2 - 4)(x2 + x + 2 + 4)
= (x2 + x - 2)(x2 + x + 6)
17 tháng 10 2020
Bài 4: Phân tích đa thức thành nhân tử
a) Ta có: \(x^4-4x^2-5\)
\(=x^4+x^2-5x^2-5\)
\(=x^2\left(x^2+1\right)-5\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2-5\right)\)
b) Ta có: \(\left(2x+1\right)^4-3\left(2x+1\right)^2+2\)
\(=\left(2x+1\right)^4-2\left(2x+1\right)^2-\left(2x+1\right)^2+2\)
\(=\left(2x+1\right)^2\left[\left(2x+1\right)^2-2\right]-\left[\left(2x+1\right)^2-2\right]\)
\(=\left(4x^2+4x+1-2\right)\left[\left(2x+1\right)^2-1\right]\)
\(=\left(4x^2+4x-1\right)\left(2x+1-1\right)\left(2x+1+1\right)\)
\(=\left(4x^2+4x-1\right)\cdot2x\cdot2\cdot\left(x+1\right)\)
\(=4x\cdot\left(x+1\right)\cdot\left(4x^2+4x-1\right)\)
d) Ta có: \(\left(x^2+2x-1\right)^2-3x\left(x^2+2x-1\right)+2x^2\)
\(=\left(x^2+2x-1\right)^2-x\left(x^2+2x-1\right)-2x\left(x^2+2x-1\right)+2x^2\)
\(=\left(x^2+2x-1\right)\left(x^2+2x-1-x\right)-2x\left(x^2+2x-1-x\right)\)
\(=\left(x^2+2x-1-2x\right)\left(x^2+x-1\right)\)
\(=\left(x^2-1\right)\left(x^2+x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+x-1\right)\)
AN
18 tháng 10 2018
a)\(x^3-x^2-x+1=\left(x^3-x\right)-\left(x^2-1\right)=x\left(x^2-1\right)-\left(x^2-1\right)=\left(x-1\right)^2.\left(x+1\right)\)
b)\(x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x^2-4\right)\left(x+1\right)=\left(x+2\right)\left(x-2\right)\left(x+1\right)\)
c)\(a^5+27a^2=a^2\left(a^3+27\right)=a^2\left(a+3\right)\left(a^2-3a+9\right)\)
d)\(x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+4\right)\)
e)\(x^4-4x^3+4x^2=x^2\left(x^2-4x+4\right)=x^2\left(x-2\right)^2\)
f)\(2x^4-32=2\left(x^4-16\right)=2\left(x^2+4\right)\left(x^2-4\right)=2\left(x^2+4\right)\left(x+2\right)\left(x-2\right)\)
H
18 tháng 10 2018
a) \(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x^2-1\right)\left(x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-1\right)=\left(x-1\right)^2\left(x+1\right)\)
b) \(x^3+x^2-4x-4\)
\(=x^2\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x^2-4\right)\left(x+1\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x+1\right)\)
c) \(a^5+27a^2=a^2\left(a^3+27\right)\)
\(=a^2\left(a+3\right)\left(a^2-3a+9\right)\)
d) \(x^4-8x=x\left(x^3-8\right)\)
\(=x\left(x-2\right)\left(x^2+2x+4\right)\)
e) \(x^4-4x^3+4x^2\)
\(=\left(x^2\right)^2-2\cdot x^2\cdot2x+\left(2x\right)^2\)
\(=\left(x^2+2x\right)^2\)\(=\left[x\left(x+2\right)\right]^2=x^2\left(x+2\right)^2\)
f) \(2x^4-32=2\left(x^4-16\right)\)
\(=2\left(x^2-4\right)\left(x^2+4\right)\)
\(=2\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)
PT
2
Q
6 tháng 10 2017
a) \(4x^4 + 4x^3 + 5x^2 + 2x + 1\)
\(= 4x^4 + 4x^3 + x^2 + 4x^2 + 2x + 1\)
\(= (2x^2 + x)^2 + (2x + 1)^2\)
\(= x(2x + 1)^2 + (2x + 1)^2\)
\(= (x + 1)(2x + 1)^2\)
27 tháng 10 2018
\(1.\)
\(a.\)
\(x^2-2x=x\left(x-2\right)\)
b.
\(3y^3+6xy^2+3x^2y\)
\(=3y\left(y^2+2xy+x^2\right)\)
\(=3y\left(x+y\right)^2\)
\(c.\)
\(x^2-2xy-xy+2y^2\)
\(=x\left(x-2y\right)-y\left(x-2y\right)\)
\(=\left(x-y\right)\left(x-2y\right)\)
\(2.\)
\(a.\)
\(x^2-y^2+5x-5y\)
\(=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+5\right)\)
\(b.\)
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(c.\)
\(x^2-6xy+9y^2-16\)
\(=\left(x^2-6xy+9y^2\right)-4^2\)
\(=\left(x-3\right)^2-4^2\)
\(=\left(x-3-4\right)\left(x-3+4\right)\)
\(=\left(x-7\right)\left(x+1\right)\)
Tương tự câu \(d,e,g\)
\(3.\)
\(a.\)
\(x^3-2x=0\)
\(\Rightarrow x\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)
\(b.\)
\(x\left(x-4\right)+\left(x-4\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
\(c.\)
\(x\left(x-3\right)+4x-12=0\)
\(\Rightarrow x\left(x-3\right)+3\left(x-3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Tương tự \(d,e,g\)
2 tháng 9 2017
a) \(x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
b) \(x^4+3x^3-4x-4=\left(x^4-4\right)+\left(2x^3-4x\right)=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+2+2x\right)\)
c) \(x^2\left(1-x\right)^2-4x-4x^2=x^2\left(x^2-2x+1\right)-4x-4x^2=x^4-2x^3+x^2-4x-4x^2\)
\(x^4-2x^3-3x^2-4x=x\left(x^3-2x^2-3x-4\right)\)
Có gì sai thì nói giùm nha vì đang vội
2 tháng 9 2017
2)a)x^2-2x-4y^2-4y
=(x^2-4y^2)-(2x+4y)
=(x-2y)(x+2y)-2(x+2y)
=(x+2y)(x-2y-2)
b﴿ x^4 + 3x^3 − 4x − 4
=﴾x^4 − 4﴿ +﴾2x^3 − 4x﴿
=﴾x^2 + 2﴿﴾x^2 − 2﴿ + 2x﴾x^2 − 2﴿
=﴾x^2 -2)(x^2+2+2x)
c)x^2(1 − x)^2 − 4x − 4x^2
= x^2 (x^2 − 2x + 1) − 4x − 4x^2
= x^4 − 2x^3 + x^2 − 4x − 4x^2x^4 − 2x^3 − 3x^2 − 4x
= x (x^3 − 2x^2 − 3x − 4)
đề bài là tìm x à