Bài 1:

a. Ta có: $1953=3^2.7.31$

$777=3.7.37$

$\Rightarrow UCLN(1953, 777) = 3.7=21$

b. Đề khó hiểu quá. Bạn xem lại/

Bài 2: Thiếu dấu giữa các số. Bạn xem lại.

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

a) x : 11 = y : 7 

=> x/7 = y/11 và  x + y = -54 Thay vào ta có :

     x/7 = y/11 = (x+y)/(7+11) = -54/18= -3

=> x = -3.7 = -27

=> y = -3.11 = -33

a) x+3=11

    x=11-3=8

b) 58-x=54

        x=58-54=4

c) 67+3=x

   x=60 

a) 8

b) 4

c) 70

a) x=11-3=8

b) x=58-54=4

c) x=67+3=70

mk xog đầu tiên k đúng cho mk nha!

bai toan nay kho

câu d vs e t nghĩ cx dễ thui nhưng mấy câu kia bó tay lun

có bn nào thông minh lm hộ mk đc k z !!!!!!

theo dãy tỉ số bằng nhau ta có:

\(\frac{x}{11}=\frac{y}{7}=\frac{x+y}{11+7}=\frac{54}{18}=3\)

=> \(\frac{x}{11}=3\Rightarrow x=3.11=33\)

=>\(\frac{y}{7}=3\Rightarrow y=3.7=21\)

C1:

a)    135  x   54    +    45    x  135    +   135

= 135 x ( 54 + 45 + 1)

=135x100=13500

C2

a)    135  x   54    +    45    x  135    +   135

= 7290 + 6075 +135

= 13500

a, x + 26 = 39 + (-5)

   x + 26 = 39 - 5

   x + 26 = 34

   x        = 34 - 26

   x        = 8

b, |x| + 50 = (-|-26|) +76

    |x| +50 = (-26) +76

    |x| +50 = 76 - 26

    |x| +50 = 50

    |x|        = 50 -50

    |x|        = 0

c, |x| + 17 = (-21)+19

    |x| + 17 = - (21-19)

    |x| + 17 = - 2

    |x| không tồn tại

d. |x-21|= 2.(|-11| + |-3|)

    |x-21|= 2.(11+3)

    |x-21|= 2.14

    |x-21|= 28

     x     = 28+21 hoặc 21-28

     x     = 49      hoặc -7

e. hình như lộn đề vì |x-54|-7|

a) 40 - x = 54

x = 40 - 54

x = -14

b) 36 - x = -11

x = 36 - (-11)

x = 47

c) 2.x - 3 = -14

2x = -14 + 3

2x = -11

x = -11/2

a) => \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}=\frac{24}{54}=\frac{4}{9}\)

=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\) => \(x=\frac{6}{5}.\left(\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\right)\)

b) \(\frac{1}{3}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\) => \(\left(\frac{1}{2}x-1\right)^4=\frac{3}{48}=\frac{1}{16}\)

=> \(\frac{1}{2}x-1=\frac{1}{2}\) hoặc  \(\frac{1}{2}x-1=-\frac{1}{2}\)

=> \(\frac{1}{2}x=\frac{3}{2}\) hoặc \(\frac{1}{2}x=\frac{1}{2}\) => x = 3 hoặc x = 1

c) \(\left(1+5\right).\left(\frac{3}{5}\right)^{x-1}=\frac{54}{25}\) => \(\left(\frac{3}{5}\right)^{x-1}=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)

=> x - 1= 2 => x = 3

d) \(\left(1+\left(\frac{2}{3}\right)^2\right).\left(\frac{2}{3}\right)^x=\frac{101}{243}\) => \(\frac{13}{9}.\left(\frac{2}{3}\right)^x=\frac{101}{243}\)

=> \(\left(\frac{2}{3}\right)^x=\frac{101}{243}:\frac{13}{9}=\frac{101}{351}\) (có lẽ đề sai)

2) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\); \(\frac{1}{81^8}=\frac{1}{\left(3^4\right)^8}=\frac{1}{3^{32}}\)

Vì 3³³ > 3³² => \(\frac{1}{3^{33}}\) < \(\frac{1}{3^{32}}\) => \(\frac{1}{27^{11}}\) < \(\frac{1}{81^8}\)

b) \(\frac{1}{3^{99}}=\frac{1}{\left(3^3\right)^{33}}=\frac{1}{27^{33}}