a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)

\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)

\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)

\(\Rightarrow x+3=-3\)

\(\Rightarrow x=-6\)

b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)

\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)

\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)

\(\Rightarrow2x+2=-2\)

\(\Rightarrow2x=-4\)

\(\Rightarrow x=-2\)

c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)

\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)

\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)

\(\Rightarrow x+5=-4\)

\(\Rightarrow x=-9\)

d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)

\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)

\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)

e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)

\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)

\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)

\(\Rightarrow2x=18\)

\(\Rightarrow x=9\)

f)\(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)

\(5^{x+3}\cdot5-3\cdot5^{x+3}=2\cdot5^{11}\)

\(5^{x+3}\left(5-3\right)=2\cdot5^{11}\)

\(5^{x+3}\cdot2=2\cdot5^{11}\)

\(\Rightarrow5^{x+3}=5^{11}\)

\(\Rightarrow x+3=11\)

\(\Rightarrow x=8\)

r)\(4\cdot3^{x-1}+2\cdot3^{x+2}=4\cdot3^6+2\cdot3^9\)

\(4\cdot3^x:3+2\cdot3^x\cdot9=4.3^7:3+2\cdot3^7\cdot9\)

\(3^x\left(4:3+2\cdot9\right)=3^7\left(4:3+2\cdot9\right)\)

\(\Rightarrow3^x=3^7\)

\(\Rightarrow x=7\)

a) \(\left(x-1\right)^3=8=2^3\)

\(x-1=2\)

\(x=2+1=3\)

b) \(7^{2x-6}=49=7^2\)

\(2x-6=2\)

\(2x=6+2=8\)

\(x=8:2=4\)

c) \(\left(2x-14\right)^7=128=2^7\)

\(2x-14=2\)

\(2x=14+2=16\)

\(x=16:2=8\)

d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)

\(x=5\)

e) \(3\cdot\left(x+2\right):7\cdot4=120\)

\(x+2=120:3\cdot7:4\)

\(x+2=70\)

\(x=70-2=68\)

Lời giải:

a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$

$\Rightarrow x=3$

b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$

$\Rightarrow 2x=8$

$\Rightarrow x=4$

c. $(2x-14)^7=128=2^7$

$\Rightarrow 2x-14=2$

$\Rightarrow 2x=16$

$\Rightarrow x=18$

d.

$x^4.x^5=5^3.5^6$

$x^9=5^9$

$\Rightarrow x=5$

e. 

$3(x+2):7=120:4=30$

$3(x+2)=30.7=210$

$x+2=210:3=70$

$x=70-2=68$

x=2009 dễ mà

mk làm câu c cho nó dễ

c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010

=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010

=1-1/x+1=2009/2010

=1/x+1=1-2009/2010

=1/x+1=1/2010

=) x+1=2010

x         =2010-1

x         =2009

Đề cho dài :v. Lần sau đăng từ từ nhé bạn, hôm qua đến giờ mình giải không hết đó =(((

a) \(\frac{1}{2}.x-\frac{3}{4}.x-\frac{7}{3}=-\frac{5}{6}=\frac{-5}{6}\)

\(\frac{1}{2}.x-\frac{3}{4}.x=\frac{-5}{6}+\frac{7}{3}=\frac{3}{2}\)

\(\Leftrightarrow x\left(\frac{1}{2}-\frac{3}{4}\right)=\frac{3}{2}\Leftrightarrow x.\frac{-1}{4}=\frac{3}{2}\)

\(x=\frac{3}{2}:\frac{-1}{4}=-6\)

b) \(\frac{4}{5}.x-x-\frac{3}{2}.x+\frac{4}{3}=\frac{1}{2}-\frac{6}{5}=-\frac{7}{10}\)

\(\Leftrightarrow x\left(\frac{4}{5}-\frac{3}{2}.\frac{4}{3}\right)=x\left(\frac{4}{5}-2\right)=-\frac{7}{10}\)

\(\Leftrightarrow x.\frac{-6}{5}=-\frac{7}{10}\)

\(x=-\frac{7}{10}:\frac{-6}{5}=\frac{7}{12}\)

c) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{2010}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}\)

\(=1-\frac{1}{x+1}=\frac{2009}{2010}\)

\(\frac{1}{x+1}=1-\frac{2009}{2010}=\frac{1}{2010}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2010-1}=\frac{1}{2009}\). Vậy x= 2009

d) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}=\frac{4023}{2015}\)

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4023}{2015}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{4023}{2015}\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4023}{2015}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{4023}{2015}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{4023}{2015}:2=\frac{4023}{4030}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{4023}{4030}=\frac{-1004}{2015}=\frac{1004}{-2015}\)

