7x3:2+1
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\(\dfrac{2}{5}+\dfrac{3}{7}\)=\(\dfrac{14}{35}\)+\(\dfrac{15}{35}\)=\(\dfrac{29}{35}\)
\(\dfrac{-6x^4+7x^3+5x+2}{3x+1}\)
\(=\dfrac{-6x^4-2x^3+9x^3+3x^2-3x^2-x+6x+2}{3x+1}\)
\(=\dfrac{-2x^3\left(3x+1\right)+3x^2\left(3x+1\right)-x\left(3x+1\right)+2\left(3x+1\right)}{3x+1}\)
\(=-2x^3+3x^2-x+2\)
\(Sửa:\left(2x^4-7x^3-7x^2-6x-2\right):\left(2x^2+x-1\right)\\ =\left(2x^4+x^3-x^2-8x^3-4x^2+4x-2x^2-x+1-9x-3\right):\left(2x^2+x-1\right)\\ =\left[x^2\left(2x^2+x-1\right)-4x\left(2x^2+x-1\right)-\left(2x^2+x-1\right)-9x-3\right]:\left(2x^2+x-1\right)\\ =x^2-4x-1\left(\text{dư }-9x-3\right)\)
1+2+3+4+5+6 = 21
2-1x9-7x3-5:1+13=(-20)
\(-\frac{23}{7}.\frac{3}{10}+\frac{13}{7}.\frac{3}{10}\)
\(=\frac{3}{10}.\left(\left(-\frac{23}{7}\right)+\frac{13}{7}\right)\)
\(=\frac{3}{10}.\left(\frac{-10}{7}\right)\)
\(=\frac{-3}{7}\)
7x3+7x2+7x3+7
= 7x (3+2+3+7)
= 7x 15
= 105
a: Ta có: \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)^3=-16\)
\(\Leftrightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)
\(\Leftrightarrow-10x^2-10x=0\)
\(\Leftrightarrow-10x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
c: Ta có: \(x^3+3x^2+3x+28=0\)
\(\Leftrightarrow\left(x+1\right)^3=-27\)
\(\Leftrightarrow x+1=-3\)
hay x=-4
7x3:2+1
=11,5 ai tích mìh tích lại
7 x 3 : 2 + 1
= 21 : 2 + 1
=21 : ( 2 + 1 )
=21 : 3
= 7
ai k mik mik k lại gấp đôi