Tìm X:
a) 18,4 - 2 x ( X + 75) = 2,2
b) 2,5 + 0,5 x ( 18,1 - 2 x X ) = 2
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\(x\times12,5=15\\ x=15:12,5=1,2\\ x-1,27=3\\ x=3+1,27=4,27\)
\(a,\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow x^3-27-x^3+4x=1\\ \Leftrightarrow4x=28\Leftrightarrow x=7\\ c,\Leftrightarrow4x^2-4x-8=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow2x^2+6x+x+3=0\\ \Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
a) \(1=\left(2x+0,5\right)^{600}\)
\(\Rightarrow1^{600}=\left(2x+0,5\right)^{600}\)
\(\Rightarrow\left[{}\begin{matrix}2x+0,5=1\\2x+0,5=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=0,5\\2x=-1,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,25\\x=-0,75\end{matrix}\right.\)
b) \(\left(x-0,125\right)^2=0,25\)
\(\Rightarrow\left(x-0,125\right)^2=0,5^2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
c) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)
\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)
\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-3=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`1 = (2x + 0,5)^600`
`=> (2x+0,5)^600 = (+-1)^600`
`=> \text {TH1: } 2x + 0,5 = 1`
`=> 2x = 1 - 0,5`
`=> 2x = 0,5`
`=> x = 0,5 \div 2`
`=> x = 0,25`
`\text {TH2: } 2x + 0,5 = -1`
`=> 2x = -1 - 0,5`
`=> 2x = -1,5`
`=> x = -1,5 \div 2`
`=> x = -0,75`
Vậy, `x \in {-0,75; 0,25}.`
`b)`
`(x - 0,125)^2 = 0,25`
`=> (x - 0,125)^2 = (+-0,5)^2`
`=> `\(\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,5+0,125\\x=-0,5+0,125\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
Vậy, `x \in {-0,375; 0,625}.`
`c)`
`(x - 3)^11 = (x - 3)^41`
`=> (x - 3)^11 - (x - 3)^41 = 0`
`=> (x - 3)^11 * [ 1 - (x - 3)^30] = 0`
`=>`\(\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\1-\left(x-3\right)^{30}=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy, `x \in {3; 4}.`
tui cũng hỏi giống bạn nhưng không ai giai cho tui cả huhu
a)\(1,5-2\left|x\right|=-0,5\)
\(\Leftrightarrow2\left|x\right|=1,5+0,5\)
\(\Leftrightarrow\left|x\right|=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) Ta có: \(\left|x-\dfrac{2}{3}\right|+2.25=\dfrac{3}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{3}\right|=\dfrac{3}{4}-\dfrac{9}{4}=\dfrac{-3}{2}\)(vô lý)
Vậy: \(x\in\varnothing\)
a) Ta có: \(x+\dfrac{-3}{7}=\dfrac{4}{7}\)
\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{4}{7}\)
hay x=1
Vậy: x=1
a) 32 : (3.x - 2) = 8
3x - 2 = 32 : 8
3x - 2 = 4
3x = 4 + 2
3x = 6
x = 6 : 3
x = 2
b) 75 : (x - 18) = 25
x - 18 = 75 : 25
x - 18 = 3
x = 3 + 18
x = 21
c) (15 - 6.x) . 243 = 729
15 - 6x = 729 : 243
15 - 6x = 3
6x = 15 - 3
6x = 12
x = 12 : 6
x = 2
d) 4.(x - 12) + 9 = 17
4(x - 12) = 17 - 9
4(x - 12) = 8
x - 12 = 8 : 4
x - 12 = 2
x = 2 + 12
x = 14
e) 20 - 2.(x + 4) = 4
2(x + 4) = 20 - 4
2(x + 4) = 16
x + 4 = 16 : 2
x + 4 = 8
x = 8 : 2
x = 4
`32: ( 3xx x -2)=8`
`3xx x-2=32:8`
`3xx x-2=4`
`3 xx x=4+2`
`3xx x=6`
`x=6:3`
`x=2`
__
`75 : (x-18) =25`
`x-18=75:25`
`x-18= 3`
`x=3+18`
`x=21`
__
`(15-6 xx x ) xx 243 =729`
`15-6 xx x = 729 : 243`
`15-6 xx x = 3`
`6 xx x=15-3`
`6 xx x=12`
`x=12:6`
`x=2`
__
`4 xx (x-12)+9=17`
`4 xx (x-12)=17-9`
`4 xx (x-12)= 8`
`x-12=8:4`
`x-12=2`
`x=2+12`
`x=14`
__
`20-2xx(x+4)=4`
`2xx(x+4)=20-4`
`2xx(x+4)=16`
`x+4=16:2`
`x+4=8`
`x=8-4`
`x=4`
sai đề bn nha ra âm