Dùng tính chất cơ bảm của phân thức, hãy điền một đa thức thích hợp vào các chỗ trống trong mỗi đẳng thức sau :
a) x2+82x−1=3x3+25x..........x2+82x−1=3x3+25x..........
c) −x2+2xy−y2x+y=.........y2−x2
Bài 1: Phân tích đa thức thành nhân tử
a. 5x - 20 y
b. 5.(x-1) - 3x.(x-1)
c. x.(x+1) -5x - 5y
d. ( x+y)^2-(x-y)^2
e. (3x+1)^2- (x+1)^2
Bài 2 . Tìm x biết
a. x + 5x^2=0
b. x+1=( x+1)^2
c. x^3 + x= 0
d. x^3 -0.25x = 0
e. x^2 - 10x = -25
Lần sau ghi tách ra tí bạn ơi ;v
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1. a) \(5x-20y=5\left(x-4y\right)\)
b) \(5\left(x-1\right)-3x\left(x-1\right)=\left(x-1\right)\left(5-3x\right)\)
c) \(x\left(x+1\right)-5x-5=x\left(x+1\right)-5\left(x+1\right)\)
\(=\left(x+1\right)\left(x-5\right)\)
d) \(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=4xy\)
e) \(\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)
\(=2x\left(4x+2\right)\)
2. a) \(x+5x^2=0\)
\(\Leftrightarrow x\left(1+5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1+5x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)
Vậy...
b) \(x+1=\left(x+1\right)^2\)
\(\Leftrightarrow x+1-x^2-2x-1=0\)
\(\Leftrightarrow-x^2-x=0\)
\(\Leftrightarrow-x\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-x=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy...
c) \(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
Vì \(x^2+1>0\Rightarrow x=0\)
Vậy...
d) \(x^3-0,25x=0\)
\(\Leftrightarrow x\left(x^2-0,25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)
Vậy..
e) \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
Vậy...
bài 1 phân tích đa thức thành phần tử
a) 5(x+4)-2x(4+x)
b) (x-2017)x-5(2017-x)
c) (x+1)^2-(x+1)
d) 9x^2(y-1)-18x(1-y)
e) 100x^2y-25xy^2-5xy
f) (n+1).n-(n+1).3
làm hết đó nha thank mấy bạn làm
a) \(5\left(x+4\right)-2x\left(4+x\right)\)
\(=\left(x+4\right)\left(5-2x\right)\)
b) \(\left(x-2017\right)x-5\left(2017-x\right)\)
\(=\left(x-2017\right)x+5\left(x-2017\right)\)
\(=\left(x-2017\right)\left(x+5\right)\)
c) \(\left(x+1\right)^2-\left(x+1\right)\)
\(=\left(x+1\right)\left(x+1-1\right)\)
= \(x\left(x+1\right)\)
d) \(9x^2\left(y-1\right)-18x\left(1-y\right)\)
\(=9x^2\left(y-1\right)+18x\left(y-1\right)\)
\(=\left(y-1\right)\left(9x^2+18x\right)\)
\(=9x\left(y-1\right)\left(x+2\right)\)
e) \(100x^2y-25xy^2-5xy\)
\(=5xy\left(20x-5y-1\right)\)
f) \(\left(n+1\right)n-\left(n+1\right)3\)
\(=\left(n+1\right)\left(n-3\right)\)
tìm x
a) 5(x+3)-2x(3+X)=0
b) 4x(x-2017)-x+2017=0
c) (x+1)^2=x^2+1
làm hết nha thank
a) 5(x+3)-2x(3+x)=0
(x+3)(5-2x)=0
\(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
b) 4x(x-2017)-x+2017=0
4x(x-2017)-(x-2017)=0
(x-2017)(4x-1)=0
\(\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
c) (x+1)2 = x2 + 1
x2+2x+1-x2-1=0
2x=0
Pt có vô số nghiệm
a) \(5\left(x+3\right)-2x\left(3+x\right)=0\) (1)
\(\Leftrightarrow5\left(x+3\right)-2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{-3;\dfrac{5}{2}\right\}\)
b) \(4x\left(x-2017\right)-x+2017=0\)
cách làm hơi khó, cho đáp án thôi nhé: \(x=2017;x=\dfrac{1}{4}\)
c) \(\left(x+1\right)^2=x^2+1\) (3)
\(\Leftrightarrow x^2+2x+1=x^2+1\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{0\right\}\)
tính giá trị biểu thức
A=X(Y-Z)+2(Z-Y)VỚI X=2; Y=1,007, Z=-0,006
B=2X(Y-Z)+(Z-Y)(X+M) VỚI X=18,3 ; Y =24,6; Z= 10,6 ; M = -31,7
a)\(A=x\left(y-z\right)+2\left(z-y\right)\)
\(=2\left(z-y\right)-x\left(z-y\right)\)
