\(a,3^4:3^x=3^3\)

\(3^{4-x}=3^3\)

=> 4 -x = 3

=> x = 1

Nguyễn Tấn Tài bn làm hết nhé

a, Ta có \(3^4:3^x=27\)

\(\Rightarrow3^4:3^x=3^3\Rightarrow3^x=3^4:3^3\Rightarrow3^x=3\Rightarrow x=1\)

Vậy x = 1

b, \(4^x.\frac{1}{3}+4^x.3=\frac{5}{3}.2^7\)

\(\Rightarrow4^x.\left(\frac{1}{3}+3\right)=\frac{5}{3}.2.4^3\)

\(\Rightarrow4^x.\frac{10}{3}=\frac{10}{3}.4^3\)

\(\Rightarrow4^x=4^3\)

\(\Rightarrow x=3\)

c,\(2^x.\frac{1}{4}+2.2^x=9.2^6\)

\(\Rightarrow2^x.\left(\frac{1}{4}+2\right)=9.2^6\)

\(\Rightarrow2^x.\frac{9}{2}=\frac{9}{2}.2^7\)

\(\Rightarrow2^x=2^7\)

\(\Rightarrow x=7\)

d,\(\left(8x-1\right)^{2x+1}=7^{2x+1}\)

\(\Rightarrow8x-1=7\)

\(\Rightarrow8x=8\)

\(\Rightarrow x=1\)

Bài 1:

a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)

\(=\dfrac{9}{20}-\dfrac{2}{9}\)

\(=\dfrac{41}{180}\)

b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)

\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)

\(=\dfrac{5}{6}\times\dfrac{12}{7}\)

\(=\dfrac{10}{7}\)

c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)

\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)

\(=\dfrac{7}{9}\times1\)

\(=\dfrac{7}{9}\)

Bài 2:

a) \(2\times\left(x-1\right)=4026\)

\(\left(x-1\right)=4026\div2\)

\(x-1=2013\)

\(x=2014\)

Vậy: \(x=2014\)

b) \(x\times3,7+6,3\times x=320\)

\(x\times\left(3,7+6,3\right)=320\)

\(x\times10=320\)

\(x=320\div10\)

\(x=32\)

Vậy: \(x=32\)

c) \(0,25\times3< 3< 1,02\)

\(\Leftrightarrow0,75< 3< 1,02\) ( S )

=> \(0,75< 1,02< 3\)

b) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{4}.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}-\frac{1}{2}\\x=\left(-\frac{1}{4}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}.\)

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

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

\(\Rightarrow3x+2=-3\)

\(\Rightarrow3x=\left(-3\right)-2\)

\(\Rightarrow3x=-5\)

\(\Rightarrow x=\left(-5\right):3\)

\(\Rightarrow x=-\frac{5}{3}\)

Vậy \(x=-\frac{5}{3}.\)

Chúc bạn học tốt!

Bạn ơi, gõ Công thức trực quan cho dễ nhìn đi bạn! :)

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

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

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

\(\Rightarrow x=\frac{1}{81}\)

Vậy \(x=\frac{1}{81}.\)

d) \(27^x:3^x=9\)

\(\Rightarrow\left(27:3\right)^x=9\)

\(\Rightarrow9^x=9\)

\(\Rightarrow9^x=9^1\)

\(\Rightarrow x=1\)

Vậy \(x=1.\)

e) \(\frac{16}{2^x}=2\)

\(\Rightarrow2^x=16:2\)

\(\Rightarrow2^x=8\)

\(\Rightarrow2^x=2^3\)

\(\Rightarrow x=3\)

Vậy \(x=3.\)

Chúc bạn học tốt!

           Bài làm :

b)\(\left(x-1\right)^3=-125\)

\(\Leftrightarrow\left(x-1\right)^3=\left(-5\right)^3\)

\(\Leftrightarrow x-1=-5\)

\(\Leftrightarrow x=-5+1=-4\)

c) \(2^{4-x}=32\)

\(\Leftrightarrow2^{4-x}=2^5\)

\(\Leftrightarrow4-x=5\)

\(\Leftrightarrow x=4-5=-1\)

d)\(\frac{1}{4}.2^x+2.2^x=9.2^6\)

\(\Leftrightarrow2^x.\left(\frac{1}{4}+2\right)=576\)

\(\Leftrightarrow2^x=256\)

