a, 7x+7y
b,2x2y-6xy2
c,3x(x-1)+7x2(x-1)
d, 3x(x-a)+5a(a-x)
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TT
3
IN
27 tháng 6 2019
a, 7x - 7y
= 7.(x - y)
=> vì đặt 7 ra làm nhân tử chung nên ta có 7.(x - y)
PN
28 tháng 7 2020
đề là phân tích đa thức thành nhân tử ?
a,\(7x-7y=7\left(x-y\right)\)
b,\(2x^2y-6xy^2=2xy\left(x-3y\right)\)
c,\(3x\left(x-1\right)+7x^2\left(x-1\right)=\left(x-1\right)\left(3x+7x^2\right)\)
T
13 tháng 9 2023
1.
\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)
\(=2x^3y^2-3x^2y^2+7x^2y\)
\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)
\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)
\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3\)
\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
2.
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3+y^3\)
\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)
\(=24xy+4x-6y-1-24xy-4x\)
\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)
\(=-6y-1\)
#Toru
9 tháng 1
`#3107.101107`
`A(x) = 3x - 9x^2 + 4x + 5x^3 + 7x^2 + 1`
`= (3x + 4x) - (9x^2 - 7x^2) + 5x^3 + 1`
`= 7x - 2x^2 + 5x^3 + 1`
`B(x) = 5x^3 - 3x^2 + 7x + 10`
`A(x) - B(x) = 7x - 2x^2 + 5x^3 + 1 - (5x^3 - 3x^2 + 7x + 10)`
`= 7x - 2x^2 + 5x^3 + 1 - 5x^3 + 3x^2 - 7x - 10`
`= (7x - 7x) + (3x^2 - 2x^2) + (5x^3 - 5x^3) - (10 - 1)`
`= x^2 - 9`
`=> C(x) = x^2 - 9`
`C(x) = 0`
`=> x^2 - 9 = 0`
`=> x^2 = 9 => x^2 = (+-3)^2 => x = +-3`
Vậy, nghiệm của đa thức `C(x)` là `x \in {3; -3}.`
NX
2
20 tháng 8 2019
Câu a bạn haphuong01 làm sai kìa
2xy(x-3y) mới đúng
Hoc24h vẫn cho đúng là sao
Hy vọng GV hoc24h chấm cẩn thận hơn nha
2 tháng 3 2022
A = \(4x^2-3x+7x^2+2x-5\)
\(11x^2-3x+2x-5\)
\(11x^2-x-5\)
B = \(3x+7y-6x-8+y-2\)
\(3x+7y-6x-10+y\)
\(- 3x+7y-10+y\)
\(3x+8y-10\)
C = chịu
D= \(6x^4-3x^2+x^2-4x+3.4-x+2\)
\(6x^4-3x^2+x^2-4x;12-x+2\\ \)
\(6x^4-3x^2+x^2-4x+14-x\)
\(6x^4-2x^2-4x+14-x\)
\(6x^4-2x^2-5x+14\)
KN
5 tháng 3 2019
a. \(\frac{1}{3}x-\frac{2}{5}=\frac{2}{3}x+1\)
\(\Leftrightarrow\frac{1}{3}x-\frac{2}{3}x=1+\frac{2}{5}\)
\(\Leftrightarrow-\frac{1}{3}x=\frac{7}{5}\)
\(\Leftrightarrow x=\frac{7}{5}\div\frac{-1}{3}\)
\(\Leftrightarrow x=\frac{21}{-5}\)
b. \(\frac{-4}{5}x+\frac{1}{3}=\frac{2}{3}x+\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{2}=\frac{2}{3}x+\frac{4}{5}x\)
\(\Leftrightarrow\frac{-1}{6}=\frac{22}{15}x\)
\(\Leftrightarrow x=\frac{-1}{6}\div\frac{22}{15}\)
\(\Leftrightarrow x=\frac{-15}{132}\)
ND
Nguyễn Đức Trí
VIP
27 tháng 8 2023
a) \(\left(x+2y\right)^2-\left(x-y\right)^2=\left(x+2y+x-y\right)\left(x+2y-x+y\right)\)
\(=\left(2x+y\right).3y\)
b) \(\left(x+1\right)^3+\left(x-1\right)^3\)
\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)
\(=2x\left[\left(x+1\right)^2-\left(x^2-1\right)+\left(x-1\right)^2\right]\)
c) \(9x^2-3x+2y-4y^2\)
\(=9x^2-4y^2-3x+2y\)
\(=\left(3x-2y\right)\left(3x+2y\right)-\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left[3x+2y-1\right]\)
d) \(4x^2-4xy+2x-y+y^2\)
\(=4x^2-4xy+y^2+2x-y\)
\(=\left(2x-y\right)^2+2x-y\)
\(=\left(2x-y\right)\left(2x-y+1\right)\)
e) \(x^3+3x^2+3x+1-y^3\)
\(=\left(x+1\right)^3-y^3\)
\(=\left(x+1-y\right)\left[\left(x+1\right)^2+y\left(x+1\right)+y^2\right]\)
g) \(x^3-2x^2y+xy^2-4x\)
\(=x\left(x^2-2xy+y^2\right)-4x\)
\(=x\left(x-y\right)^2-4x\)
\(=x\left[\left(x-y\right)^2-4\right]\)
\(=x\left(x-y+2\right)\left(x-y-2\right)\)
KV
27 tháng 8 2023
a) (x + 2y)² - (x - y)²
= (x + 2y - x + y)(x + 2y + x - y)
= 3y(2x + y)
b) (x + 1)³ + (x - 1)³
= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]
= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)
= 2x(x² + 3)
c) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) x³ + 3x² + 3x + 1 - y³
= (x³ + 3x² + 3x + 1) - y³
= (x + 1)³ - y³
= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]
= (x - y + 1)(x² + 2x + 1 + xy + y + y²)
g) x³ - 2x²y + xy² - 4x
= x(x² - 2xy + y² - 4)
= x[(x² - 2xy + y²) - 4]
= x[(x - y)² - 2²]
= x(x - y - 2)(x - y + 2)
22 tháng 1 2021
a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)
\(\Leftrightarrow2x^2+2-2x^2-2x=0\)
\(\Leftrightarrow-2x+2=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1(nhận)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)
\(\Leftrightarrow6x^2-3x+4x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow6x^2-6x+7x-7=0\)
\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)
d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)
Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)
\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)
\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)
a, 7x+7y
= 7(x+y)
b,2x2y-6xy2
= 2xy(x-3y)
c,3x(x-1)+7x2(x-1)
= (3x + 7x^2) (x-1)
= x(3+7x)(x-1)
d, 3x(x-a)+5a(a-x)
= 3x(x-a) - 5x(x-a)
= (3x - 5x) (x-a)
= (-2x) (x-a)