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a: \(\Leftrightarrow6x=30\)
hay x=5
b: \(\Leftrightarrow6x=25+12-1=36\)
hay x=6
a: \(\Leftrightarrow8x=108+12=120\)
hay x=15
b: \(\Leftrightarrow6x=60\)
hay x=10
a) \(5x+x=39-3^{11}:3^9\)
\(\Leftrightarrow6x=39-3^2\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)
b) \(2^x:2^5=16\)
\(\Leftrightarrow2^x:2^5=2^4\)
\(\Leftrightarrow2^x=2^4.2^5\)
\(\Leftrightarrow2^x=2^9\)
\(\Leftrightarrow x=9\)
c) \(7x-x=5^{21}:5^{19}+3.2^2-7^0\)
\(\Leftrightarrow6x=5^2+3.4-1\)
\(\Leftrightarrow6x=36\)
\(\Leftrightarrow x=6\)
d) \(7x-2x=6^{17}:6^{15}+44:11\)
\(\Leftrightarrow5x=6^2+4\)
\(\Leftrightarrow5x=40\)
\(\Leftrightarrow x=8\)
a)⇔6x=39-32
⇔6x=30
⇔ x=5
b)2x:25=16
⇔2x=24.25
⇔ 2x=29
⇔ x=9
c)⇔6x=52+3.22-1
⇔ 6x= 36
⇔ x=6
d)⇔5x=62+4
⇔ 5x=40
⇔ x=8
\(a,\Rightarrow\left(4x-1\right)^2=25=5^2=\left(-5\right)^2\\ \Rightarrow\left[{}\begin{matrix}4x-1=5\\4x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\\ b,\Rightarrow2^x\left(1+2^3\right)=144\\ \Rightarrow2^x=144:9=16=2^4\Rightarrow x=4\\ c,\Rightarrow3^{2x+3}=3^{2\left(x+3\right)}\\ \Rightarrow2x+3=2x+6\Rightarrow0x=3\left(vô.lí\right)\\ \Rightarrow x\in\varnothing\)
a: 3x=81
nên x=27
b: \(5\cdot4^x=80\)
\(\Leftrightarrow4^x=16\)
hay x=2
c: \(2^x=4^5:4^3\)
\(\Leftrightarrow2^x=2^4\)
hay x=4
a) 89-(73-x) = 20
=> 73+x =89-20
=> 73+x=69
=> x=69-73
=> x=-4
b) (x+7)-25=13
=> x+7=13+25
=> x+7=38
=> x=38-7
=> x=31
c) 98-(x+4)=20
=> x+4=98-20
=> x+4=78
=> x=78-4
=> x=73
d) 140:(x-8)=7
=> x-8=140:7
=> x-8=20
=> x=20+8
=> x=28
e) 4(x+41)=400
=> x+41=400:4
=> x+41=100
=> x=100-41
=> x=59
f) x-[42+(-28)]=-8
=> x-14=-8
=> x=-8+14
=> x=6
a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
a) 32 + 3x = 90
<=> 32 ( 1 + 3x-2 )= 90
<=> 1+ 3x-2 = 10
<=> 3x-2 = 9 = 32
<=> x - 2 = 2
<=> x = 0
b) 2x + 2x+3 = 144
<=> 2x ( 1 + 23 ) = 144
<=> 2 x = 16 = 24
<=> x = 4
d) ta có 1 + 3 + 5 +..+ 99 = 2500 = 502
<=> x = 50