tìm x
ccau ơi, giúp tớ vs, nhanh nha, chi tiết nx
ccau ơi, giúp tớ vs, nhanh nha, chi tiết nx
\(a,5^x+5^{x+2}=650\\ \Rightarrow5^x+5^x.5^2=650\\ \Rightarrow5^x\left(1+5^2\right)=650\\ \Rightarrow5^x.26=650\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)
\(b,\left(4x+1\right)^2=25.9\\\Rightarrow\left(4x+1\right)^2=225\\ \Rightarrow\left[{}\begin{matrix}4x+1=15\\4x+1=-15\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-4\end{matrix}\right.\)
\(c,2^x+2^{x+3}=144\\ \Rightarrow2^x+2^x.2^3=144\\ \Rightarrow2^x\left(1+2^3\right)=144\\ \Rightarrow2^x=144:\left(1+2^3\right)\\ \Rightarrow2^x=16\\ \Rightarrow2^x=2^4\\ \Rightarrow x=4\)
\(d,3^{x+2}=9^{x+3}\\ \Rightarrow3^{x+2}=\left(3^2\right)^{x+3}\\ \Rightarrow3^{x+2}=3^{2x+6}\\ \Rightarrow x+2=2x+6\\ \Rightarrow x-2x=6-2\\ \Rightarrow-x=4\\ \Rightarrow x=-4\)
\(e,x^{15}=x^2\\ \Rightarrow x^{15}-x^2=0\\ \Rightarrow x^2\left(x^{13}-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\x^{13}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
a: =>5^x+5^x*25=650
=>5^x*26=650
=>5^x=25
=>x=2
b: =>4x+1=15 hoặc 4x+1=-15
=>4x=-16 hoặc 4x=14
=>x=7/2 hoặc x=-8
c: =>2^x*9=144
=>2^x=16
=>x=4
d: =>2x+2=2x+6
=>2=6(loại)
e: =>x^2(x^13-1)=0
=>x=0 hoặc x=1
ccau giúp tớ vs, nhanh nha, chi tiết nx
`@` `\text {Ans}`
`\downarrow`
`a)`
\(5^x+5^{x+2}=650\)
`\Rightarrow 5^x + 5^x . 5^2 = 650`
`\Rightarrow 5^x . (1 + 5^2) = 650`
`\Rightarrow 5^x . 26 = 650`
`\Rightarrow 5^x = 650 \div 26`
`\Rightarrow 5^x = 25`
`\Rightarrow 5^x = 5^2`
`\Rightarrow x = 2`
Vậy, `x = 2`
`b)`
`(4x + 1)^2 = 25.9`
`\Rightarrow (4x + 1)^2 = 225`
`\Rightarrow (4x + 1)^2 = (+-15^2)`
`\Rightarrow`\(\left[{}\begin{matrix}4x-1=15\\4x-1=-15\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}4x=16\\4x=-14\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}x=4\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-\dfrac{7}{2};4\right\}\)
`c)`
\(2^x+2^{x+3}=144\)
`\Rightarrow 2^x + 2^x . 2^3 = 144`
`\Rightarrow 2^x . (1 + 2^3) = 144`
`\Rightarrow 2^x . 9 = 144`
`\Rightarrow 2^x = 144 \div 9`
`\Rightarrow 2^x = 16`
`\Rightarrow 2^x = 2^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
`d)`
\(3^{2x+2}=9^{x+3}\)
`\Rightarrow `\(3^{2x+2}=\left(3^2\right)^{x+3}\)
`\Rightarrow `\(3^{2x+2}=3^{2x+6}\)
`\Rightarrow 2x + 2 = 2x + 6`
`\Rightarrow 2x - 2x = 6 - 2`
`\Rightarrow 0 = 4 (\text {vô lý})`
Vậy, `x` không có giá trị nào thỏa mãn.
`e)`
\(x^{15}=x^2\)
`\Rightarrow `\(x^{15}-x^2=0\)
`\Rightarrow `\(x^2\cdot\left(x^{13}-1\right)=0\)
`\Rightarrow `\(\left[{}\begin{matrix}x^2=0\\x^{13}-1=0\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}x=0\\x^{13}=1\end{matrix}\right.\)
`\Rightarrow `\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy, `x \in`\(\left\{0;1\right\}.\)
tìm x
ccau ơi, nhanh nha, giúp tớ vs, chi tiết nx
\(f,\left(2x+1\right)^3=343\\\Rightarrow \left(2x+1\right)^3=7^3\\ \Rightarrow2x+1=7\\ \Rightarrow2x=6\\ \Rightarrow x=3\\ g,\left(x-1\right)^3=125\\ \Rightarrow\left(x-1\right)^3=5^3\\ \Rightarrow x-1=5\\ \Rightarrow x=6\\ h,2^{x+2}-2^x=96\\ \Rightarrow2^x.2^2-2^x=96\\ \Rightarrow2^x.\left(2^2-1\right)=96\\ \Rightarrow2^x=96:\left(2^2-1\right)\\ \Rightarrow2^x=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5\)
\(i,\left(x-5\right)^4=\left(x-5\right)^6\\ \Rightarrow\left(x-5\right)^4\left(1-\left(x-5\right)^2\right)=0\\\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{matrix}\right. \\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\\x-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(t/m\right)\\x=6\left(t|m\right)\\x=4\left(loại\right)\end{matrix}\right.\)
