x+x+1/2.2/5+x+8/10=136
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X+X+1/2x2/5+X+8/10=136
3X+1/5+4/5=136
3X+1=136
3X =136-1=135
X =135:3=45
X+X+\(\frac{1}{2}x\frac{2}{5}\)+X+\(\frac{8}{10}\)=136
X.(1+1+1)+\(\frac{1}{5}+\frac{8}{10}\)=136
3X+1=136
3X =136-1
3X =135
X =135:3
X =45
Vậy X=45
Chúc bn học tốt
1: 7-x=8+(-7)
=>7-x=8-7=1
=>x=7-1=6
2: \(x-8=\left(-3\right)-8\)
=>x-8=-11
=>\(x=-11+8=-3\)
3: \(2-x=10-9+23\)
=>\(2-x=33-9=24\)
=>x=2-24=-22
4: \(-2-x=15\)
=>\(x=-2-15=-17\)
5: \(-7+x-8=-3-1+13\)
=>x-14=13-4=9
=>x=9+14=23
6: 100-x+7=-x+3
=>107-x=3-x
=>107=3(vô lý)
7: \(23+x=8-2x\)
=>\(x+2x=8-23\)
=>3x=-15
=>x=-15/3=-5
`#3107`
b)
`2.3^x = 162`
`\Rightarrow 3^x = 162 \div 2`
`\Rightarrow 3^x = 81`
`\Rightarrow 3^x = 3^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
c)
`(2x - 15)^5 = (2 - 15)^3`
\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`
\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)
`d)`
\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!
`e)`
\(7\cdot4^{x-1}+4^{x-1}=23?\)
\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)
Bạn xem lại đề!
`f)`
\(2\cdot2^{2x}+4^3\cdot4^x=1056\)
\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
_____
\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)
\(\Rightarrow\left(x\div3+17\right)\div10=2\)
\(\Rightarrow x\div3+17=20\)
\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)
Vậy, `x = 9.`
`136 -4^2 : x=134`
`=> 136-16:x=134`
`=> 16:x= 136-134`
`=> 16:x=2`
`=>x=16:2`
`=>x=8`
__
`19-5(x-1)=4`
`=> 5(x-1)=19-4`
`=> 5(x-1) =15`
`=> x-1=15:5`
`=>x-1=3`
`=>x=3+1`
`=>x=4`
__
`257-(x^2-1) =177`
`=>x^2-1=257-177`
`=>x^2-1=80`
`=>x^2=80+1`
`=>x^2=81`
`=>x^2=(+- 9)^2`
`=> x= +- 9`
__
`(x^2-17) : 8=2^3`
`=> x^2 -17 = 8 * 8`
`=> x^2 -17= 64`
`=> x^2 =64+17`
`=>x^2=81`
`=>x^2=(+- 9)^2`
`=> x= +- 9`
136 - 4² : x = 134
4² : x = 136 - 134
8 : x = 2
x = 8 : 2
x = 4
--------
129 - 5(x - 1) = 4
5(x - 1) = 129 - 4
5(x - 1) = 125
x - 1 = 125 : 5
x - 1 = 25
x = 25 + 1
x = 26
------------
257 - (x² - 1) = 177
x² - 1 = 257 - 177
x² - 1 = 80
x² = 80 + 1
x² = 81
x = -9 hoặc x = 9
--------
(x² - 17) : 8 = 8
x² - 17 = 8 . 8
x² - 17 = 64
x² = 64 + 17
x² = 81
x = -9 hoặc x = 9
=> 3x + 1/5 = 136 - 8/10
=> 3x + 1/5 = 125,2
=> 3x = 125
=> x = 125/3
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