\(\left(x^2-x+3\right)\left(-2x^2+3x+5\right)\)

\(=-2x^4+3x^3+5x^2+2x^3-3x^2-5x-6x^2+9x+15\)

\(=-2x^4+5x^3-4x^2+4x+15\)

\(x^3+x=0\)

\(\Leftrightarrow x\left(x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x=0\)

Câu hỏi của nguyen khanh li - Toán lớp 6 - Học toán với OnlineMath

\(\left(x-\dfrac{1}{2}y\right)\left(x-\dfrac{1}{2}y\right)\)

\(=\left(x-\dfrac{1}{2}y\right)^2\)

\(=x^2-xy+\dfrac{1}{4}y^2\)

cái dấu cạnh \(\dfrac{1}{\sqrt{x}+1}\) với \(\dfrac{2}{x-1}\) là dấu j

a) \(\left(3x^{n+1}-y^{n-1}\right)-3\left(x^{n+1}+5y^{n-1}\right)-4\left(x^{n+1}+2y^{n-1}\right)\)

\(=3x^{n+1}-y^{n-1}-3x^{n+1}-15y^{n-1}+4x^{n+1}+8y^{n-1}\)

\(=-8y^{n-1}+4x^{n+1}\)

b) \(\left(\dfrac{3}{4}x^{n+1}-\dfrac{1}{2}y^n\right)\cdot2xy-\left(\dfrac{2}{3}x^{n+1}-\dfrac{5}{6}y^n\right)\cdot7xy\)

\(=\dfrac{3}{2}x^{n+2}y-xy^{n+1}+\left(-\dfrac{2}{3}x^{n+1}-\dfrac{5}{6}y^n\right)\cdot7xy\)

\(=\dfrac{3}{2}x^{n+2}y-xy^{n+1}-\dfrac{14}{3}x^{n+2}y+\dfrac{35}{6}xy^{n+1}\)

\(=-\dfrac{19}{6}x^{n+2}y+\dfrac{29}{6}xy^{n+1}\)

24,8       5,6      9,57       0,01

tick nha

Ta có 3 trường hợp:

+) Nếu \(1\le n\le2016\) thì ta có:

\(S_{\left(n\right)}=n^2-2017n+10< n^2-2017n+2016\)

\(=\left(n-1\right)\left(n-2016\right)\ge0\) (loại)

+) Nếu \(n=2017\) thì ta có:

\(S_{\left(n\right)}=S_{\left(2017\right)}=10=n^2-2017n+10\) (nhận)

+) Nếu \(n>2017\) thì ta có:

\(S_{\left(n\right)}=n^2-2017n+10>n\left(n-2017\right)>n\) (loại)

Vậy \(n=2017\)