a) =7
b) =16
c) =69

189 : (3 x 9)                    672 : (6 x 7)

= 189 : 27                      = 672 : 42

= 7                                 = 16            

1656 : (3 x 8) 

= 1656 : 24

= 69

a. 189 : (3 x 9) 

= 189 : 27 = 7.

b. 672 : (6 x 7)

= 672 : 42 = 16.

c. 1656 : (3 x 8)

= 1656 : 24 = 69.

\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)

=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)

\(\Rightarrow7x< 36< 7x+7\)

\(\Rightarrow x< \dfrac{36}{7}< x+1\)

\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)

\(\Rightarrow x=5\)

tik cho mình nhé

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\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)

\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)

7\(x\) < 36 < 63\(x\) + 7

⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)

\(\dfrac{29}{63}\)<  \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}

⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)

\(2x\cdot\dfrac{-4}{9}+2x\cdot\dfrac{-5}{9}=\dfrac{8}{11}\)

\(2x\cdot\left(\dfrac{-4}{9}+\dfrac{-5}{9}\right)=\dfrac{8}{11}\)

\(2x\cdot\left(-1\right)=\dfrac{8}{11}\)

2x = \(\dfrac{8}{11}:\left(-1\right)=\dfrac{-8}{11}\)

x = \(\dfrac{-8}{11}:2=\dfrac{-4}{11}\)

       4/3 x 5/4 x 6/5 x 7/6 x  8/7 x 9/8 

   =  4/9   

a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)

b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)

c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)

d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)

Mong bạn xem lại đề bài.

Đặt \(A=2^{x+9}-2^{x+8}-2^{x+7}-...-2^{x+1}-2^x\)

\(\Rightarrow2A=2\left(2^{x+9}-2^{x+8}-...-2^x\right)\)

\(\Rightarrow2A=2^{x+9}.2^1-2^{x+8}.2^1-...-2^x.2^1\)

\(\Rightarrow2A=2^{x+10}-2^{x+9}-...-2^{x+1}\)

\(\Rightarrow A=2A-A=2^{x+10}-2^{x+9}-...-2^{x+1}-\left(2^{x+9}-2^{x+8}-...-2^{x+1}-2^x\right)=2^{x+10}-2^{x+9}-2^{x+9}+2^x\)

\(\Rightarrow A=2^{x+10}-2.2^{x+9}+2^x=2^{x+10}-2^{x+10}+2^x=2^x\)

\(\Rightarrow2^x=1024\Rightarrow x=10\)

Bạn ơi bạn ghi rõ lại đề cho mik đi ạ khó đọc lắm ạ

#)Giải :

\(\left(\left|x\right|-\frac{1}{8}\right)\left(-\frac{1}{8}\right)^5=\left(-\frac{1}{8}\right)^7\)

\(\Leftrightarrow\left(\left|x\right|-\frac{1}{8}\right)=\left(-\frac{1}{8}\right)^2\)

\(\Leftrightarrow\left(\left|x\right|-\frac{1}{8}\right)=\frac{1}{64}\)

\(\Leftrightarrow\left|x\right|=\frac{9}{64}\)

\(\Leftrightarrow\hept{\begin{cases}x>0\\x< 0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{9}{64}\\x=-\frac{9}{64}\end{cases}}}\)

\(9x-5x+3x=56\)

<=>  \(7x=56\)

<=> \(x=8\)

Vậy...

\(9x-x+3x=165\)

<=> \(11x=165\)

<=> \(x=15\)

Vậy....

\(7x+8x-x=168\)

<=> \(14x=168\)

<=> \(x=12\)

Vậy...

\(\frac{5}{8}x-\frac{1}{3}x-\frac{1}{6}x=15\)

<=> \(\frac{1}{8}x=15\)

<=> \(x=120\)

Vậy...

Dấu . là dấu nhân nha

---

a, 9 . x - 5 . x + 3 . x = 56

x . (9 - 5 + 3) = 56

x . 7 = 56

x = 56 : 7

x = 8

b, 9 . x - x + 3 . x = 165

x . (9 - 1 + 3) = 165

x . 11 = 165

x = 165 : 11

x = 15

c, 7 . x + 8 . x - x = 168

x . (7 + 8 - 1) = 168

x . 14 = 168

x = 168 : 14

x = 12

d, 5/8 . x - 1/3 . x - 1/6 . x = 15

x . (5/8 - 1/3 - 1/6) = 15

x . 1/8 = 15

x = 15 : 1/8

x = 120