?

:>?

?

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.

b) 2.x + 3 = 9 

⇒2.x = 9 -3 

⇒  2.x = 6

⇔ x = 3

Vậy x = 3.

\(\frac{3}{3.6.9}+\frac{3}{6.9.12}+\frac{3}{9.12.15}+\frac{3}{12.15.18}=\frac{3}{6}\left(\frac{6}{3.6.9}+\frac{6}{6.9.12}+\frac{6}{9.12.15}+\frac{6}{12.15.18}\right)\)

= \(\frac{1}{2}\left(\frac{1}{3.6}-\frac{1}{6.9}+\frac{1}{6.9}-\frac{1}{9.12}+\frac{1}{9.12}-\frac{1}{12.15}+\frac{1}{12.15}-\frac{1}{15.18}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3.6}-\frac{1}{15.18}\right)=\frac{1}{2}.\frac{14}{270}=\frac{7}{270}\)

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...

d, x=-110

\(a,3\left(x-5\right)-4\left(x-3\right)=-12\)

\(\Leftrightarrow3x-15-4x+12=-12\)

\(\Leftrightarrow-x=-9\)

\(\Leftrightarrow x=9\)

Vậy x=9

\(c,20-\left(x-15\right)=x-15\)

\(\Leftrightarrow20-x+15=x-15\)

\(\Leftrightarrow2x=50\)

\(\Leftrightarrow x=25\)

Vậy x=25

x=-36

x=11

x=100

x=14

x=-36

x=11

x=100

x=14

2: \(\Leftrightarrow\left(\dfrac{x+1}{12}-1\right)+\left(\dfrac{x+2}{13}-1\right)=\left(\dfrac{x+3}{14}-1\right)+\left(\dfrac{x+4}{15}-1\right)\)

=>x-11=0

=>x=11

3: \(\Leftrightarrow\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)

=>x-100=0

=>x=100

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 \(x\in\left\{-535;-534;...;-64\right\}\)

75 nhé bạn

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_

1423828.125 

        - ht -