\(\dfrac{1}{5}\)=0,8:(x+1)

2|3x+1|-27 = -32

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22 tháng 12 2021

b: =>2|3x+1|=18

=>|3x+1|=9

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=9\\3x+1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

a) 27x : 3x = 9

(27 : 3)x = 9

9x = 91

x = 1

b) 25 : 5x =5

5x = 25 : 5

5x = 51

x = 1

c) 2 : (x + 2)2 = \(\dfrac{1}{18}\)

(x + 2)2 = 2 : \(\dfrac{1}{18}\)

(x + 2)2 = 36

\(\Rightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)

d) (5x - 1)2 = \(\dfrac{36}{49}\)

(5x - 1)2 = \(\left(\dfrac{6}{7}\right)^2\)

Bạn làm tiếp nha, mình có việc bận :v

23 tháng 9 2017

a.\(3^{x-1}=243\)

\(3^x:3^1=243\)

\(3^x=729\)

\(\Leftrightarrow3^6=729\)

\(\Leftrightarrow x=6\)

b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)

\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)

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

Câu b tính đến đây rồi không mò đc x nữa.

5 tháng 7 2017

1) Tính

a) 253 : 52 = (52)3 : 52 = 56 : 52 = 54 = 625

\(b)\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^6=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^6=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^9\) d) 9 . 32 . \(\dfrac{1}{81}\) . 32 = 32 . 32 . \(\dfrac{1}{3^4}\) . 32 = 9

2) Tìm x thuộc Q, biết:

a) 3x + 2 = 27

=> 3x + 2 = 33

x + 2 = 3

x = 3 - 2

x = 1

b) \(\left(\dfrac{1}{2}x-3\right)^4=81\)

\(\Rightarrow\left(\dfrac{1}{2}x-3\right)^4=3^4\)

\(\dfrac{1}{2}x-3=3^{ }\)

\(\dfrac{1}{2}x=3+3\)

\(\dfrac{1}{2}x=9\)

\(x=9:\dfrac{1}{2}\)

\(x=18\)

c) \(\left(x-\dfrac{1}{2}\right)^3=-27\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(-3\right)^3\)

\(x-\dfrac{1}{2}=-3\)

\(x=-3+\dfrac{1}{2}\)

\(x=\dfrac{-5}{2}\)

d) 5 . 5x + 1 = 125

5x + 1 = 125 : 5

5x + 1 = 25

5x + 1 = 52

x + 1 = 2

x = 2 - 1

x = 1.

14 tháng 6 2017

a.

| x | = 5,6

=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)

Vậy \(x\in\left\{-5,6;5,6\right\}\)

b, \(\left|x-3,5\right|=5\)

=>\(\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)

Vậy \(x\in\left\{-1,5;8,5\right\}\)

c,\(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

=> \(\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{4};\dfrac{5}{4}\right\}\)

d,\(\left|4x\right|-\left(\left|-13,5\right|\right)=\left|\dfrac{1}{4}\right|\)

=> \(\left|4x\right|-13,5=\dfrac{1}{4}\)

=> \(\left|4x\right|=13,75\)

=>\(\left[{}\begin{matrix}4x=13,75\\4x=-13,75\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3,4375\\x=-3,4375\end{matrix}\right.\)

Vậy \(x\in\left\{-3,4375;3,4375\right\}\)

14 tháng 6 2017

e, ( x - 1 ) 3 = 27

=> x - 1 = 3

=> x = 4

Vậy x = 4

f, ( 2x - 3)2 = 36

=> \(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=4,5\\x=-1,5\end{matrix}\right.\)

Vậy x\(\in\left\{-1,5;4,5\right\}\)

g, \(5^{x+2}=625\)

=> \(5^{x+2}=5^4\)

=> x + 2 = 4

=> x = 2

Vậy x = 2

h, ( 2x - 1)3 = -8

=> 2x - 1 = -2

=> x = \(\dfrac{-1}{2}\)

Vậy x = \(\dfrac{-1}{2}\)

i, \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)

=> \(\dfrac{1.2.3.4.5...30.31}{4.6.8.10.12...62.64}=2^x\)

=>\(\dfrac{1.2.3.4.5...30.31}{\left(2.3.4.5...30.31.32\right)\left(2.2.2.2...2.2_{ }\right)}=2^x\)(có 31 số 2)

=> \(\dfrac{1}{32.2^{31}}=2^x\)

=> \(\dfrac{1}{2^{36}}=2^x\)

=> x = -36

Vậy x = -36

16 tháng 11 2017

a) \(\left(x-3\right)^2=1\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=1^2\\\left(x-3\right)^2=-1^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

b) \(\left(x-\dfrac{1}{7}\right)^2=0\)

\(\Rightarrow x-\dfrac{1}{7}=0\)

\(\Rightarrow x=0+\dfrac{1}{7}\)

\(\Rightarrow x=\dfrac{1}{7}\)

c) \(\left(2x+3\right)^3=-27\)

\(\Rightarrow\left(2x+3\right)^3=\left(-3\right)^3\)

\(\Rightarrow2x+3=-3\)

\(\Rightarrow2x=-6\)

\(\Rightarrow x=-3\)

d) \(-\left(5+35x\right)^2=36\)

