2x2+3-1:3=?
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C
1
8 tháng 1 2022
a: \(=6x^3-10x^2+6x\)
b: \(=-2x^4-10x^3+6x^2\)
c: \(=-x^5+2x^3-\dfrac{3}{2}x^2\)
d: \(=2x^3+10x^2-8x-x^2-5x+4=2x^3+9x^2-13x+4\)
QN
15 tháng 3 2021
\(\left(2x^2+1\right)\left(4x-3\right)=\left(2x^2+1\right)\left(x-13\right)\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+13\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(3x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+1=0\\3x+10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-\dfrac{1}{2}\left(VN\right)\\x=-\dfrac{10}{3}\end{matrix}\right.\)
\(S=\left\{-\dfrac{10}{3}\right\}\)
19 tháng 5 2022
\(\left(2x^2+1\right)\left(4x-3\right)=\left(2x^2+1\right)\left(x-12\right)\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3\right)-\left(2x^2+1\right)\left(x-12\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+12\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\)
\(\Leftrightarrow3x+9=0\) (do \(2x^2+1>0\forall x\in R\))
\(\Leftrightarrow x=-3\)
-Vậy \(S=\left\{-3\right\}\)
5 tháng 6 2018
a) \(M=\frac{2\times2}{1\times5}+\frac{2\times2}{5\times9}+\frac{2\times2}{9\times13}+...+\frac{2\times2}{45\times40}\)
\(M=\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{45\times49}\)
\(M=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}-\frac{1}{49}\)
\(M=1-\frac{1}{49}\)
\(M=\frac{48}{49}\)
b) \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+5+...+10}\)
= \(\frac{2}{2\times\left(1+2\right)}+\frac{2}{2\times\left(1+2+3\right)}+...+\frac{2}{2\times\left(1+2+3+...+10\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{110}\)
\(=\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{10\times11}\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=2\times\frac{9}{22}\)
\(=\frac{9}{11}\)
5 tháng 6 2018
Mình trả lời câu a nha M= 4/1*5+4/5*9+4/9*13+...+4/45*49 M=1-1/5+1/5-1/9+1/9-1/13+...+1/45-1/49 M=1-1/49=48/49
TT
12 tháng 2 2023
Đặt \(t=2x^2-3x-1\)
\(\Rightarrow t^2-3\left(t-4\right)-16=0\)
\(\Rightarrow t^2-3t+12-16=0\)
\(\Rightarrow t^2-3t-4=0\)
\(\Rightarrow\left\{{}\begin{matrix}t_1=-1\\t_2=4\end{matrix}\right.\)
\(TH_1:t=-1\)
\(\Leftrightarrow2x^2-3x-1=-1\)
\(\Leftrightarrow2x^2-3x=0\)
\(\Leftrightarrow x\left(2x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(TH_2:t=4\)
\(\Leftrightarrow2x^2-3x-1=4\)
\(\Leftrightarrow2x^2-3x-5=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=-1\\x_2=\dfrac{5}{2}\end{matrix}\right.\)
MH
1 tháng 11 2021
1.
a) \(2x^4-4x^3+2x^2\)
\(=2x^2\left(x^2-2x+1\right)\)
\(=2x^2\left(x-1\right)^2\)
b) \(2x^2-2xy+5x-5y\)
\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)
\(=2x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(2x+5\right)\)
KJ
1 tháng 11 2021
2 .
a,
\(4x\left(x-3\right)-x+3=0\)
⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)
⇒\(\left(x-3\right)\left(4x-1\right)=0\)
⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)
vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)
b,
\(\)\(\left(2x-3\right)^2-\left(x+1\right)^2=0\)
⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0
⇒\(\left(x-4\right)\left(3x-2\right)=0\)
⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)
vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)
có ngoặc không nhỉ
KQ là 6,(6)7 nhé
có ngoặc thì KQ là 2