Tìm x:
a) (32+23).x+42.2.x=98
b) 846:[41-(2x-5) = 22+52
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a, 2 3 x + 5 2 x = 2 5 2 + 2 3 - 33
8x+25x = 33
33x = 33
x = 1
b, 260 : x + 4 = 5 2 3 + 5 - 3 3 2 + 2 2
260:(x+4) = 5.13–3.13
x+4 = 260:26
x+4 = 10
x = 6
c, 720 : [ 41 - 2 x - 5 ] = 2 3 . 5
41–(2x–5) = 720:40
2x–5 = 41–18
2x = 28
x = 14
d, 3 2 - 2 x - 12 + 35 = 5 2 + 279 : 3 2
7(x–12)+35 = 56
7(x–12) = 21
x–12 = 3
x = 15
1: =>\(5^{2x-3}=5^2\cdot3+5^2\cdot2=5^2\cdot5=5^3\)
=>2x-3=3
=>2x=6
=>x=3
2: \(41-2^{x+1}=9\)
=>\(2^{x+1}=32\)
=>x+1=5
=>x=4
3: =>\(4^{x+2}=65-1=64\)
=>x+2=3
=>x=1
\(5^{2x-3}-2.5^2=5^2.3\\ 5^{2x-3}=5^2.3+5^2.2\\ 5^{2x-3}=5^2.\left(3+2\right)\\ 5^{2x-3}=5^2.5\\ 5^{2x-3}=5^3\\ \Rightarrow2x-3=3\\ 2x=3+3\\ 2x=6\\ x=\dfrac{6}{2}\\ Vậy:x=3\)
a) x = 2.
b) x = 7.
c) x= 12.
d) x= 45.
e) x = 18.
f) x = 10.
a, 36:(x–5) = 2 2
(x–5) = 9
x = 14
b, [3.(70–x)+5]:2 = 46
[3.(70–x)+5] = 92
70–x = 29
x = 41
c, 450:[41–(2x–5)] = 3 2 .5
41–(2x–5) = 10
2x–5 = 31
2x = 36
x = 18
d, 230+[ 2 4 +(x–5)] = 315. 2018 0
16+(x–5) = 315–230
x–5 = 85–16
x = 69+5
x = 74
e, 2 x + 2 x + 1 = 48
2 x .(2+1) = 48
2 x = 16 = 2 4
x = 4
f, 3 x + 2 + 3 x = 2430
3 x . 3 2 + 1 = 2430
3 x = 2430:10 = 243 = 3 5
x = 5
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}.`
\(5x+2x\cdot\left(2^3\cdot5-3^2\cdot4\right)+5^2=4^3\)
\(\Rightarrow5x+2x\cdot\left(8\cdot5-9\cdot4\right)+25=64\)
\(\Rightarrow5x+2x\cdot\left(40-36\right)=64-25\)
\(\Rightarrow5x+2x\cdot4=39\)
\(\Rightarrow5x+8x=39\)
\(\Rightarrow x\cdot\left(5+8\right)=39\)
\(\Rightarrow13x=39\)
\(\Rightarrow x=\dfrac{39}{13}\)
\(\Rightarrow x=3\)
Vậy: ...
(3^2+2^3).x+4^2.2.x=98
(9+8).x+16.2.x=98
17.x+16.2.x=98
17.x+32.x=98
(17+32).x=98
49.x=98
x=98:49
x=2