Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
`a, M(x)+N(x)=(3x^2+5x-x^3+4)+(x^3-5+4x^2+6x)`
`M(x)+N(x)= 3x^2+5x-x^3+4+x^3-5+4x^2+6x`
`M(x)+N(x)= (3x^2+4x^2)+(5x+6x)-(x^3-x^3)+(4-5)`
`M(x)+N(x)= 7x^2+11x-1`
`b, M(x)-N(x)=(3x^2+5x-x^3+4)-(x^3-5+4x^2+6x)`
`M(x)-N(x)= 3x^2+5x-x^3+4-x^3+5-4x^2-6x`
`M(x)-N(x)=(-x^3-x^3)+(3x^2-4x^2)+(5x-6x)-(x^3+x^3)+(4+5)`
`M(x)-N(x)= -2x^3-x^2-x+9`
Lời giải:
a.
$M(x)+N(x)=(3x^2+5x-x^3+4)+(x^3-5+4x^2+6x)$
$=3x^2+5x-x^3+4+x^3-5+4x^2+6x$
$=(-x^3+x^3)+(3x^2+4x^2)+(5x+6x)+(4-5)$
$=7x^2+11x-1$
b.
$M(x)-N(x)=(3x^2+5x-x^3+4)-(x^3-5+4x^2+6x)$
$=3x^2+5x-x^3+4-x^3+5-4x^2-6x$
$=(-x^3-x^3)+(3x^2-4x^2)+(5x-6x)+(4+5)$
$=-2x^3-x^2-x+9$
Câu 5:
\(\dfrac{x}{y}=a\Rightarrow\dfrac{x}{a}=\dfrac{y}{1}=\dfrac{x-y}{a-1}=\dfrac{x+y}{a+1}\)
\(\Rightarrow\dfrac{x+y}{x-y}=\dfrac{a+1}{a-1}\)
Câu 6:
\(9x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}\)
\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{3x}{15}=\dfrac{2y}{18}=\dfrac{3x-2y}{15-18}=\dfrac{12}{-3}=-4\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-4\right).5=-20\\y=\left(-4\right).9=-36\end{matrix}\right.\)
Câu 7:
\(\dfrac{x}{-5}=\dfrac{y}{7}=\dfrac{x+y}{-5+7}=\dfrac{-10}{2}=-5\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-5\right).\left(-5\right)=25\\y=\left(-5\right).7=-35\end{matrix}\right.\)
\(a,=\dfrac{3}{4}-\dfrac{7}{2}-5=-\dfrac{31}{4}\\ b,=-\dfrac{1}{15}+\dfrac{4}{9}\cdot\dfrac{3}{8}-\dfrac{5}{6}=-\dfrac{9}{10}+\dfrac{1}{6}=-\dfrac{11}{15}\\ c,=\dfrac{1}{12}-\dfrac{4}{15}\cdot\dfrac{5}{6}+\left(-\dfrac{2}{3}\right)^3=\dfrac{1}{12}-\dfrac{2}{9}-\dfrac{8}{27}=-\dfrac{47}{108}\\ d,=\left[2\left(-\dfrac{1}{2}\right)\right]^5-\left[3\cdot\left(-\dfrac{1}{3}\right)\right]^3+\dfrac{2}{3}:\left(\dfrac{5}{3}-\dfrac{13}{6}\right)=-1-\left(-1\right)+\dfrac{2}{3}:\left(-\dfrac{1}{2}\right)=-\dfrac{4}{3}\)
a: \(=\dfrac{5}{6}\cdot10=\dfrac{50}{6}=\dfrac{25}{3}\)
g: \(\Leftrightarrow\left[{}\begin{matrix}x-3=-6\\x-3=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=9\end{matrix}\right.\)
Ta có: \(\left|2x-4\right|\ge0\forall x\)
\(\left|3y+9\right|\ge0\forall y\)
\(\Rightarrow C\le-15-0-0=-15\)
Dấu '=' xảy ra <=> \(\hept{\begin{cases}2x-4=0\\3y+9=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)
Cái dấu này là gì ạ?