\(x+1=\hept{\begin{cases}2015\\-2015\end{cases}}\Rightarrow x=\hept{\begin{cases}2014\\-2016\end{cases}}\)

e) Bạn tự làm, nhiều quá mình làm không hết

a, => 2^x = (2^3)^4/(2^4)^3 = 2^12/2^12 = 1 = 2^0

=> x = 0

c, => 4^x = 4^10.(4-3) = 4^10

=> x=10

d, => 2^2.3^x-1 + 2.3^x.9 = 2^2.3^6+2.3^9

=> 2.3^x-1 . (2+3.9) = 2.3^6.(2+3^3)

=> 2.3^x-1 . 27 = 2.3^6 . 27

=> 3^x-1 = 3^6

=> x-1 = 6

=> x = 7

e, => 2^x.(1/3+1/6+2) = 2^11.(2+1/2)

=> 2^x. 5/2 = 2^11. 5/2

=> 2^x = 2^11

=> x = 11

Tk mk nha

câu b) chưa có ai làm thì mình làm nốt vậy 

\(\left(-2\right)^2=-4^6-8^5\)

\(\left(-2\right)^x=-4096-32768\)

\(\left(-2\right)^x=-36864\)

\(\Rightarrow x\) sẽ 1 số thập phân nào đó 

\(5^{x+4}-3.5^{x+3}=2.5^{11}\)

\(5^{x+3}\left(5-3\right)=2.5^{11}\)

\(5^{x+3}.2=2.5^{11}\)

\(5^{x+3}=5^{11}\)

\(x+3=11\)

\(x=8\)

\(4^{x+3}-3.4^{x+1}=13.4^{11}\)

\(4^{x+1}\left(4^2-3\right)=13.4^{11}\)

\(4^{x+1}.13=13.4^{11}\)

\(4^{x+1}=4^{11}\)

\(x+1=11\)

\(x=10\)

a) 32 - 6 . (8 - 2³) + 18 =  32 - 6 . (8 - 8) + 18

= 32 - 6 . 0 + 18 = 32 + 18 = 50

b) (3 . 5 - 9)³ . (1 + 2 . 3)² + 4²

= (15 - 9)³ . (1 + 6)² + 4²

= 6³ . 7² + 4² = 216 . 49 + 16 = 10 584 + 16 = 10 600

a) 32 - 6 . (8 - 2³) + 18 =  32 - 6 . (8 - 8) + 18

= 32 - 6 . 0 + 18 = 32 + 18 = 50

b) (3 . 5 - 9)³ . (1 + 2 . 3)² + 4²

= (15 - 9)³ . (1 + 6)² + 4²

= 6³ . 7² + 4² = 216 . 49 + 16 = 10 584 + 16 = 10 600

\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)

\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)

\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

Mình làm tiếp câu b nha !

b,                                               Bài giải

\(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)

\(\left(\frac{x-2018}{2}+1\right)+\left(\frac{x-2020}{4}+1\right)=\left(\frac{x-2040}{8}+1\right)+\left(\frac{x-2030}{14}+1\right)\)

\(\frac{x-2016}{2}+\frac{x-2016}{4}=\frac{x-2032}{8}+\frac{x-2016}{14}\)

\(\left(x-2016\right)\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{x-2016}{8}-2+\frac{x-2016}{14}\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\left(\frac{1}{8}+\frac{1}{14}\right)-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\cdot\frac{11}{56}-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}-\left(x-2016\right)\cdot\frac{11}{56}=-2\)

\(\left(x-2016\right)\left(\frac{3}{4}-\frac{11}{56}\right)=-2\)

\(\left(x-2016\right)\cdot\frac{31}{56}=-2\)

\(x-2016=-2\text{ : }\frac{31}{56}\)

\(x-2016=-\frac{112}{31}\)

\(x=-\frac{112}{31}+2016\)

\(x=\frac{62384}{31}\)

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

a: 3-x=x+1,8

=>-2x=-1,2

=>x=0,6

b: 2x-5=7x+35

=>-5x=40

=>x=-8

c: 2(x+10)=3(x-6)

=>3x-18=2x+20

=>x=38

d; 8(x-3/8)+1=6(1/6+x)+x

=>8x-3+1=1+6x+x

=>8x-2=7x+1

=>x=3

e: =>-3x+x=4/3-2/9

=>-2x=12/9-2/9=10/9

=>x=-5/9

g: =>3/4x-1/2x=5/6+1/2

=>1/4x=5/6+3/6=8/6=4/3

=>x=4/3*4=16/3

h: =>x-4=-x+5

=>2x=9

=>x=9/2