\(\Leftrightarrow x=8\)

a. ( x - 1 )³ = - 125

<=> ( x - 1 )³ = - 5³

<=> x - 1 = - 5

<=> x = - 4

b. 2^{4 - x} = 32

2^{4 - x} = 2⁵

<=> 4 - x = 5

<=> x = - 1

c. \(\frac{1}{4}.2^x+2.2^x=9.2^6\)

\(\Leftrightarrow2^x\left(\frac{1}{4}+2\right)=9.64\)

\(\Leftrightarrow2^x.\frac{9}{4}=576\)

\(\Leftrightarrow2^x=256\)

<=> 2^x = 2⁸

<=> x = 8

\(b,\left(x-1\right)^3=-125\)

\(\left(x-1\right)^3=-\left(5\right)^3\)

\(\Rightarrow x-1=-5\)

\(\Rightarrow x=-5+1\)

\(x=-4\)

\(c,2^{4-x}=32\)

\(2^{4-x}=2^5\)

\(\Rightarrow4-x=5\)

\(\Rightarrow x=4-5\)

\(\Rightarrow x=-1\)

\(d,\frac{1}{4}.2^x+2.2^x=9.2^6\)

\(\left(\frac{1}{4}+2\right).2^x=576\)

\(\frac{9}{4}.2^x=576\)

\(2^x=576:\frac{9}{4}\)

\(2^x=256\)

\(2^x=2^8\)

\(\Rightarrow x=8\)

Học tốt

a) \(x+\left(-7\right)=-20\)

\(\Rightarrow x=-20+7\)

\(\Rightarrow x=-13\)

Vậy \(x=-13\)

b) \(8-x=-12\)

\(\Rightarrow x=8-\left(-12\right)\)

\(\Rightarrow x=20\)

Vậy \(x=20\)

c) \(|x|-7=-6\)

\(\Rightarrow|x|=-6+7\)

\(\Rightarrow|x|=1\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy \(x\in\left\{1;-1\right\}\)

d) \(5^2.2^2-7.|x|=65\)

\(\Rightarrow\left(5.2\right)^2-7.|x|=65\)

\(\Rightarrow10^2-7.|x|=65\)

\(\Rightarrow100-7.|x|=65\)

\(\Rightarrow7.|x|=35\)

\(\Rightarrow|x|=5\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

Vậy \(x\in\left\{5;-5\right\}\)

e) \(37-3.|x|=2^3-4\)

\(\Rightarrow37-3.|x|=8-4\)

\(\Rightarrow37-3.|x|=4\)

\(\Rightarrow3.|x|=33\)

\(\Rightarrow|x|=11\)

\(\Rightarrow\orbr{\begin{cases}x=11\\x=-11\end{cases}}\)

Vậy \(x\in\left\{11;-11\right\}\)

f) \(|x|+|-5|=|-37|\)

\(\Rightarrow|x|+5=37\)

\(\Rightarrow|x|=32\)

\(\Rightarrow\orbr{\begin{cases}x=32\\x=-32\end{cases}}\)

Vậy \(x\in\left\{32;-32\right\}\)

g)\(5.|x+9|=40\)

\(\Rightarrow|x+9|=8\)

\(\Rightarrow\orbr{\begin{cases}x+9=8\\x+9=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-17\end{cases}}\)

Vậy \(x\in\left\{-1;-17\right\}\)

h) \(-\frac{5}{6}+\frac{8}{3}+\frac{-29}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)

\(\Rightarrow\frac{-5}{6}+\frac{16}{6}+\frac{-29}{6}\le x\le\frac{-1}{2}+\frac{4}{2}+\frac{5}{2}\)

\(\Rightarrow-3\le x\le4\)

Vậy \(-3\le x\le4\)

câu a

x+(-7)=-20

x=-20-(-7)

x=-13

Hai bài đầu mình vừa làm cho bạn đầu kia rồi !

Làm tiếp nha ^^

\(2\cdot2^2+2^{x+2}=2^8+2^5\)

\(2^{2+1}\cdot2^{x+2}=2^{13}\)

\(2^{2+1+x+2}=2^{13}\)

\(2^{x+5}=2^{13}\)

\(\Rightarrow\text{ }x+5=13\)

\(x=13-5\)

\(x=8\)

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)

\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)

\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)

\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)

Vậy A = \(\frac{1}{2}\)

\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(B=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)

\(B=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+2^{20}.3^{10}}\)

\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)

Vậy B = \(\frac{1}{2}\)