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b)4.|x|-25=39:38
4.|x|-25=31(=3)
4.|x|=3+25
4.|x|=28
|x|=28:4
|x|=7
TH1: TH2:
x=7 x=-7
vậy x E(dấu thuộc nha) (7;-7)
a) \(2^2.2^4.x=16^2\)
\(2^6x=\left(2^4\right)^2\)
\(2^6.x=2^8\)
\(x=2^8:2^6\)
\(x=2^2\)
\(x=4\)
vay \(x=4\)
b) \(5^{2x+1}=125\)
\(5^{2x+1}=5^3\)
\(\Rightarrow2x+1=3\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
vay \(x=1\)
c) \(\left(2x-1\right)^3=64\)
\(\left(2x-1\right)^3=4^3\)
\(\Rightarrow2x-1=4\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=\frac{5}{2}\)
vay \(x=\frac{5}{2}\)
125(28+72)-25(3^2.4+64)
=125.100-25(9.4+64)
=125.100-25.(36+64)
=125.100-25.100
=12500-2500
=10000
a) (123-4x)-67=8
123-4x =8+67
123-4x =73
4x =123-73
4x =60
x =60:4
x =15
a)(123-4x)-67= 8 =(123-4x) =8+67 = 4x =123-75 4x=48 x=48:4 x=12
\(\left(x-5\right).4=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\4=0\left(ktm\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\4=0\left(ktm\right)\end{cases}}}\)
vậy x=5
\(8x-4x=1208\)
\(x\left(8-4\right)=1208\)
\(x4=1208\)
\(\Rightarrow x=1208:4=302\)
\(\left(x-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}}\)
vậy x=4 hoặc x=3
1) \(\Leftrightarrow x+11-15+x+20=0\)
\(\Leftrightarrow2x+16=0\)
\(\Leftrightarrow x=-8\)
2) \(\Leftrightarrow2x-16+x-13=16\)
\(\Leftrightarrow3x-45=0\)
\(\Leftrightarrow x=15\)
Những câu dưới bạn làm tương tự như vậy nhé
1)(x+11)–(15–x) =–20
x+11 - 15 + x = -20
x + ( 11 -15 ) = -20
x + ( -4 ) = -20
x = -20 - ( -4 )
x = -16
a)\(3^{x-2}+3^x=810\)
\(\Leftrightarrow3^x\left(3^{-2}+1\right)=810\)
\(\Leftrightarrow3^x\cdot\frac{10}{9}=810\)
\(\Leftrightarrow3^x=729\)
\(\Leftrightarrow3^x=3^6\)
\(\Leftrightarrow x=6\)
b)402240223.(x2-1)=804480443
\(\Leftrightarrow x^2-1=80448044^3:40224022^3\)
\(\Leftrightarrow x^2-1=\left(\frac{80448044}{40224022}\right)^3\)
\(\Leftrightarrow x^2-1=2^3\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow x=\pm3\)
`#3107`
a)
`8x^2 - 5 = 67 `
`=> 8x^2 = 67 + 5`
`=> 8x^2 = 72`
`=> x^2 = 9`
`=> x^2 = (+-3)^2`
`=> x = +-3`
Vậy, `x \in {-3;3}`
b)
`(4x - 2)^4=16`
`=> (4x - 2) = (+-2)^4`
`=> ` TH1: `4x - 2 = 2`
`=> 4x = 4`
`=> x =1`
TH2: `4x - 2 = -2`
`=> 4x = 0`
`=> x=0`
Vậy, `x \in {0; 1}.`
a) \(8x^2-5=67\)
\(8x^2=72\)
\(x^2=9\)
\(x=3\)
Vậy x = 3
b) \(\left(4x-2\right)^4=16\)
\(\left(4x-2\right)^2=4\)
\(4x-2=2\)
\(4x=4\)
\(x=1\)
Vậy x = 1