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a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
a) 0 : x = 0
=> x \(\in\)N ( x \(\ne\)0 )
b) 3x = 9
=> x = 2
c) 4x = 64
=> x = 3
d) 2x = 16
=> x = 4
e) 9x-1 = 9
9 = 91
=> x = 1 + 1 = 2
x4 = 16
=> x = 2
g) 2x : 25 = 1
Quy ước a0 = 1 , ta suy ra : x = 5
Giải ko giải thích , bỏ qua nhá :))
A, (2x+1)3=343
=> (2x+1)3=73
=> 2x + 1 = 7
=> 2x = 6
=> x = 3
B, 2x+2x+3=144
=> 2x+2x . 23 =144
=> 2x ( 1 + 23 ) =144
=> 2x ( 1 + 8 ) =144
=> 2x . 9 =144
=> 2 x = 16
=> 2 x = 2 4
=> x = 4
C, 3x+3x+2=2430
=> 3x+3x . 32 =2430
=> 3x . ( 1 + 32 ) =2430
=> 3x . ( 1 + 9 ) =2430
=> 3x . 10 =2430
=> 3x = 243
=> 3x = 3 5
=> x = 5
\(2^{10}.2^{x+4}=64^5\)
\(\Leftrightarrow2^{x+14}=2^{30}\)
\(\Leftrightarrow x+14=30\)
\(\Leftrightarrow x=16\)
\(5^x+5^{x+3}=630\)
\(\Rightarrow5^x.1+5^x.125=630\)
\(\Rightarrow5^x.126=630\)
\(\Rightarrow5^x=5\)
\(\Rightarrow x=1\)
\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+..............+\left(x+100\right)=7450\)
\(\Rightarrow\left(x+x+x+x+.........+x\right)+\left(1+2+3+..........+100\right)=7450\)
\(101x+5050=7450\)
Đến đây tự tính
a,A=|x-7|+12
Vì \(\left|x-7\right|\ge0\forall x\)nên \(\left|x-7\right|+12\ge12\forall x\)
Ta thấy A=12 khi |x-7| = 0 => x-7 = 0 => x = 7
Vậy GTNN của A là 12 khi x = 7
b,B=|x+12|+|y-1|+4
Vì \(\left|x+12\right|\ge0\forall x\)
\(\left|y-1\right|\ge0\forall y\)
nên \(\left|x+12\right|+\left|y-1\right|\ge0\forall x,y\)
\(\Rightarrow\left|x+12\right|+\left|y-1\right|+4\ge4\forall x,y\)
Ta thấy B = 4 khi \(\hept{\begin{cases}\left|x+12\right|=0\\\left|y-1\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x+12=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-12\\y=1\end{cases}}\)
Vậy GTNN của B là 4 khi x = -12 và y = 1
ko ghi đề
\(=25,97+\left(6,54+103,46\right)\)
\(=25,97+110\)
\(=135,97\)
a: x^3=7^3
=>x^3=343
=>\(x=\sqrt[3]{343}=7\)
b: x^3=27
=>x^3=3^3
=>x=3
c: x^3=125
=>x^3=5^3
=>x=5
d: (x+1)^3=125
=>x+1=5
=>x=4
e: (x-2)^3=2^3
=>x-2=2
=>x=4
f: (x-2)^3=8
=>x-2=2
=>x=4
h: (x+2)^2=64
=>x+2=8 hoặc x+2=-8
=>x=6 hoặc x=-10
j: =>x-3=2 hoặc x-3=-2
=>x=1 hoặc x=5
k:
9x^2=36
=>x^2=36/9
=>x^2=4
=>x=2 hoặc x=-2
l:
(x-1)^4=16
=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
a) 3 + x - ( 3x - 1 ) = 6 - 2x
\(\Rightarrow3+x-3x+1=6-2x\)
\(\Rightarrow-2x+4=6-2x\)
\(\Rightarrow-2x+2x=-4+6\)
\(\Rightarrow0x=-2\)
\(\Rightarrow x=\varnothing\)
Vậy: \(x=\varnothing\)
b) -12 . (x - 5 ) + 7.(3 - x ) = 5
\(\Rightarrow-12x+60+21-7x-5=0\)
\(\Rightarrow-19x+76=0\)
\(\Rightarrow-19x=-76\)
\(\Rightarrow x=\frac{76}{19}\)
Vậy: \(x=\frac{76}{19}\)
c) 30. ( x + 2 ) - 6 . ( x - 5 ) - 24x = 100
\(\Rightarrow30x+60-6x+30-24x-100=0\)
\(\Rightarrow0x-10=0\)
\(\Rightarrow x=\varnothing\)
Vậy: \(x=\varnothing\)
1.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172