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6) \(\left(2x+\dfrac{1}{2}\right)^3=8x^3+4x^2+\dfrac{3}{2}x+\dfrac{1}{8}\)
7) \(\left(x-3\right)^3=x^3-9x^2+27x-27\)
\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)
\(\left(x-4\right)^2=\left(2x+1\right)^2\)
\(\Leftrightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\3\left(x-1\right)=0\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}}\)
11)11) 3x(x-5)2-(x+2)3+2(x-1)3-(2x+1)(4x2-2x+1)=3x(x2-10x+25)-(x3+6x2+12x+8)+2(x3-3x2+3x-1)-(8x3+1)=3x3-30x2+75x-x3-6x2-12x-8+2x3-6x2+6x-2-8x3-1=-4x3-42x2+63x-11
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 \(n\in\left\{0;-1;1\right\}\)
Bài 2:
a: =>(x+5)(4-x)=0
=>x=4 hoặc x=-5
b: =>2x(2x-1)=0
=>x=0 hoặc x=1/2
c: =>2x(x^2+1)+x^2+1=0
=>(x^2+1)(2x+1)=0
=>2x+1=0
=>x=-1/2
d: Δ=(-3)^2-4*1*4=9-16=-7<0
=>PTVN
3x3 - 5x2 + 5x - 2
= 3x3 - 2x2 - 3x2 +2x + 3x-2
= x2(3x -2) - x(3x -2) + (3x-2)
= (3x-2)(x2 -x +1)
đề bài là j ạ?