\(\Rightarrow\left[{}\begin{matrix}\left(-5-35x\right)^2=6^2\\\left(-5-35x\right)^2=\left(-6\right)^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-5-35x=6\\-5-35x=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}35x=-11\\35x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{35}\\x=\dfrac{1}{35}\end{matrix}\right.\)

16 tháng 11 2017

a) (x-3)mũ 2 = 1

Vậy x-3 = 1( vì 1 mũ 2 sẽ bằng 1)

=> x = 1+3 = 4

b) (x - 1/7) mũ 2 = 0

Vậy x - 1/7 = 0 ( vì 0 mũ 2 sẽ bằng 0)

=> x = 0 + 1/7 = 1/7

c) (2x + 3 ) mũ 3 = -27

vậy 2x + 3 = -3 ( vì -3 mũ 3 sẽ bằng -27)

=> 2x = -3-3 = -6

=> x = -6/2 = -3

22 tháng 7 2018

\(3x\left(x-1\right)+5\left(2-x\right)=3x^2-7x+6\) \(6\)

<=> \(3x^2-3x+10-5x=3x^2-7x+6\)

<=> \(-x=-4\)

<=> \(x=4\)

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

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

<=> \(\hept{\begin{cases}x+2=\sqrt{\frac{1}{6}}\\x+2=-\sqrt{\frac{1}{6}}\end{cases}}\)

<=> \(\hept{\begin{cases}x=\sqrt{\frac{1}{6}}-2\\x=-\sqrt{\frac{1}{6}}-2\end{cases}}\)

19 tháng 11 2022

a: =>1/6x=-49/60

=>x=-49/60:1/6=-49/60*6=-49/10

b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2

=>x=17/15 hoặc x=-13/15

c: =>1,25-4/5x=-5

=>4/5x=1,25+5=6,25

=>x=125/16

d: =>2^x*17=544

=>2^x=32

=>x=5

i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5

=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2

=>x=14,4 hoặc x=9,6

j: =>(2x-1)(2x+1)=0

=>x=1/2 hoặc x=-1/2

18 tháng 7 2017

Câu 1 :

\(\text{a) }B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\\ B=\dfrac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot\left(2^3\cdot3\cdot5\right)}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-\left(2\cdot3\right)^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\\ B=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\\ B=\dfrac{2\cdot6}{3\cdot5}\\ B=\dfrac{4}{5}\\ \)

\(\text{b) }C=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\\ C=\dfrac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\\ C=\dfrac{2^{29}\cdot3^{18}\left(10-9\right)}{2^{28}\cdot3^{18}\left(15-14\right)}\\ C=\dfrac{2^{29}\cdot3^{18}}{2^{28}\cdot3^{18}}\\ C=2\\ \)

\(\text{c) }D=\dfrac{49^{24}\cdot125^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot4^5}{5^{29}\cdot16^2\cdot7^{48}}\\ D=\dfrac{\left(7^2\right)^{24}\cdot\left(5^3\right)^{10}\cdot2^8-5^{30}\cdot7^{49}\cdot\left(2^2\right)^5}{5^{29}\cdot\left(2^4\right)^2\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{5^{29}\cdot2^8\cdot7^{48}}\\ D=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1-28\right)}{5^{29}\cdot2^8\cdot7^{48}}\\ D=5\cdot\left(-27\right)\\ D=-135\)

18 tháng 7 2017

Câu 2 :

\(\text{a) }9^{x+1}-5\cdot3^{2x}=324\\ \Leftrightarrow9^x\cdot9-5\cdot9^x=81\cdot4\\ \Leftrightarrow9^x\left(9-5\right)=9^2\cdot4\\ \Leftrightarrow9^x\cdot4=9^2\cdot4\\ \Leftrightarrow9^x=9^2\\ \Leftrightarrow x=2\\ \text{Vậy }x=2\\ \)

Sorry . Mình chỉ biết đến đây thôi

8 tháng 7 2018

câu a) \(A=3x^3+7x^2+3x-\left(\dfrac{1}{4}+3x^3\right)-3\dfrac{3}{4}\)

\(\Leftrightarrow A=3x^3+7x^2+3x-\dfrac{1}{4}-3x^3-\dfrac{15}{4}\)

\(\Leftrightarrow A=7x^2+3x-4\)

\(B=x\left(x^2-x+1\right)-\dfrac{1}{2}x^2\left(2x-4\right)-2\)

\(\Leftrightarrow B=x^3-x^2+x-x^3+2x^2-2\)

\(\Leftrightarrow B=x^2+x-2\)

câu b) chỉ cần thế \(x=-1\) vào biểu thức \(A\) \(\Rightarrow\) tính

và thế \(x=\dfrac{1}{2}\) vào biểu thức \(B\) \(\Rightarrow\) tính

câu c) ta có \(B+M=A\Leftrightarrow x^2+x-2+M=7x^2+3x-4\)

\(\Leftrightarrow M=7x^2+3x-4-\left(x^2+x-2\right)=6x^2+2x-2\)

câu d) ta có : \(\dfrac{x+5}{-3}=\dfrac{x}{2}\Leftrightarrow2\left(x+5\right)=-3x\Leftrightarrow2x+10=-3x\)

\(\Leftrightarrow5x=-10\Leftrightarrow x=-2\)

thế \(x=-2\) vào \(M=6x^2+2x-2=6.\left(-2\right)^2+2\left(-2\right)-2=18\)