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a) 6 . x - 15 = 15
6 . x = 15 + 15
6 . x = 30
x = 30 : 6
x = 5
b) 2 . x + 3 = 9
2 . x = 9 - 3
2 . x = 6
x = 6 : 2
x = 3
c) 17 - 6 : x = 15
6 : x = 17 - 15
6 : x = 2
x = 6 : 2
x = 3
d) x : 3 + 12 = 14
x : 3 = 14 - 12
x : 3 = 2
x = 2 . 3
x = 6
a) 6 . x -15 = 15
⇒ 6 .x = 15 + (-15)
⇒ 6 .x = 0
⇔ x = 0
Vậy x = 0.
\(\frac{3}{3.6.9}+\frac{3}{6.9.12}+\frac{3}{9.12.15}+\frac{3}{12.15.18}=\frac{3}{6}\left(\frac{6}{3.6.9}+\frac{6}{6.9.12}+\frac{6}{9.12.15}+\frac{6}{12.15.18}\right)\)
= \(\frac{1}{2}\left(\frac{1}{3.6}-\frac{1}{6.9}+\frac{1}{6.9}-\frac{1}{9.12}+\frac{1}{9.12}-\frac{1}{12.15}+\frac{1}{12.15}-\frac{1}{15.18}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3.6}-\frac{1}{15.18}\right)=\frac{1}{2}.\frac{14}{270}=\frac{7}{270}\)
Áp dụng bất đẳng thức AM - GM:
\(\sqrt{\left(x^2-15\right)\left(x-3\right)}\le\dfrac{x^2-15+x-3}{2}=\dfrac{x^2+x-18}{2};\sqrt{x^2-15}\le\dfrac{x^2-15+1}{2}=\dfrac{x^2-14}{2};\sqrt{x-3}\le\dfrac{x-3+1}{2}=\dfrac{x-2}{2}\).
Do đó \(F\ge x^2+x-\dfrac{x^2+x-18}{2}-\dfrac{x^2-14}{2}-\dfrac{x-2}{2}-38=-21\).
Đẳng thức xảy ra khi x = 4.
Vậy...
\(a,3\left(x-5\right)-4\left(x-3\right)=-12\)
\(\Leftrightarrow3x-15-4x+12=-12\)
\(\Leftrightarrow-x=-9\)
\(\Leftrightarrow x=9\)
Vậy x=9
1: =>3^x=81
=>x=4
2: =>2^x=8
=>x=3
3: =>x^3=2^3
=>x=2
4: =>x^20-x=0
=>x(x^19-1)=0
=>x=0 hoặc x=1
5: =>2^x=32
=>x=5
6: =>(2x+1)^3=9^3
=>2x+1=9
=>2x=8
=>x=4
7: =>x^3=115
=>\(x=\sqrt[3]{115}\)
8: =>(2x-15)^5-(2x-15)^3=0
=>(2x-15)^3*[(2x-15)^2-1]=0
=>2x-15=0 hoặc (2x-15)^2-1=0
=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1
=>x=15/2 hoặc x=8 hoặc x=7
1. Tìm số tự nhiên x biết:
1) \(3^x.3=243\)
\(3^x=243:3\)
\(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
_____
2) \(7.2^x=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
_____
3) \(x^3=8\)
\(x^3=2^3\)
\(\Rightarrow x=3\)
_____
4) \(x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x=1\)
5) \(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
_____
6) \(\left(2x+1\right)^3=9.81\)
\(\left(2x+1\right)^3=729=9^3\)
\(\rightarrow2x+1=9\)
\(2x=9-1\)
\(2x=8\)
\(x=8:2\)
\(\Rightarrow x=4\)
_____
7) \(x^6:x^3=125\)
\(x^3=125\)
\(x^3=5^3\)
\(\Rightarrow x=5\)
_____
8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)
_____
9) \(3^{x+2}-5.3^x=36\)
\(3^x.\left(3^2-5\right)=36\)
\(3^x.\left(9-5\right)=36\)
\(3^x.4=36\)
\(3^x=36:4\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
_____
10) \(7.4^{x-1}+4^{x+1}=23\)
\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)
\(4^{x-1}.\left(7+4^2\right)=23\)
\(4^{x-1}.\left(7+16\right)=23\)
\(4^{x-1}.23=23\)
\(4^{x-1}=23:23\)
\(4^{x-1}=1\)
\(4^{x-1}=4^1\)
\(\rightarrow x-1=0\)
\(x=0+1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt
\(\frac{15}{4}+\frac{15}{28}+\frac{15}{70}+...+\frac{15}{x.\left(x+3\right)}=\frac{99}{20}\)
=> \(5.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{99}{20}\)
=> \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{99}{20}:5\)
=> \(1-\frac{1}{x+3}=\frac{99}{20}.\frac{1}{5}=\frac{99}{100}\)
=> \(\frac{1}{x+3}=1-\frac{99}{100}\)
=> \(\frac{1}{x+3}=\frac{1}{100}\)
=> x + 3 = 100
=> x = 100 - 3
=> x = 97
a) Ta có: \(\dfrac{-11}{15}< \dfrac{x}{15}< \dfrac{-8}{15}\)
nên -11<x<-8
hay \(x\in\left\{-10;-9\right\}\)
b) Ta có: \(\dfrac{3}{7}< \dfrac{x}{21}< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{9}{21}< \dfrac{x}{21}< \dfrac{14}{21}\)
Suy ra: 9<x<14
hay \(x\in\left\{10;11;12;13\right\}\)
c) Ta có: \(\dfrac{-67}{21}< \dfrac{x}{168}< \dfrac{-3}{8}\)
nên \(\dfrac{-536}{168}< \dfrac{x}{168}< \dfrac{-63}{168}\)
Suy ra: -536<x<-63
hay \(x\in\left\{-535;-534;...;-64\right\}\)
TL :
( 15 x 15 ) : 2 x ( 15 x 15 ) : 2 x ( 15 x 15 ) : 2 + ( 0 + 1 + 2 + 2 + 3 + 45 ) x 0
= 0
=> vì : tất cả số nào nhân vs 0 đều = 0
_HT_
3|x| =|-15|
3|x|=15
|x| =15:3
|x|=5
=> x= 5 hoặc x = -5
Vậy x = 5 hoặc x =-5
\(3.\left|x\right|=\left|-15\right|\)
\(\Leftrightarrow3.\left|x\right|=15\)
\(\Leftrightarrow\left|x\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
Vậy \(x\in\left\{5;-5\right\}\)