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c) \(5x-7=3x+9\)
d) \(5x-\left|9-7x\right|=3\)
e) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
h) \(5^{-1}.25^x=125\)
\(\Rightarrow\frac{1}{5}.25^x=125\)
\(\Rightarrow25^x=125:\frac{1}{5}\)
\(\Rightarrow25^x=625\)
\(\Rightarrow25^x=25^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
g) \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.\left[1-\left(x-1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=1+1\\x=\left(-1\right)+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;0\right\}.\)
i) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\) \(\forall x.\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0.\)
Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)
\(\Rightarrow x+1+x+2+x+3=4x\)
\(\Rightarrow\left(x+x+x\right)+\left(1+2+3\right)=4x\)
\(\Rightarrow3x+6=4x\)
\(\Rightarrow6=4x-3x\)
\(\Rightarrow6=1x\)
\(\Rightarrow x=6\left(TM\right).\)
Vậy \(x=6.\)
Chúc bạn học tốt!
a) \(5x-7=3x+9\)
\(\Rightarrow5x-3x=9+7\)
\(\Rightarrow2x=16\)
\(\Rightarrow x=16:2\)
\(\Rightarrow x=8\)
Vậy \(x=8.\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{2}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{2}{5}-\frac{1}{2}\\x=\left(-\frac{2}{5}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}.\)
c) \(5x-\left|9-7x\right|=3\)
\(\Rightarrow\left|9-7x\right|=5x-3\)
\(\Rightarrow\left[{}\begin{matrix}9-7x=5x-3\\9-7x=3-5x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}9+3=5x+7x\\9-3=-5x+7x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}12=12x\\6=2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=12:12\\x=6:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{1;3\right\}.\)
d) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Rightarrow-5+\left|3x-1\right|+6=4\)
\(\Rightarrow-5+\left|3x-1\right|=4-6\)
\(\Rightarrow-5+\left|3x-1\right|=-2\)
\(\Rightarrow\left|3x-1\right|=\left(-2\right)+5\)
\(\Rightarrow\left|3x-1\right|=3.\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=3\\3x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=4\\3x=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4:3\\x=\left(-2\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{4}{3};-\frac{2}{3}\right\}.\)
Chúc bạn học tốt!
a) \(5^{-1}.25^x=125\)
\(\Rightarrow5^{-1}.5^{2x}=5^3\)
\(\Rightarrow5^{2x-1}=5^3\)
\(\Rightarrow2x-1=3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) \(|x+1|+|x+2|+|x+3|=4x\)
Vì \(\hept{\begin{cases}|x+1|\ge0\forall x\\|x+2|\ge0\forall x\\|x+3|\ge0\forall x\end{cases}}\)
\(\Rightarrow|x+1|+|x+2|+|x+3|\ge0\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\hept{\begin{cases}x+1>0\\x+2>0\\x+3>0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}|x+1|=x+1\\|x+2|=x+2\\|x+3|=x+3\end{cases}}\)
\(\Rightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)
\(\Rightarrow3x+6=4x\)
\(\Rightarrow x=6\)
Vậy \(x=6\)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
Tìm X
a) \(2x+\dfrac{3}{24}=3x-\dfrac{1}{32}\)
\(\Leftrightarrow\left(2x+\dfrac{3}{24}\right)-\left(3x-\dfrac{1}{32}\right)=0\)
\(\Leftrightarrow2x+\dfrac{3}{24}-3x+\dfrac{1}{32}=0\)
\(\Leftrightarrow\left(\dfrac{3}{24}+\dfrac{1}{32}\right)+\left(2x-3x\right)=0\)
\(\Leftrightarrow\dfrac{5}{32}-x=0\)
\(\Leftrightarrow x=\dfrac{5}{32}\)
a,100-x-2x-3x-4x=90
100-10x=90
10.(10-x)=90
10-x=9
x=10-9=1
Vậy....
b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Leftrightarrow\left|3x+1\right|+1=4\)
\(\Leftrightarrow\left|3x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)
c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)
\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;2\right\}\)
d) Ta có: \(5^{-1}\cdot25^x=125\)
\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)
\(\Leftrightarrow5^{2x-1}=5^3\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
hay x=2
Vậy: x=2
cảm ơn nhìu ak