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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) Sửa đề \(x\left(x+y\right)-3y\left(x+y\right)\)
\(=\left(x+y\right)\left(x-3y\right)\)
b) \(x^2+2019x-xy-2019y\)
\(=x\left(x+2019\right)-y\left(x+2019\right)\)
\(=\left(x+2019\right)\left(x-y\right)\)
c) \(x^2-9y^2-4x+4\)
\(=\left(x^2-4x+4\right)-9y^2\)
\(=\left(x-2\right)^2-\left(3y\right)^2\)
\(=\left(x-2-3y\right)\left(x-2+3y\right)\)
d) \(3x^2-5x+2\)
\(=3x^2-3x-2x+2\)
\(=3x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-2\right)\)
Bài 2:
a) \(\left(6x^3y^3-27xy^2\right):\left(3x^2y\right)-2xy^2\)
\(=6x^3y^3:3x^2y-27xy^2:3x^2y-2xy^2\)
\(=2xy^2-\dfrac{9y}{x}-2xy^2\)
\(=-\dfrac{9y}{x}\)
b) \(\dfrac{2}{x-2}+\dfrac{1-2x}{x+2}+\dfrac{3x+2}{4-x^2}\)
\(=\dfrac{2}{x-2}+\dfrac{1-2x}{x+2}-\dfrac{3x+2}{x^2-4}\)
\(=\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(1-2x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2\left(x+2\right)+\left(1-2x\right)\left(x-2\right)-3x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x+4+x-2-2x^2+4x-3x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x}{x+2}\)
Bài 3:
a) \(3x\left(2x-3\right)-x\left(6x+4\right)=7-12x\)
\(\Rightarrow6x^2-9x-6x^2-4x=7-12x\)
\(\Rightarrow-13x=7-12x\)
\(\Rightarrow-13x+12x-7=0\)
\(\Rightarrow-x-7=0\)
\(\Rightarrow-x=7\)
\(\Rightarrow x=-7\)
b) \(3\left(x-5\right)-2x^2+10x=0\)
\(\Rightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
a) \(N=\left(x-5\right)\left(x+2\right)+3\left(x-2\right)\left(x+2\right)-\left(3x-\dfrac{1}{2}x^2\right)+5x^2\)
\(=x^2+2x-5x-10+3x^2-12-3x+\dfrac{1}{2}x^2+5x^2\)
\(=\dfrac{19}{2}x^2-6x-22\)
Vậy biểu thức trên phụ thuộc vào biến x.
b) \(\left(y-1\right)\left(y^2+y+1\right)=y^3-1\)
Giải:
VT = \(\left(y-1\right)\left(y^2+y+1\right)\)
\(=y^3+y^2+y-y^2-y-1\)
\(=y^3-1\)
Vậy \(\left(y-1\right)\left(y^2+y+1\right)=y^3-1\).
Giải:
a) \(N=\left(x-5\right)\left(x+2\right)+3\left(x-2\right)\left(x+2\right)-\left(3x-\dfrac{1}{2}x^2\right)+5x^2\)
\(\Leftrightarrow N=x^2-3x-10+3\left(x^2-4\right)-3x+\dfrac{1}{2}x^2+5x^2\)
\(\Leftrightarrow N=x^2-3x-10+3x^2-12x-3x+\dfrac{1}{2}x^2+5x^2\)
\(\Leftrightarrow N=-10-18x+\dfrac{19}{2}x^2\)
Vậy biểu thức trên phụ thuộc vào biễn x
b) \(\left(y-1\right)\left(y^2+y+1\right)\)
\(=y^3-y^2+y^2-y+y-1\)
\(=y^3-\left(y^2-y^2\right)-\left(y-y\right)-1\)
\(=y^3-1\)
Vậy ...
\(A=x^2-20x+100=\left(x-10\right)^2\)
Với \(x=10\Rightarrow A=\left(10-10\right)^2=0\)
\(B=4x^2-4xy+y^2=\left(2x-y\right)^2\)
Với \(x=\dfrac{1}{2};y=1\Rightarrow B=\left(2.\dfrac{1}{2}-1\right)^2=0\)
\(C=4x^2-20x+25=\left(2x-5\right)^2\)
Với \(x=\dfrac{5}{2}\Rightarrow\left(2.\dfrac{5}{2}-5\right)^2=0\)
d, ko có x you ạ
D là với y = \(\dfrac{2}{3}\) nha bạn. Mình nhầm đề bài.
$a)$ \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^{12-6}\)
\(=x^6\)
$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!
\(H=\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\ge\dfrac{\left(1+1+1\right)^2}{3+xy+yz+xz}=\dfrac{9}{3+xy+yz+xz}\)
Mặt khác,theo AM-GM: \(xy+yz+xz\le x^2+y^2+z^2=3\)
\(\Rightarrow\dfrac{9}{3+xy+yz+xz}\ge\dfrac{9}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\)
Dấu "=" khi: \(x=y=z=1\)
Bài 1:
\(A=\left(x-y\right)\left(x^2+xy+y^2\right)+2y^3\)
\(A=x^3-y^3+2y^3\)
\(A=x^3+y^3\)
Thay \(x=\dfrac{2}{3},y=\dfrac{1}{3}\) vào A, ta có:
\(A=\left(\dfrac{2}{3}\right)^3+\left(\dfrac{1}{3}\right)^3=\dfrac{8}{27}+\dfrac{1}{27}=\dfrac{9}{27}=\dfrac{1}{3}\)
viết đầu bài rõ ràng 1 chút chả hiểu gì cả
chứng minh biểu thức M có giá trị không phụ thuộc x,y =)) Giúp mk vs ạ