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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
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 \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)
c, Ta có: \(2.|\dfrac{1}{2}x - \dfrac{1}{3}| - \dfrac{3}{2} = \dfrac{1}{4}\)
\(\Rightarrow\) \(2.|\dfrac{1}{2}x - \dfrac{1}{3}| = \dfrac{7}{4}\)
\(\Rightarrow\) \(|\dfrac{1}{2}x - \dfrac{1}{3}| = \dfrac{7}{8}\)
\(\Rightarrow\) \(\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{7}{8}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{29}{24}\\\dfrac{1}{2}x=-\dfrac{13}{24}\end{matrix}\right.\)
\(\Rightarrow\) \(\left[{}\begin{matrix}x=\dfrac{29}{24}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
a,10x+65=125
10x =125-65
x =60:10
x =6
Vậy:...
b,45-(5-2x)^3=2.3^2
45-(5-2x)^3=18
(5-2x)^3 =45-18
(5-2x)^3 =27=3^3
=> 5-2x = 3
2x = 5-3
x= 2:2
x=1
Vậy:....
-Hok tốt ~ K mk nhak.
a, 10x + 65 = 125
10x = 125 - 65
10x = 60
x = 60 : 10
x = 6
b,45-(5-2x)^3=2.3^2
45 - ( 5 - 2x )^3 =18
(5 - 2x)^3 = 45 - 18
(5 - 2x )^3= 3^3
5 - 2x = 3
2x = 2
x = 1
c,2(x-3)-12=(-10)
2x - 6 -12 =-10
2x = -10 + 6 + 12
2x = 8
x = 4
d,x-12=(-13)+1+|-13|
x - 12 = -13 + 1 + 13
x - 12 = 1
x = 13
hok tốt!!!
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
Bài 1:
a, 2x + 35 = x - 27
2x - x = -27 - 35
x = -62
b, 2x - 41 = 3x + (-23)
2x - 3x = -23 + 41
-x = 18
x = -18
c, 4 . (x - 3) - 3 . (x - 5) = 45 . (-2) - 34
4x - 12 - 3x + 15 = -90 - 34
4x - 3x = -90 - 34 + 12 - 15
x = -127
Bài 2:
- ( -1234 + 345 - 29 ) - ( 1234 + 135 - 59 )
= 1234 - 345 + 29 - 1234 - 135 + 59
= (1234 - 1234) - (345 + 135) + (29 + 59)
= 0 - 480 + 88
= -392
Bài 2:
a, |x-1| -x +1=0
|x-1| = 0-1+x
|x-1| = -1 + x
\(\orbr{\begin{cases}x-1=-1+x\\x-1=1-x\end{cases}}\)
\(\orbr{\begin{cases}x=-1+x+1\\x=1-x+1\end{cases}}\)
\(\orbr{\begin{cases}x=x\\x=2-x\end{cases}}\)
x = 2-x
2x = 2
x = 2:2
x=1
b, |2-x| -2 = x
|2-x| = x+2
\(\orbr{\begin{cases}2-x=x+2\\2-x=2-x\end{cases}}\)
2-x = x+2
x+x = 2-2
2x = 0
x = 0
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a) x + ( x+1) + (x+2)+...+(x+9)=345
<=> 10x + (1 + 2 + 3 + ..... + 9) = 345
=> 10x + 45 = 345
=> 10x = 300
=> x = 30