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$a)$ \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^{12-6}\)
\(=x^6\)
$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!
\(e,\)
\(\left(\dfrac{1}{3}a^3b+\dfrac{1}{3}a^2b^2-\dfrac{1}{4}ab^3\right):5ab\)
\(=\dfrac{1}{15}a^2+\dfrac{1}{15}ab-\dfrac{1}{20}b^2\)
\(f,\)
\(\left(-\dfrac{2}{3}x^5y^2+\dfrac{3}{4}x^4y^3-\dfrac{4}{5}x^3y^4\right):6x^2y^2\)
\(=-\dfrac{1}{9}x^3+\dfrac{1}{8}x^2y-\dfrac{2}{15}xy^2\)
\(g,\)
\(\left(\dfrac{3}{4}a^6b^3+\dfrac{6}{5}a^3b^4-\dfrac{5}{10}ab^5\right):\left(\dfrac{3}{5}ab^3\right)\)
\(=\dfrac{5}{4}a^5+2a^2b-\dfrac{5}{6}b^2\)
b) \(\dfrac{3}{4}xy+\dfrac{3}{4}x^2y-\dfrac{3}{4}xy^2\Leftrightarrow\dfrac{3}{4}xy+\dfrac{3}{4}xy\left(x-y\right)\Leftrightarrow\dfrac{3}{4}xy\left(x-y+1\right)\)
c) \(x\left(x-2\right)+y\left(2-x\right)\Leftrightarrow x\left(x-2\right)-y\left(x-2\right)=\left(x-y\right)\left(x-2\right)\)
d) \(x\left(3-2x\right)+6-4x\Leftrightarrow x\left(3-2x\right)+2\left(3-2x\right)\Leftrightarrow\left(x+2\right)\left(3-2x\right)\)
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)\(\left(-xy\right)^{10}:\left(-xy\right)^5=\left(-xy\right)^{10-5}=\left(-xy\right)^5\)
\(a\)) \(-2,5ab\left(-2a^2+3b^2\right)=5a^3b-7,5ab^3\)
b) \(-2x^3\left(3x+0,5x^2-7x^3-2\right)\)
\(=-6x^4-1x^5+14x^6+4x^3\)
c/ \(\left(x^3-2x^2+3x-5\right)\left(-xy\right)\)
\(=-x^4y+2x^3y-3x^2y+5xy\)
d/ \(\left(-\dfrac{1}{2}x^2y\right)\left(3x^3-\dfrac{2}{7}x^2-\dfrac{4}{5}x+8\right)\)
\(=-\dfrac{3}{2}x^5y+\dfrac{1}{7}x^4y+\dfrac{2}{5}x^3y-4x^2y\)
\(-2,5ab\left(-2a^2+3b^2\right)=5a^3b-7,5ab^3\)
\(-2x^3\left(3x+0,5x^2-7x^3-2\right)=-6x^4-x^5+14x^6+4x^3\)
\(\left(x^3-2x^2+3x-5\right)\left(-xy\right)=-x^4y+2x^3y-3x^2y+5xy\)
\(\left(\dfrac{1}{2}x^2y\right)\left(3x^3-\dfrac{2}{7}x^2-\dfrac{4}{5}x+8\right)\)
\(=\dfrac{-3}{2}x^5y+\dfrac{1}{7}x^4y+\dfrac{2}{5}x^3y-4x^2y\)
\(B=\dfrac{1}{x}+\dfrac{1}{y}\\ =\dfrac{x+y}{xy}=\dfrac{5}{6}\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\\ =5^3-3.6.5\\ =125-90\\ =35\)
A = x2 + y2
= (x2 + 2xy + y2) - 2xy
= (x + y)2 - 2xy
= 52 - 2.6
= 25 - 12
= 13
F = x3 + y3
= (x + y)3 - 3xy(x + y)
= 53 - 3.6.5
= 125 - 90
= 35
Bài 1 :
\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
Vậy \(MIN_A=-36\) . Dấu \("="\) xảy ra khi \(x^2+5x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Bài 2 :
a ) \(x+y=5\Rightarrow\left(x+y\right)^2=25\)
\(\Leftrightarrow x^2+2xy+y^2=25\)
\(\Leftrightarrow x^2+y^2=25-2.6=13\)
\(B=x^2-4x+1\)
\(B=x^2-4x+4-3\)
\(B=\left(x-2\right)^2-3\ge-3\)
"="<=>x=2
\(C=\dfrac{-4}{x^2-4x+10}\)
Ta có:\(x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)
\(\Rightarrow\dfrac{-4}{x^2-4x+10}\ge-\dfrac{4}{6}=-\dfrac{2}{3}\)
"="<=>x=2
D\(\ge-\dfrac{8}{3}\)<=>x=0,5(tương tự)
a, \(xy\left(x+y\right)-x^2\left(x+y\right)-y^2\left(x-y\right)\)
\(=x^2y+xy^2-x^3-x^2y-xy^2+y^3\)
\(=y^3-x^3\)
b, \(x^2-x^2\left(5x+1\right)+x\left(x-3\right)\)
\(=x^2-5x^3-x^2+x^2-3x\)
\(=-5x^3+x^2-3x\)
Chúc bạn học tốt!!!
c, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2-5x^2-8x^2\right)+\left(-6x-5x\right)+24\)
\(=-10x^2-11x+24\)
d, \(\dfrac{1}{2}\left(x+4\right)+\dfrac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\dfrac{1}{2}\right)\)
\(=\dfrac{1}{2}x+2+3x^3-\dfrac{3}{2}x^2-x^3-\dfrac{1}{2}x\)
\(=-x^3+\left(3x^2-\dfrac{3}{2}x^2\right)+\left(\dfrac{1}{2}x-\dfrac{1}{2}x\right)+2\)
\(=-x^3+\dfrac{3}{2}x^2+2\)
\(=-\left(x^3-\dfrac{3}{2}x^2-2\right)=-\left(x^3-2x^2+\dfrac{1}{2}x^2-x+x-2\right)\)
\(=-\left[\left(x^3-2x^2\right)+\left(\dfrac{1}{2}x^2-x\right)+\left(x-2\right)\right]\)
\(=-\left[x^2.\left(x-2\right)+\dfrac{1}{2}x.\left(x-2\right)+\left(x-2\right)\right]\)
\(=-\left[\left(x-2\right).\left(x^2+\dfrac{1}{2}x+1\right)\right]\)
Chúc bạn học tốt!!!
\(\dfrac{4}{3}xy\left(\dfrac{3}{4}x^2y-\dfrac{3}{2}xy^2-\dfrac{5}{6}y^3\right)\)
\(=x^3y^2-2x^2y^3-\dfrac{10}{9}xy^4\)
Chúc bạn học tốt!!!