\(j,720:\left[41-\left(2x-5\right)\right]=2^3.5\\ \Rightarrow720:\left(41-2x+5\right)=40\\ \Rightarrow\left(46-2x\right)=720:40\\ \Rightarrow46-2x=18\\ \Rightarrow2x=46-18=28\\ \Rightarrow x=28:2=14\)
@seven
f: =>2x+1=7
=>2x=6
=>x=3
g: =>x-1=5
=>x=6
h: =>2^x*3=96
=>2^x=32
=>x=5
i: =>(x-5)^4*[(x-5)^2-1]=0
=>(x-5)(x-4)(x-6)=0
=>x=5;x=4;x=6
j: =>41-(2x-5)=720:40=18
=>2x-5=23
=>2x=28
=>x=14
bài 18: so sánh ( đưa về cùng cơ số hoặc cùng số mũ )
a) \(2^{100}\) và \(1024^9\)
b) \(5^{30}\) và \(6.5^{29}\)
c) \(2^{98}\) và \(9^{49}\)
d) \(10^{30}\) và \(2^{100}\)
ccau giúp tớ vs, nhanh nha, chi tiết nx
a, 2100 và 10249
10249 = (210)9 = 290
2100 > 290
Vậy 2100 > 290
b, 530 và 6.529
6.529 > 5.529 = 530
vậy 530 < 6.529
c, 298 và 949
(22)49 = 449 < 949
vậy: 298 < 949
d, 1030 và 2100
(103)10 = 100010
2100 = (210)10 = 102410
Vì 100010 < 102410
Nên 1030 < 2100
ccau ơi, giúp tớ vs, nhanh nha
\(\left(2x-1\right)^5=x^5\)
\(2x-1=x\)
\(2x-x=1\)
\(x=1\)
_________________________
\(2x^2+2=20\)
\(2x^2=20-2\)
\(2x^2=18\)
\(x^2=\dfrac{18}{2}\)
\(x^2=9\)
\(x=\pm3\)
tìm x
a) \(x^2\) = \(x^3\)
e) \(3^{2x+1}\) = 27
g) \(6^2\) = \(6^{x-3}\)
ccau giúp tớ vs, nhanh nha
a, \(x^2\) = \(x^3\)
\(x^3\) - \(x^2\) = 0
\(x^2\)( \(x\) -1) = 0
\(\left[{}\begin{matrix}x^2=0\\x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(x\) \(\in\) { 0; 1}
e, 32\(x+1\) = 27
\(3^{2x}\)+1 = 33
2\(x\) + 1 = 3
2\(x\) = 2
\(x\) = 1
g, 62 = 6\(x-3\)
2 = \(x-3\)
\(x\) = 3 + 2
\(x\) = 5
\(a,x^2=x^3\\ \Rightarrow x^2-x^3=0\\ \Rightarrow x^2\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\1-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(b,3^{2x+1}=27\\ \Rightarrow3^{2x+1}=3^3\\ \Rightarrow2x+1=3\\ \Rightarrow2x=3-1\\ \Rightarrow2x=2\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)
\(c,6^2=6^{x-3}\\ \Rightarrow6^{x-3}=6^2\\ \Rightarrow x-3=2\\ \Rightarrow x=2+3\\ \Rightarrow x=5\)
bài 19: tìm x
l) \(2^x\) = 1
m) \(3^x\) = 81
n) \(3^x\) = 27
o) \(9^x\) = \(3^4\)
mn giúp tớ vs, nhanh và chi tiết nha, cảm ơn trc
\(l,\\ 2^x=1=2^0\\ Vậy:x=0\\ m,\\ 3^x=81=3^4\\ Vậy:x=4\\ n,\\ 3^x=37=3^3\\ Vậy:x=3\\ o,\\ 9^x=3^4=\left(3^2\right)^2=9^2\\ Vậy:x=2\)
l) \(2^x=1\)
\(\Rightarrow2^x=2^0\)
\(\Rightarrow x=0\)
m) \(3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
n) \(3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
o) \(9^x=3^4\)
\(\Rightarrow9^x=\left(3^2\right)^2\)
\(\Rightarrow9^x=9^2\)
\(\Rightarrow x=2\)
tìm số nguyên x, biết:
nhanh nhé ccau ơi
đún + chi tiết + nhanh = tick
y) \(\dfrac{x+2}{4}=\dfrac{3}{6}\)
\(\Leftrightarrow6.\left(x+2\right)=4.3\\ \Leftrightarrow6x+12=12\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\)
Vậy x = 0
z) \(\dfrac{2x}{49}=\dfrac{-2}{7}\)
\(\Leftrightarrow2x.7=\left(-2\right).49\\ \Leftrightarrow14x=-98\\ \Leftrightarrow x=-7\)
Vậy x = -7
y) (x + 2)/4 = 3/6
x + 2 = 1/2 . 4
x + 2 = 2
x = 2 - 2
x = 0
z) 2x/49 = -2/7
2x = -2/7 . 49
2x = -14
x = -14 : 2
x = -7
y) \(\dfrac{x+2}{4}=\dfrac{3}{6}\)
\(\Rightarrow6\cdot\left(x+2\right)=3\cdot4\)
\(\Rightarrow6\cdot\left(x+2\right)=12\)
\(\Rightarrow x+2=12:6\)
\(\Rightarrow x+2=2\)
\(\Rightarrow x=2-2=0\left(tm\right)\)
Vậy: \(x=0\)
z) \(\dfrac{2x}{49}=\dfrac{-2}{7}\)
\(\Rightarrow7\cdot2x=-2\cdot49\)
\(\Rightarrow14x=-98\)
\(\Rightarrow x=-98:14=-7\left(tm\right)\)
Vậy: \(x=-7\).
tìm x
nhanh nha ,chi tiết nx
Lời giải:
$3^{2x+2}=9^{x+3}$
$3^{2x+2}=(3^2)^{x+3}=3^{2(x+3)}=3^{2x+6}$
$\Rightarrow 2x+2=2x+6$
$\Rightarrow 2=6$ (vô lý)
Vậy không có x thỏa mãn đề.