Làm tính chia :
a) \(\left[5\left(a-b\right)^3+2\left(a-b\right)^2\right]:\left(b-a\right)^2\)
b) \(5\left(x-2y\right)^3:\left(5x-10y\right)\)
c) \(\left(x^3+8y^3\right):\left(x+2y\right)\)
a)\(\left(\dfrac{5}{7}x^2y\right)^3:\left(\dfrac{1}{7}xy\right)^3\)
b) \(\left[5\left(a-b\right)^3+2\left(a-b\right)^2\right]:\left(b-a\right)^2\)
c) \(5\left(x-2y\right)^3:\left(5x-10y\right)\)
d) \(\left(x^3+8y^3\right):\left(x+2y\right)\)
a)\((\dfrac{5}{7}x^2y)^3:(\dfrac{1}{7}xy)^3\)
=\((\dfrac{5}{7}x^2y:\dfrac{1}{7}:x:y)^3\)
=(\(\dfrac{5}{7}.7.x^2:x.y:y)^3\)
=(5x)\(^3\)
=5\(^3\).x\(^3\)
=125.x\(^3\)
Làm tính chia: A)\(5\left(x-2y\right)^3:\left(5x-10y\right)\)
B)\(\left(x^3+8y^3\right):\left(x+2y\right)\)
Làm tính chia:
a) \(5x^2y^4:10x^2y\)
b)\(\dfrac{3}{4}x^3y^3:\left(-\dfrac{1}{2}x^2y^2\right)\)
c)\(\left(-xy\right)^{10}:\left(-xy\right)^5\)
a: \(5x^2y^4:10x^2y=\dfrac{1}{2}y^3\)
c: \(\left(-xy\right)^{10}:\left(-xy\right)^5=-x^5y^5\)
Thực hiện phép tính:
a) \(\dfrac{2}{5}xy\left(x^2y-5x+10y\right)\)
b) \(\left(x^2-1\right)\left(x^2+2x+y\right)\)
c) \(\left(x+3y\right)^2\)
d) \(\left(4x-y\right)^3\)
e) \(\left(x^2-2y\right)\left(x^2+2y\right)\)
g) \(18x^4y^2z:10x^4y\)
h) \(\left(x^3y^3+\dfrac{1}{2}x^2y^3-x^3y^2\right):\dfrac{1}{3}x^2y^2\)
i) \(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\)
k) \(\dfrac{5x-1}{3x^2y}+\dfrac{x+1}{3x^2y}\)
l) \(\dfrac{3x+1}{x^2-3x+1}+\dfrac{x^2-6x}{x^2-3x+1}\)
m) \(\dfrac{2x+3}{10x-4}+\dfrac{5-3x}{4-10x}\)
n) \(\dfrac{x}{x^2+2x+1}+\dfrac{3}{5x^2-5}\)
o) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
p) \(\dfrac{4x+2}{15x^3y}\dfrac{5y-3}{9x^2y}+\dfrac{x+1}{5xy^3}\)
q) \(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)
r) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
x) \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)\)
y) \(\dfrac{x^2-4}{3x+12}.\dfrac{x+4}{2x-4}\)
z) \(\left(x^2-25\right):\dfrac{2x+10}{3x-7}\)
t) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
w) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
c: \(=x^2+6xy+9y^2\)
e: \(=x^4-4y^2\)
Tính
a) (3x+y-z)-(-x-2y+6z)
b)$\left(x^3+6x^2+5y^3\right)-\left(-x^3-5x+7y^3\right)$(x3+6x2+5y3)−(−x3−5x+7y3)
c)$\left(5.7x^2y-3,2xy+8y^3\right)-\left(6,9xy-2,3x^2y-8y^3\right)$(5.7x2y−3,2xy+8y3)−(6,9xy−2,3x2y−8y3)
d)$\left(3x^2y-x^3-2xy^2+5\right)+\left(2x^3-3xy^2-x^2y+xy+6\right)$
Tính
a) (3x+y-z)-(-x-2y+6z)
b)\(\left(x^3+6x^2+5y^3\right)-\left(-x^3-5x+7y^3\right)\)
c)\(\left(5.7x^2y-3,2xy+8y^3\right)-\left(6,9xy-2,3x^2y-8y^3\right)\)
d)\(\left(3x^2y-x^3-2xy^2+5\right)+\left(2x^3-3xy^2-x^2y+xy+6\right)\)
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)
=3(x-y)^2+9
b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)
c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)
1.tìm x
a) \(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)\)
b) \(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)=3\left(x-1\right)\left(x+2\right)\)
2. CMR
a) \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
b)\(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5+y^5\)
c)\(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)
giúp mik nha
chiều nay nộp r
2. CMR:
a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
Ta có: VT=\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5=x^5-y^5=VP\)=> đpcm.
b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5+y^5\)
Ta có: VT=\(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5=VP\)
=> đpcm.
c. \(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)
\(\Leftrightarrow x^2+bx+ax+ab=x^2+ax+bx+ab\) (đúng)
=> đpcm.
1.
b. \(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)=3\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow4\left(x^2+5x-x-5\right)-\left(x^2+5x+2x+10\right)=3\left(x^2+2x-x-2\right)\)
\(\Leftrightarrow4x^2+20x-4x-20-x^2-5x-2x-10=3x^2+6x-3x-6\)
\(\Leftrightarrow4x^2+20x-4x-x^2-5x-2x-3x^2-6x+3x=20+10-6\)
\(\Leftrightarrow6x=24\)
\(\Leftrightarrow x=4\)
Vậy ....
\(a\text{)}.\: \left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow8x-5x^2+16-10x+4x^2-4x-8+2x^2-8=0\\ \Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
1)rút gọn
a) (x+5)(\(x^2\) - 5x + 25) - \(\left(x+3\right)^3\) + (x-2)(\(x^2\) + 2x + 4) - \(\left(x-1\right)^3\)
b)\(\left(x+3y\right)^3+\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)-\left(2y-3x\right)^3\)
c)\(\left(3x+y\right)^3-\left(5x-y\right)\left(25x^2+5xy+y^2\right)+\left(x+2y\right)^3\)
2)tìm x,biết
a)\(\left(x-1\right)^2-\left(x-2\right)\left(x+3\right)+\left(x+2\right)^3=\left(x-3\right)\left(x^2+3x+9\right)+6x\left(x+2\right)\)
Cảm ơn các bạn ^^