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85+(-35)+(-12)+(-35)=50+(-12)+(-35)=38+(-35)=3
53.42+85.53-57:54=125.16+85.125-53=125.16+85.125-125
=125.16+85.125-125.1=(16+85-1).125=100.125=12500
\(2^3-5^3:5^2+12\cdot2^2=8-5+48=51\\ 5\left[\left(85-35:7\right):8+90\right]-5\cdot2^2\\ =5\left(80:8+90\right)-20=5\cdot100-20=480\)
b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5^2\cdot2\)
\(=5\cdot\left[\left(85-5\right):8+90\right]-25\cdot2\)
\(=5\cdot\left(10+90\right)-50\)
=450
\(2^x:2^5=1\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(7^2-\left(13+4x\right)=5.2^3\)
\(\Rightarrow13+4x=5.2^3+7^2=89\)
\(\Rightarrow4x=89-13=76\)
\(\Rightarrow x=19\)
a) \(\left(3^4.57-9^2.21\right):3^5\)
\(=\left(3^4.57-3^4.21\right):3^5\)
\(=\left[3^4\left(57-21\right)\right]:3^5\)
\(=3^4.36:3^5\)
\(=3^4.2^2.3^2:3^5\)
\(=3.4\)
\(=12\)
b) Ta có; \(1^3+2^3+...+9^3=2025\)
\(\Leftrightarrow2^3.\left(1^3+2^3+....+9^3\right)=2^3.2025\)
\(\Leftrightarrow2^3+4^5+...+18^3=16200\)
A = 1+2+22+...+210
=> 2A = 2+22+23+...+211
=> 2A - A = (2+22+23+...+211) - (1+2+22+...+210)
=> A = 211 - 1
B = 1+3+32+...+310
=> 3B = 3+32+33+...+311
=> 3B - B = (3+32+33+...+311) - (1+3+32+...+310)
=> 2B = 311 - 1
=> B = \(\frac{3^{11}-1}{2}\)
A = 1 + 2 1 + 2 2 + 2 3 + ... + 2 9 + 2 10
2A = 2 + 2 2 + 2 3 + 2 4 + ... + 2 10 + 2 11
2A - A = ( 2 + 2 2 + 2 3 + 2 4 + ... + 2 10 + 2 11 )
- ( 1 + 2 1 + 2 2 + 2 3 + ... + 2 9 + 2 10 )
A = 2 11 - 1
A = 2047
B = 1 + 3 1 + 3 2 + 3 3 + ... + 3 9 + 3 10
3B = 3 1 + 3 2 + 3 3 + 3 4 + ... + 3 10 + 3 11
3B - B= ( 3 1 + 3 2 + 3 3 + 3 4 + ... + 3 10 + 3 11 )
- ( 1 + 3 1 + 3 2 + 3 3 + ... + 3 9 + 3 10 )
2B = 3 11 - 1
B = \(\frac{3^{11}-1}{2}\)
B = 88573
\(5\left[\left(85-35:7\right):2^3+3^2.10\right]-50\)
\(=5\left[\left(85-5\right):8+9.10\right]-50\)
\(=5\left[80:8+90\right]-50\)
\(=5\left[10+90\right]-50\)
\(=5.100-50\)
\(=500-50\)
\(=450\)
5.[ (85-35:7):2^3+3^2.10 ] - 50
=5.[(85-5):8+9.10 ]-50
=5.[80:8+9.10]-50
=5.[10+90]-50
=5.100-50
=5.2.50-50
=10.50-50.1
=50.(10-1)
=50.9
=450