tim x
\(^{ }5^x.5^{\left(x+1\right)}.5^{\left(x+2\right)}_{ }< ,=100...0:2^{18}_{_{ }}\)
100...0=18 chu so 0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
16x<1284
=>(24)x<(27)4
=>24x<228
=>4x<28
=>x<7
=>x=0;1;2;3;4;5;6
vậy x=0;1;2;3;4;5;6
b,=>5x.5x+1.5x+2<1010:218
=>53x+3<510.210:218
=>53x+3<510.28
=>53x+3:510<28
=>53(x+1)-10<256<54
=>3(x+1)-10<4
=>3(x+1)<4+10
=>x+1<14/3<5
=>x<4
=>x=0;1;2;3
vậy x=0;1;2;3
16^x < 128^4
=> 2^4x < 2^7.4
=> 2 ^4x < 2^28
=> 4x < 28
=> x < 7
b/100x+(1+2+3+...+100)=205550
100x+5050=205550
100x=205550-5050
100x=200500
x=200500/100
x=2005
d/(3x-24).75=2.76.1/20090
(3x-24).75=2.76.1
(3x-24)=2.76:75
(3x-24)=2.7
3x-16=14
3x=14+16
3x=30
x=30/10=3
Sửa đề : Số 100...00 là 18 chữ số 0
Ta có : x + x + 1 + x + 2 và 100...00 = 1018
< = > 53x + 3 \(\le\)1018 : 218
< = > 53x + 3 \(\le\)518
= > 3x + 3 \(\le\)18
= > 3x \(\in\)15 = > x \(\in\){ 0;1;...;5}
Vậy x = { 0;1;...;5}
a: \(\sqrt{\left(x+1\right)^2}=5\)(ĐKXĐ: \(x\in R\))
=>|x+1|=5
=>\(\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)
b: Sửa đề: \(5\sqrt{9x-9}-\sqrt{4\left(x-1\right)}+\sqrt{36\left(x-1\right)}-18=0\)
ĐKXĐ: x>=1
\(PT\Leftrightarrow5\cdot3\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}-18=0\)
=>\(15\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}=18\)
=>\(19\sqrt{x-1}=18\)
=>\(\sqrt{x-1}=\dfrac{18}{19}\)
=>\(x-1=\left(\dfrac{18}{19}\right)^2=\dfrac{324}{361}\)
=>\(x=\dfrac{324}{361}+1=\dfrac{324+361}{361}=\dfrac{685}{361}\)
Lời giải:
a. PT $\Leftrightarrow |x+1|=5$
$\Leftrightarrow x+1=\pm 5\Leftrightarrow x=4$ hoặc $x=-6$
b. ** Sửa $x-9$ thành $x-1$
ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow 5\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}-18=0$
$\Leftrightarrow (5-2+6)\sqrt{x-1}=18$
$\Leftrightarrow 9\sqrt{x-1}=18$
$\Leftrightarrow \sqrt{x-1}=2$
$\Leftrightarrow x-1=4$
$\Leftrightarrow x=5$ (tm)
Ta có
\(5^x.5^{x+1}.5^{x+2}\le10^{18}:2^{18}=>5^{3x+3}\le5^{18}=>3x+3\le18=>x\le5\)
ta có : \(5^x.5^{\left(x+1\right)}.5^{\left(x+2\right)}\le\dfrac{100...0}{2^{18}}\) \(\Leftrightarrow5^{3x+3}\le\dfrac{10^{18}}{2^{18}}=5^{18}\)
\(\)\(\Leftrightarrow3x+3\le18\Leftrightarrow x\le5\) vậy \(x\le5\)
cảm ơn nha