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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: \(\left[5\cdot36-4\cdot\left(82-7\cdot11^2\right)\right]+11\cdot4\)
\(=\left[180-4\cdot\left(82-7\cdot121\right)\right]+44\)
\(=\left(180-328+3388\right)+44\)
\(=3284\)
\(A=5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9\)
\(A=5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9\)
\(A=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}\)
\(A=5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}\)
\(A=2^{29}\cdot3^{18}\cdot\left(5\cdot2^1\cdot1-1\cdot3^2\right)\)
\(A=2^{29}\cdot3^{18}\cdot\left(5-9\right)\)
\(A=-2^2\cdot2^{29}\cdot3^{18}\)
\(A=-2^{31}\cdot3^{18}\)
_______________
\(B=5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6\)
\(B=5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6\)
\(B=5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)
\(B=2^{28}\cdot3^{18}\cdot\left(5\cdot1\cdot3-7\cdot2\cdot1\right)\)
\(B=2^{28}\cdot3^{18}\cdot\left(15-14\right)\)
\(B=2^{28}\cdot3^{18}\)
Ta có: \(A:B\)
\(=\left(-2^{31}\cdot3^{18}\right):\left(2^{28}\cdot3^{18}\right)\)
\(=\left(-2^{31}:2^{28}\right)\cdot\left(3^{18}:3^{18}\right)\)
\(=-2^3\cdot1\)
\(=-8\)
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) \(5^{25}\cdot5^{x-1}=5^{25}\)
\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)
\(\Rightarrow5^{x-1}=1\)
\(\Rightarrow5^{x-1}=5^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
\(a,\left|x\right|-2=7-\left(-8\right)\)
\(\Rightarrow\left|x\right|-2=15\)
\(\Rightarrow\left|x\right|=17\)
\(\Rightarrow\orbr{\begin{cases}x=-17\\x=17\end{cases}}\)
\(b,5^{2x-3}-2.5^2=5^2.3\)
\(\Rightarrow5^{2x-3}-50=75\)
\(\Rightarrow5^{2x-3}=125\)
\(\Rightarrow5^{2x-3}=5^3\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
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 ....
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