Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2:
Ta có: \(3n^3+10n^2-5⋮3n+1\)
\(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)
\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow3n\in\left\{0;-3;3\right\}\)
hay \(n\in\left\{0;-1;1\right\}\)
\(2,\\ a,=2x^2+4x-3x-6-2x^2-4x-2=-3x-8\\ b,=\left[x-2+2\left(x+1\right)\right]^2=\left(x-2+2x+2\right)^2=9x^2\)
\(\left(x^3-8\right):\left(x^2+2x+4\right)\\ =\left(x-2\right)\left(x^2+2x+4\right):\left(x^2+2x+4\right)\\ =x-2\)
bài 2
a)
\(2xy^2-4y\\ =2y\left(xy-2\right)\)
b)
\(x^2-6xy+9y^2\\ =\left(x-3y\right)^2\)
c)
\(x^2+x-y^2+y\\ =\left(x^2-y^2\right)+\left(x+y\right)\\ =\left(x-y\right)\left(x+y\right)+\left(x+y\right)\\ =\left(x+y\right)\left(x-y+1\right)\)
d)
\(x^2+4x+3\\ =x^2+3x+x+3\\ =x\left(x+3\right)+\left(x+3\right)\\ =\left(x+3\right)\left(x+1\right)\)
Bài 10:
e: \(\Leftrightarrow x\left(x+2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Ta có: ΔABC∼ΔDEF
nên DE/AB=EF/BC=DF/AC
=>9/6=EF/10=DF/14
=>EF/10=DF/14=3/2
=>EF=15cm; DF=21cm
a, \(2x=5\Leftrightarrow x=\dfrac{5}{2}\)
b, \(2x-1=4x-8\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\)
c, \(3x+9-6=2x+4\Leftrightarrow x=1\)
d, \(\left[{}\begin{matrix}2x+1=0\\-3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
e, đk : x khác 0 ; 3
\(2x+8x-24=16\Leftrightarrow10x=40\Leftrightarrow x=4\left(tm\right)\)
bẹn ghi ra đc hông
mik ko đọc đc chữ bẹn