Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
$a)$ \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^{12-6}\)
\(=x^6\)
$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!
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) \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)
b) \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2\)
\(=x^4-\dfrac{4}{25}y^2\)
c) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=\left(x-3y\right)\left[x^2+3y.x+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3=x^3-27y^3\)
d) \(\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)
e) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)=\left(x^2-3\right)\left[\left(x^2\right)^2+3.x^2+3^2\right]\)
\(=\left(x^2\right)^3-3^3=x^6-27\)
Bài 1:
a) \(25\left(x+2y\right)^2-16\left(2x-y\right)^2\)
\(=\left[5\left(x+2y\right)\right]^2-\left[4\left(2x-y\right)\right]^2\)
\(=\left[5\left(x+2y\right)-4\left(2x-y\right)\right]\left[5\left(x+2y\right)+4\left(2x-y\right)\right]\)
\(=\left(5x+10y-8x+4y\right)\left(5x+10y+8x-4y\right)\)
\(=\left(14y-3x\right)\left(13x+6y\right)\)
b) \(0,25\left(x-2y\right)^2-4\left(x+y\right)^2\)
\(=\left[\dfrac{1}{2}\left(x-2y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)
\(=\left[\dfrac{1}{2}\left(x-2y\right)-2\left(x+y\right)\right]\left[\dfrac{1}{2}\left(x-2y\right)+2\left(x+y\right)\right]\)
\(=\left(\dfrac{1}{2}x-y-2x-2y\right)\left(\dfrac{1}{2}x-y+2x+2y\right)\)
\(=\left(-\dfrac{3}{2}x-3y\right)\left(\dfrac{5}{2}x+y\right)\)
\(=-3\left(\dfrac{1}{2}x+y\right)\left(\dfrac{5}{2}x+y\right)\)
c) \(\dfrac{4}{9}\left(x-3y\right)^2-0,04\left(x+y\right)^2\)
\(=\left[\dfrac{2}{3}\left(x-3y\right)\right]^2-\left[\dfrac{1}{5}\left(x+y\right)\right]^2\)
\(=\left[\dfrac{2}{3}\left(x-3y\right)-\dfrac{1}{5}\left(x+y\right)\right]\left[\dfrac{2}{3}\left(x-3y\right)+\dfrac{1}{5}\left(x+y\right)\right]\)
\(=\left(\dfrac{2}{3}x-2y-\dfrac{1}{5}x-\dfrac{1}{5}y\right)\left(\dfrac{2}{3}x-2y+\dfrac{1}{5}x+\dfrac{1}{5}y\right)\)
\(=\left(\dfrac{7}{15}x-\dfrac{11}{5}y\right)\left(\dfrac{13}{15}x-\dfrac{9}{5}y\right)\)
\(=\dfrac{1}{5}\left(\dfrac{7}{3}x-11y\right).\dfrac{1}{5}\left(\dfrac{13}{3}x-9y\right)\)
\(=\dfrac{1}{25}\left(\dfrac{7}{3}x-11y\right)\left(\dfrac{13}{3}x-9y\right)\)
d) \(-25x^2+30x-9\)
\(=-\left(25x^2-30x+9\right)\)
\(=-\left[\left(5x\right)^2-2.5x.3+3^2\right]\)
\(=-\left(5x-3\right)^2\)
Bài 2:
a) \(x^3y^2-x^2y^3-2x+2y\)
\(=x^2y^2\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2y^2-2\right)\)
Thay x = -1 và y = -2 vào ta được
\(=\left[-1-\left(-2\right)\right]\left[\left(-1\right)^2\left(-2\right)^2-2\right]\)
\(=1\left(4-2\right)\)
\(=2\)
b) \(5x^2-3x+3y-5y^2\)
\(=5\left(x^2-y^2\right)-3\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
Thay x = 3 và y = 1 vào ta được
\(=5\left(3-1\right)\left(3+1\right)-3\left(3-1\right)\)
\(=5.2.4-3.2\)
\(=34\)
\(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)
bài 2 áp dụng hằng đẳng thức bạn nhé
bài 3\(A=\left(x^3+3x^2+3x+1\right)+5\)
\(=\left(x+1\right)^3+5\) thay x=19 vào ta được
\(A=20^3+5=8005\)
\(B=\left(x^3-3x^2+3x-1\right)+1\)
\(=\left(x-1\right)^3+1\)
thay x=11 vào ta được
\(B=\left(11-1\right)^3+1=10^3+1=1001\)
a: \(=\dfrac{5}{3}x^2-x+\dfrac{1}{3}\)
b: \(=-5y-9+xy\)