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a, x=-505
b, x=35/8 hoac -37/8
nhung cau con lai thi tong tu
|\(x-\dfrac{1}{2}\)| + 2\(x\) = 6
|\(x-\dfrac{1}{2}\)| = 6 - 2\(x\); 6 - 2\(x\) > 0 ⇒ 6 > 2\(x\) ⇒ \(x\) < 3
\(\left[{}\begin{matrix}x-\dfrac{1}{2}=6-2x\\x-\dfrac{1}{2}=-6+2x\end{matrix}\right.\)
\(\left[{}\begin{matrix}x+2x=6+\dfrac{1}{2}\\2x-x=6-\dfrac{1}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=\dfrac{13}{2}\\x=\dfrac{11}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=\dfrac{11}{2}\end{matrix}\right.\)
\(x=\dfrac{11}{2}\) > 3 (loại)
Vậy \(x\) = \(\dfrac{13}{6}\)
a) \(\left|3x-4\right|+\left|3y+5\right|=0\)
\(\Rightarrow\hept{\begin{cases}3x-4=0\\3y+5=0\end{cases}\Rightarrow\hept{\begin{cases}3x=4\\3y=-5\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{4}{3}\\y=\frac{-5}{3}\end{cases}}\)
b) \(\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\)
\(\Rightarrow\hept{\begin{cases}x-y=0\\y+\frac{9}{25}=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\y=\frac{-9}{25}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{-9}{25}\\y=\frac{-9}{25}\end{cases}}}\)
c) \(\left|3-2x\right|+\left|4y+5\right|=0\)
\(\Rightarrow\hept{\begin{cases}3-2x=0\\4y+5=0\end{cases}\Rightarrow\hept{\begin{cases}2x=3\\4y=-5\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{-5}{4}\end{cases}}}\)
d) \(\left|5-\frac{3}{4}x\right|+\left|\frac{2}{7}y-3\right|=0\)
\(\Rightarrow\hept{\begin{cases}5-\frac{3}{4}x=0\\\frac{2}{7}y-3=0\end{cases}\Rightarrow\hept{\begin{cases}\frac{3}{4}x=5\\\frac{2}{7}y=3\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{20}{3}\\y=\frac{21}{2}\end{cases}}\)
e) \(\left(x-1\right)^2+\left(y+3\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=-3\end{cases}}}\)
a: 2x-5=0
=>2x=5
hay x=5/2
b: =>x(3x+1)=0
=>x=0 hoặc x=-1/3
c: =>(x+4+3)(x+4-3)=0
=>(x+7)(x+1)=0
=>x=-7 hoặcx=-1
a) \(x^2=\frac{4}{9}\)
\(\Rightarrow x=\frac{2}{3}\)
b)\(x=0,6\)
a) \(x=\frac{2}{3};x=-\frac{2}{3}\)
b) \(x=0,6;x=-0,6\)
c) \(x=0,5;x=-0,5\)
d)\(x=-1\)
a) \(\left(x-4\right)\left(x^2+1\right)=0\)
\(\Rightarrow\) Có 2 trường hợp:
1) x - 4 = 0 \(\Rightarrow\)x = 4
2) \(x^2+1=0\Rightarrow x^2=-1\) .Mà \(x^2\ge0\forall x\Rightarrow x\in\varnothing\)
Vậy x =4
b) \(3.x^2-4x=0\)
\(\Rightarrow x\left(3x-4\right)=0\Rightarrow\) Có 2 trường hợp:
1) x = 0
2) 3x - 4 = 0 \(\Rightarrow\) 3x = 4 \(\Rightarrow x=\frac{4}{3}\)
Vậy \(x\in\left\{0;\frac{4}{3}\right\}\)
c) \(x^2+9=0\)
\(\Rightarrow x^2=-9\) . Mà \(x^2\ge0\forall x\Rightarrow x\in\varnothing\)
Vậy \(x\varnothing\in\)
Đề yêu cầu gì thế?!