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mk chi lam duoc cau c thoi
suy ra 2x-1=5 hoac 2x-1=-5
2x=6 2x=-4
x=3 x=-2
vay x=3 hoac x=-2
a. 3/4.x+1/3=-1/2
=>3/4.x=-1/2-1/3=-5/6
=>x=-5/6 chia 3/4=-10/9
b. -x/4=-9/x =>-x*x=4*-9
=>-2x=-36 =>x=18
c./2x-1/=5
=> 2x-1=5 =>2x=5+1=6 =>x=3
hoặc 2x-1=-5 =>2x=-5+1=-4 =>x=-2
d,e: Sai đề rồi
`#040911`
`a,`
`15 + 25 \div (2x - 1) = 20`
`\Rightarrow 25 \div (2x - 1) = 20 - 15`
`\Rightarrow 25 \div (2x - 1) = 5`
`\Rightarrow 2x - 1 = 25 \div 5`
`\Rightarrow 2x - 1 = 5`
`\Rightarrow 2x = 6`
`\Rightarrow x = 3`
Vây, `x = 3.`
`b,`
\(3^{x-1}+2\cdot3^x=21\)
`\Rightarrow 3^x \div 3 + 2. 3^x = 21`
`\Rightarrow 3^x . \frac{1}{3} + 2. 3^x = 21`
`\Rightarrow 3^x . (\frac{1}{3} + 2) = 21`
`\Rightarrow 3^x . \frac{7}{3} = 21`
`\Rightarrow 3^x = 21 \div \frac{7}{3}`
`\Rightarrow 3^x = 9`
`\Rightarrow 3^x = 3^2`
`\Rightarrow x = 2`
Vậy, `x = 2.`
`c,`
\(2^{x-3}+2^{x+1}=17\)
`\Rightarrow 2^x \div 2^3 + 2^x . 2 = 17`
`\Rightarrow 2^x . \frac{1}{8} + 2^x . 2 = 17`
`\Rightarrow 2^x . (\frac{1}{8} + 2) = 17`
`\Rightarrow 2^x . \frac{17}{8} = 17`
`\Rightarrow 2^x = 17 \div \frac{17}{8}`
`\Rightarrow 2^x = 8`
`\Rightarrow 2^x = 2^3`
`\Rightarrow x = 3`
Vậy, `x = 3`
`d,`
\(5^x-5^{x-1}=20\)
`\Rightarrow 5^x - 5^x \div 5 = 20`
`\Rightarrow 5^x - 5^x . \frac{1}{5} = 20`
`\Rightarrow 5^x . (1 - \frac{1}{5} = 20`
`\Rightarrow 5^x . \frac{4}{5} = 20`
`\Rightarrow 5^x = 20 \div \frac{4}{5}`
`\Rightarrow 5^x = 25`
`\Rightarrow 5^x = 5^2`
`\Rightarrow x = 2`
Vậy, `x = 2.`
\(a.25:\left(2x-1\right)=5\)
\(2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)
\(b.3^x:3+2.3^x=21\)\(\Leftrightarrow3^x.\dfrac{1}{3}+2.3^x=21\)
\(\Leftrightarrow3^x\left(\dfrac{1}{3}+2\right)=21\)
\(\Leftrightarrow3^x.\dfrac{7}{3}=21\)
\(\Leftrightarrow3^x=9\Leftrightarrow x=2\)
\(c.2^x:2^3+2^x.2=17\Leftrightarrow2^x.\dfrac{1}{8}+2^x.2=17\)
\(\Leftrightarrow2^x.\dfrac{17}{8}=17\Leftrightarrow2^x=8\Leftrightarrow x=3\)
\(d.5^x-5^x:5=20\Leftrightarrow5^x-5^x.\dfrac{1}{5}=20\)
\(\Leftrightarrow5^x\left(1-\dfrac{1}{5}\right)=20\Leftrightarrow5^x=20:\dfrac{4}{5}\Leftrightarrow5^x=25\Leftrightarrow x=2\)
Bài 1:
a) Ta có: \(-5x+32=\left(-2\right)^3\)
\(\Leftrightarrow-5x+32=-8\)
\(\Leftrightarrow-5x=-40\)
hay x=8
Vậy: x=8
b) Ta có: \(-2x+36=6\)
\(\Leftrightarrow-2x=6-36=-30\)
hay x=15
Vậy: x=15
e) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{-9;5;-5\right\}\)
b,-2x+36=6
tương đương -2x=-30
tương đương x=15
a, -5x+32=(-2)^3
tương đương -5x+32=8
tương đương -5x=-24
tương đương x=24/5
`#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.`
e) 96-5(2x-1)=41
5(2x-1)= 96-41
5(2x-1)=55
2x-1=55:5
2x-1=11
2x=11+1
2x=12
x=12:2
x=6
a) \(2^{2x}.2^4=1024\)
\(2^{2x}=1024:2^4\)
\(2^{2x}=1024:16\)
\(2^{2x}=64\)
\(2^{2x}=2^6\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
vay \(x=3\)
b) \(2.3^x=10.3^{12}+8.27^4\)
\(2.3^x=2.5.3^{12}+2^3.\left(3^3\right)^4\)
\(2.3^x=2.5.3^{12}+2^3.3^{12}\)
\(2.3^x=2.3^{12}.\left(5+2^2\right)\)
\(2.3^x=2.3^{12}.9\)
\(2.3^x=2.3^{12}.3^2\)
\(2.3^x=2.3^{14}\)
\(\Rightarrow x=14\)
vay \(x=14\)
c) \(5^8.25^x+1=5^{17}\)
\(5^8.\left(5^2\right)^x+1=5^{17}\)
\(5^8.5^{2x}+1=5^{17}\)
\(5^{8+2x}=5^{17}-1\)
e) \(\left(2x-4\right)^5=\left(2x-4\right)^3\)
\(\left(2x-4\right)^5-\left(2x-4\right)^3=0\)
\(\left(2x-4\right)\left[\left(2x-4\right)^2-1\right]=0\)
\(\left(2x-4\right)\left(2x-4-1\right)\left(2x-4+1\right)=0\)
\(\left(2x-4\right)\left(2x-5\right)\left(2x-3\right)=0\)
\(\Rightarrow2x-4=0\)hoac \(\orbr{\begin{cases}2x-5=0\\2x-3=0\end{cases}}\)
\(\Rightarrow2x=4\)hoac \(\orbr{\begin{cases}2x=5\\2x=3\end{cases}}\)
\(\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)
vay \(x=2\)hoac \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)
a) x + ( x+1) + (x+2)+...+(x+9)=345
<=> 10x + (1 + 2 + 3 + ..... + 9) = 345
=> 10x + 45 = 345
=> 10x = 300
=> x = 30