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a) \(5x^2y^4:10x^2y=\dfrac{5x^2y^4}{10x^2y}=\dfrac{5.x^2.y.y^3}{5.2.x^2.y}=\dfrac{y^3}{2}\)
Các câu khác tương tự mà làm
b) \(\dfrac{3}{4}x^3y^3:\left(-\dfrac{1}{2}x^2y^2\right)=\left[\dfrac{3}{4}:\left(-\dfrac{1}{2}\right)\right].\left(x^3:x^2\right).\left(y^3:y^2\right)\)
\(=-\dfrac{3}{2}xy\)
c)\(\left(-xy\right)^{10}:\left(-xy\right)^5=\left(-xy\right)^{10-5}=\left(-xy\right)^5\)
a) 5x - 15y = 5(x - 3y)
b) \(\dfrac{3}{5}\)x2 + 5x4 - x2 - y
= \(\dfrac{3}{5}\)x2 + 5x2.x2 - x2 - y
= x2(\(\dfrac{3}{5}\) + 5x2 -1) - y
c) 14x2y2 - 21xy2 + 28x2y
= 7xy.xy - 7xy.3y + 7xy.4x
= 7xy(xy - 3y + 4x)
= 7xy[(xy - 3y) + 4x]
= 7xy[y(x - 3) +4x]
d) \(\dfrac{2}{7}x\)(3y - 1) - \(\dfrac{2}{7}y\)(3y - 1)
= (3y - 1).(\(\dfrac{2}{7}x\) - \(\dfrac{2}{7}y\) )
= (3y - 1).[\(\dfrac{2}{7}\)(x - y)]
e) x3 - 3x2 + 3x - 1
= x2.x - 3x.x + 3.x - 1
= x(x2-3x+3) - 1
g) 27x3 + \(\dfrac{1}{8}\)
= (3x)3 + \(\left(\dfrac{1}{2}\right)^3\)
= (3x + \(\dfrac{1}{2}\)).(9x2 - \(\dfrac{3}{2}\)x + \(\dfrac{1}{4}\))
h) (x+y)3 - (x-y)3
= 2(3x2y) + 2y3
f) (x+y)2 - 4x2
= -3x2 + y(2x + y)
1.a/(x²+2x+1)(x+1)
=(x+1)(x²+2x+1)
=x(x²+2x+1)+1(x²+2x+1)
=x³+2x²+x+x²+2x+1
=x³+3x²+3x+1
c/(x-5)(x³-2x²+x-1)
=x(x³-2x²+x-1)-5(x³-2x²+x-1)
=x⁴-2x³+x²-1-5x³+10x²-5x+5
=x⁴-7x³+11x²+4-5x
=x⁴-7x³+11x²-5x+4
3.
| Giá trị của x và y | Giá trị của biểu thức(x+y) (x²-Xy+y²) |
| x=-10,y =2 | -1008 |
| x=-1,y=0 | -1 |
| x=2,y=-1 | 7 |
| x=-0,5;y=1,25 | -2,08125 |
4).
(x-5)(3x+3)-3x(x-3)+3x+7
= 3x2+3x-15x-15-3x2+9x+3x+7
=(3x2-3x2)+(3x-15x+9x+3x)-15+7
=0 + 0 -8= -8
Vậy biểu thức được chứng minh
5). Sai đề rồi bn ơi!
a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
3x^2(5x^2-7x+4)
=15x^4-21x^3+12x^2
xy^2(2x^2y-5xy+y)
=2x^3y^3-5x^2y^3+xy^3
(2x^2-5x)(3x^2-2x+1)
=6x^4-4x^3+2x^2-15x^3+10x^2-5x
=6x^4-19x^3+12x^2-5x
(x-3y)(2xy+y^2+x)
=2x^2y+xy^2+x^2-6xy^2-3y^3-3xy
=-3y^3+2x^2y-5xy^2+x^2-3xy
$a)$ \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^{12-6}\)
\(=x^6\)
$b) $ \(\left(-x\right)^7:\left(-x\right)^5\)
\(=\left(-x\right)^{7-5}\)
\(=\left(-x\right)^2\)
\(=x^2\)
$c)$ \(5x^2y^4:10x^2y\)
\(=\dfrac{1}{2}y^3\)
$e)$ \(\left(-xy\right)^{14}:\left(-xy\right)^7\)
\(=\left(-xy\right)^{14-7}\)
\(=\left(-xy\right)^7\)
Các câu còn lại tương tự nha bạn!
Bài 1:
a) \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
\(=\left(x^2-1\right)\left(x+2\right)\)
\(=x^3+x-2\)
b) \(\dfrac{1}{2}x^2y^2\left(2x+y\right)\left(2x-y\right)\)
\(=\dfrac{1}{2}x^2y^2\cdot\left(4x^2-y^2\right)\)
\(=2x^4y^2-\dfrac{1}{2}x^2y^4\)
Bài 2:
a) \(2x\cdot\left(x-5\right)-x\left(2x+3\right)=26\)
\(\Rightarrow2x^2-10x-2x^2-3x=26\)
\(\Rightarrow-13x=26\)
\(\Rightarrow x=2\)
b) \(\left(3y^2-y+1\right)\cdot\left(y-1\right)+y^2\cdot\left(4-3y\right)-\dfrac{5}{2}=0\)
\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3-\dfrac{5}{2}=0\)
\(\Rightarrow2y+\dfrac{7}{5}=0\)
\(\Rightarrow2y=-1,4\)
\(\Rightarrow y=-0,7\)
c) \(2x^2+3\left(x-1\right)\cdot\left(x+1\right)=5x\left(x+1\right)\)
\(\Rightarrow2x^2+3\left(x^2-1\right)=5x^2+5x\)
\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Rightarrow5x^2-5x^2-5x=3\)
\(\Rightarrow-5x=3\)
\(\Rightarrow x=0,6\)
a ) \(\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
b ) \(\left(x^2-2xy+y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x-y\right)=\left(x-y\right)^3\)
c ) \(\left(x^2y^2-\dfrac{1}{3}xy+3y\right)\left(x-3y\right)\)
\(=\left(x^2y^2-\dfrac{1}{3}xy+3y\right)x-3y\left(x^2y^2-\dfrac{1}{3}xy+3y\right)\)
\(=x^3y^2-\dfrac{1}{3}x^2y+3xy-3x^2y^3+xy^2-9y^2\)
d ) \(\left(\dfrac{1}{5}x-1\right)\left(x^2-5x+2\right)\)
\(=\dfrac{1}{5}x\left(x^2-5x+2\right)-x^2+5x-2\)
\(=\dfrac{1}{5}x^3-x^2+\dfrac{2}{5}x-x^2+5x-2\)
\(=\dfrac{1}{5}x^3-2x^2+\dfrac{27}{5}x-2\)
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)