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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, \(390-\left(x-7\right)=13^2:12\)
\(390-\left(x-7\right)=\) \(\dfrac{169}{12}\)
\(x-7=390-\dfrac{169}{12}\)
\(x-7=\dfrac{4511}{12}\)
\(x=\dfrac{4511}{12}+7\)
\(x=\dfrac{4595}{12}\)
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 ....
1. 2x=16\(\Rightarrow\)X=4
2. 22x-1=27
\(\Rightarrow\)27=22.4-1
Vậy x =4
\(a,\left|x\right|-2=7-\left(-8\right)\)
\(\Rightarrow\left|x\right|-2=15\)
\(\Rightarrow\left|x\right|=17\)
\(\Rightarrow\orbr{\begin{cases}x=-17\\x=17\end{cases}}\)
\(b,5^{2x-3}-2.5^2=5^2.3\)
\(\Rightarrow5^{2x-3}-50=75\)
\(\Rightarrow5^{2x-3}=125\)
\(\Rightarrow5^{2x-3}=5^3\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
a) 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