1,Tìm x
a,2x + (1/2+5/3)=(2/3)^3
mn giúp mk vs ạ......mk đg cần gấp ^^
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VM
13 tháng 11 2019
b) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\Rightarrow\frac{3}{5}x=\left(-\frac{1}{7}\right)+\frac{1}{2}\)
\(\Rightarrow\frac{3}{5}x=\frac{5}{14}\)
\(\Rightarrow x=\frac{5}{14}:\frac{3}{5}\)
\(\Rightarrow x=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}.\)
c) \(5-\left|3x-1\right|=3\)
\(\Rightarrow\left|3x-1\right|=5-3\)
\(\Rightarrow\left|3x-1\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3:3\\x=\left(-1\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{1;-\frac{1}{3}\right\}.\)
d) \(\left(1-2x\right)^2=9\)
\(\Rightarrow\left(1-2x\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow1-2x=\pm3.\)
\(\Rightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-2\right):2\\x=4:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{-1;2\right\}.\)
Chúc bạn học tốt!
20 tháng 8 2021
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
23 tháng 9 2021
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
23 tháng 9 2021
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
HL
0
N
1 tháng 10 2018
\(=>\orbr{\begin{cases}\left(2x-3\right)^2=6^2\\\left(2x-3\right)^2=\left(-6\right)^2\end{cases}}\)
\(=>\orbr{\begin{cases}2x-3=6\\2x-3=-6\end{cases}}\)
\(=>\orbr{\begin{cases}2x=9\\2x=-3\end{cases}}\)
\(=>\orbr{\begin{cases}x=\frac{9}{2}\\x=-\frac{3}{2}\end{cases}}\)
1 tháng 10 2018
( 2x - 3 ) ^2 = 36
(2x-3 ) ^2 = ( + - 6 ) ^2
-> 2x-3 = +- 6
* 2x -3 =6 * 2x -3 = -6
2x = 9 2x= -3
x= 9/2 x = -3 /2
vậy x \(\in\)( 9/2 : -3 /2 )
ý b ) tự làm nha bạn , nó còn dễ hơn 1^ 2
HL
1
DD
Đoàn Đức Hà
Giáo viên
1 tháng 8 2021
\(B=-x^2-10y^2+6xy-2x+10y-3\)
\(=-x^2-9y^2-1+6xy-2x+6y-y^2+4y-4+2\)
\(=-\left(x-3y+1\right)^2-\left(y-2\right)^2+2\le2\)
Dấu \(=\)khi \(\hept{\begin{cases}x-3y+1=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=2\end{cases}}\).
2x + 13/6 =8/27
2x = 8/27 - 13/6
2x = - 101/54
x = - 101/54 : 2
x = - 101/108