\(=\left(2-x\right)\left(z-y\right)\) với \(x=2;y=1,007;z=-0,006\) thì
\(A=\left(2-2\right)\left(-0,006-1,007\right)=0\)
b)\(B=2x\left(y-z\right)+\left(z-y\right)\left(x+m\right)\)
\(=\left(z-y\right)\left(x+m\right)-2x\left(z-y\right)\)
\(=\left(z-y\right)\left(x+m-2x\right)\)
\(=\left(z-y\right)\left(m-x\right)\) với \(x=18,3;y=24,6;z=10,6;m=-31,7\) thì
\(B=\left(10,6-24,6\right)\left(-31,7-18,3\right)=700\)
1 tính
a) (x+1)(x^2-x+1)-(x-1)(x^2+x+1)
b) x(x-4)(x+4)-(x^2+1)(x^2-1)
a,
\(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3+1-x^3+1=2\)
b, \(x\left(x-4\right)\left(x+4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^3-16x-x^4+1\)
a) \(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
= \(x^3-x^2+x+x^2-x+1-\left(x^3+x^2+x-x^2-x-1\right)\)
= \(x^3-x^2+x+x^2-x+1-x^3-x^2-x+x^2+x+1\)
= \(2\)
b) \(x\left(x-4\right)\left(x+4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
= \(x\left(x^2-16\right)-\left(x^4-1\right)\) = \(x^3-16x-x^4+1\)
= \(-x^4+x^3-16x+1\)
a,
(x+1)(x2−x+1)−(x−1)(x2+x+1)(x+1)(x2−x+1)−(x−1)(x2+x+1)
=x3+1−x3+1=2=x3+1−x3+1=2
b, x(x−4)(x+4)−(x2+1)(x2−1)x(x−4)(x+4)−(x2+1)(x2−1)
=x(x2−16)−(x4−1)=x(x2−16)−(x4−1)
=x3−16x−x4+1
tính
a) (x-3)(x+3)-(x+1)^2
b) (4x-3)(4x+3)-16x^2
c)(x+4)(x^2-4x+16)-x^3
làm hết hộ
a) \(\left(x-3\right)\left(x+3\right)-\left(x+1\right)^2\) = \(x^2-9-\left(x^2+2x+1\right)\)
\(x^2-9-x^2-2x-1\) = \(-2x-10\)
b) \(\left(4x-3\right)\left(4x+3\right)-16x^2\) = \(16x^2-9-16x^2=-9\)
c) \(\left(x+4\right)\left(x^2-4x+16\right)-x^3\) = \(x^3-4x^2+16x+4x^2-16x+64-x^3\)
= \(64\)
\(a,\left(x-3\right)\left(x+3\right)-\left(x+1\right)^2=x^2-9-x^2-2x-1=-10-2x\) \(b,\left(4x-3\right)\left(4x+3\right)-16x^2=16x^2-9-16x^2=-9\)\(c,\left(x+4\right)\left(x^2-4x+16\right)-x^3=x^3+64-x^3=64\)
Phân tích đa thức thành nhân tử:
1). (x - 3)(x - 1) - 3(x - 3)
2). (6x + 3) - (2x - 5)(2x + 1)
3). (x - 1)(2x + 1) + 3(x - 1)(x + 2)(2x + 1)
4). (3x - 2)(4x - 3) - (2 - 3x)(x - 1) - 2(3x - 2)(x + 1)
5). \(\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)-\left(5-x\right)\left(2x+1\right)\)
6). (a - b)(a + 2ab) - (b - a)(2a - b) - (a - b)(a + 3b)
\(1,\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)
\(=\left(x-3\right)\left(x-1-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
\(2,6x+3-\left(2x-5\right)\left(2x+1\right)\)
\(=3\left(2x+1\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(-2-2x\right)\)
\(3,\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)\(=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)
\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)
\(4,\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(5,\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)-\left(5-x\right)\left(2x+1\right)\)\(=\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)\)\(=\left(x-5\right)\left(x-5+x+5+2x+1\right)\)
\(=\left(x-5\right)\left(4x+1\right)\)
6, Tương tự
Phân tích đa thức thành nhân tử:
\(x^3+6x^2+11x+6\)
\(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3x+2\right)\left(x+3\right)\)
Phân tích đa thức sau thành nhân tử:
1)14x^2-21xy^2+28x^2y^2 2)8x^3+4x^2-y^3-y^2 Giúp mk nha!!!
a, \(14x^2-21x^2+28x^2y^2\)
\(=-7x^2+28x^2y^2=-7x^2\left(1-4y^2\right)\)
\(=-7x\left(1-2y\right)\left(1+2y\right)\)
b, \(8x^3+4x^2-y^3-y^2\)
\(=\left(8x^3-y^3\right)+\left(4x^2-y^2\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(2x-y\right)\left(2x+y\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2+2x+y\right)\)
