-2/3.15/-4-(-1)/5:1/-5
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H
1
11 tháng 3 2022
\(M=\dfrac{5^4\cdot50}{5^3\cdot15}=\dfrac{50}{3}>\dfrac{50}{4}=N\)
NY
1
16 tháng 4 2019
\(4n+1⋮2n-1\)
\(\Leftrightarrow2\left(2n-1\right)+3⋮2n-1\)
\(\Leftrightarrow3⋮2n-1\)
\(\Leftrightarrow2n-1\in\left\{1;3;-1;-3\right\}\)
\(\Leftrightarrow2n\in\left\{2;4;0;-2\right\}\)
\(\Leftrightarrow n\in\left\{1;2;0;-1\right\}\)
24 tháng 10 2021
Bài 2:
a: \(\left(6x-39\right):7=3\)
\(\Leftrightarrow6x-39=21\)
hay x=10
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2019
1/
\(\Leftrightarrow12.3^x+3.15^x-5.5^x-20=0\)
\(\Leftrightarrow3.3^x\left(4+5^x\right)-5\left(5^x+4\right)=0\)
\(\Leftrightarrow\left(4+5^x\right)\left(3^{x+1}-5\right)=0\)
\(\Rightarrow3^{x+1}=5\Rightarrow x+1=log_53\Rightarrow x=log_5\frac{3}{5}\)
2/ \(\Leftrightarrow2^{2x^2+2x}-2^{x^2+2x+1}+2^{1-x^2}-1=0\)
\(\Leftrightarrow2^{2x^2+2x}\left(1-2^{1-x^2}\right)-\left(1-2^{1-x^2}\right)=0\)
\(\Leftrightarrow\left(1-2^{1-x^2}\right)\left(2^{2x^2+2x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2^{1-x^2}=1\\2^{2x^2+2x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}1-x^2=0\\2x^2+2x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
3/ \(\Leftrightarrow6^x-3^x-\left(2^x-1\right)=0\)
\(\Leftrightarrow3^x\left(2^x-1\right)-\left(2^x-1\right)=0\)
\(\Leftrightarrow\left(3^x-1\right)\left(2^x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3^x=1\\2^x=1\end{matrix}\right.\) \(\Rightarrow x=0\)
LQ
4
16 tháng 2 2022
\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)
\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)
\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)
\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)
\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
DT
16 tháng 2 2022
3.15:
a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)
b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3.16
\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)
\(\Leftrightarrow-14m+35-2m^2+8=0\)
\(\Leftrightarrow-14m-2m^2+43=0\)
\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)
\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)
\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)
\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)
pt vô nghiệm
\(\dfrac{-2}{3}.\dfrac{15}{-4}-\dfrac{-1}{5}:\dfrac{1}{-5}\)
\(\dfrac{5}{2}-1=\dfrac{3}{2}\)
5/2 - 1
= 3/2