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18 tháng 10 2020

a) Ta có: \(48-3\left(x+5\right)=24\)

\(\Leftrightarrow48-3x-15-24=0\)

\(\Leftrightarrow9-3x=0\)

\(\Leftrightarrow3x=9\)

hay x=3

Vậy: x=3

b) Ta có: \(9x-2x=7^8:7^6\)

\(\Leftrightarrow7x=7^2\)

\(\Leftrightarrow x=7\)

Vậy: x=7

18 tháng 10 2020

a) Ta có: \(48\cdot19+48\cdot115+134\cdot52\)

\(=48\cdot\left(19+115\right)+134\cdot52\)

\(=48\cdot134+52\cdot134\)

\(=134\cdot\left(48+52\right)\)

\(=134\cdot100=13400\)

b) Ta có: \(7^3\cdot2+8^4:64-5\)

\(=343\cdot2+8^2-5\)

\(=686+64-5\)

\(=745\)

c) Ta có: \(\left[5\cdot36-4\cdot\left(82-7\cdot11^2\right)\right]+11\cdot4\)

\(=\left[180-4\cdot\left(82-7\cdot121\right)\right]+44\)

\(=\left(180-328+3388\right)+44\)

\(=3284\)

18 tháng 7 2023

\(A=5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9\)

\(A=5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9\)

\(A=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}\)

\(A=5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}\)

\(A=2^{29}\cdot3^{18}\cdot\left(5\cdot2^1\cdot1-1\cdot3^2\right)\)

\(A=2^{29}\cdot3^{18}\cdot\left(5-9\right)\)

\(A=-2^2\cdot2^{29}\cdot3^{18}\)

\(A=-2^{31}\cdot3^{18}\)

_______________

\(B=5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6\)

\(B=5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6\)

\(B=5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)

\(B=2^{28}\cdot3^{18}\cdot\left(5\cdot1\cdot3-7\cdot2\cdot1\right)\)

\(B=2^{28}\cdot3^{18}\cdot\left(15-14\right)\)

\(B=2^{28}\cdot3^{18}\)

Ta có: \(A:B\)

\(=\left(-2^{31}\cdot3^{18}\right):\left(2^{28}\cdot3^{18}\right)\)

\(=\left(-2^{31}:2^{28}\right)\cdot\left(3^{18}:3^{18}\right)\)

\(=-2^3\cdot1\)

\(=-8\)

20 tháng 7 2023

a) \(5\left(x+7\right)-10=2^3\cdot5\)

\(\Rightarrow5\left(x+7\right)-10=40\)

\(\Rightarrow5\left(x+7\right)=40+10\)

\(\Rightarrow x+7=\dfrac{50}{5}\)

\(\Rightarrow x+7=10\)

\(\Rightarrow x=10-7\)

\(\Rightarrow x=3\)

b) \(9x-2\cdot3^2=3^4\)

\(\Rightarrow9x-18=81\)

\(\Rightarrow9x=81+18\)

\(\Rightarrow9x=99\)

\(\Rightarrow x=\dfrac{99}{9}\)

\(\Rightarrow x=11\)

c) \(5^{25}\cdot5^{x-1}=5^{25}\)

\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)

\(\Rightarrow5^{x-1}=1\)

\(\Rightarrow5^{x-1}=5^0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

20 tháng 7 2023

a) 5(�+7)−10=23⋅5

⇒5(�+7)−10=40

⇒5(�+7)=40+10

⇒�+7=505

⇒�+7=10

⇒�=10−7

⇒�=3

b) 9�−2⋅32=34

⇒9�−18=81

⇒9�=81+18

⇒9�=99

⇒�=999

⇒�=11

c) 525⋅5�−1=525

⇒5�−1=525:525

⇒5�−1=1

⇒5�−1=50

⇒�−1=0

⇒�=1

6 tháng 12 2016

may cai bai day ma cung khong biet oc cho

6 tháng 12 2016

a) 89-(73-x)=20

=>73-x=89-20

=>73-x=69

=>x=73-69

=>x=4

19 tháng 12 2018

\(a,\left|x\right|-2=7-\left(-8\right)\)

\(\Rightarrow\left|x\right|-2=15\)

\(\Rightarrow\left|x\right|=17\)

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

\(b,5^{2x-3}-2.5^2=5^2.3\)

\(\Rightarrow5^{2x-3}-50=75\)

\(\Rightarrow5^{2x-3}=125\)

\(\Rightarrow5^{2x-3}=5^3\)

\(\Rightarrow2x-3=3\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

19 tháng 12 2018

Cảm ơn bạn

12 tháng 8 2017

a, \(390-\left(x-7\right)=13^2:12\)

\(390-\left(x-7\right)=\) \(\dfrac{169}{12}\)

\(x-7=390-\dfrac{169}{12}\)

\(x-7=\dfrac{4511}{12}\)

\(x=\dfrac{4511}{12}+7\)

\(x=\dfrac{4595}{12}\)

Vậy ...

b, \(\left(x-35.2^2\right):7=3^3-24\)

\(\left(x-35.4\right):7=27-24\)

\(\left(x-140\right):7=3\)

\(\Leftrightarrow\left(x-140\right)=3.7\)

\(\Leftrightarrow x-140=21\)

\(\Leftrightarrow x=161\)

Vậy .....

c) \(x-6:2-\left(4^2.3-24\right):2:6=3\)

\(x-3-\left(16.3-24\right):2:6=3\)

\(x-3-\left(48-24\right):2:6=3\)

\(x-3-24:2:6=3\)

\(x-3-2=3\)

\(x=3+2+3\)

\(x=8\)

Vậy ......

d) \(4x-5=5+5^2+5^3+.....+5^{99}\)

Đặt :

\(A=5+5^2+.........+5^{99}\)

\(\Leftrightarrow5A=5^2+5^3+..........+5^{100}\)

\(\Leftrightarrow5A-A=\left(5^2+5^3+......+5^{100}\right)-\left(5+5^2+....+5^{99}\right)\)

\(\Leftrightarrow4A=5^{100}-5\)

\(\Leftrightarrow A=\dfrac{5^{100}-5}{4}\)

\(\Leftrightarrow4x+5=\dfrac{5^{100}-5}{4}\)

Đến đây thì sao nữa nhỉ ?

e) \(\left(2x-1\right)^4=625\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=5\\\left(2x-1\right)^4=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

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

Vậy ....