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Tính nhanh
a) 72. (-50).(-2).(-2)
b) 125.(-5).(-20).8
c) 125.(-3).(-32).(-4)
d) (-123 – 123-123-123).125
123 - 5 . ( x + 4 ) = 38
5 . ( x + 4 ) = 123 - 38
5 . ( x + 4 ) = 85
x + 4 = 85 : 5
x + 4 = 17
x = 17 - 4
x = 13
( 3x - 24 ) . 73 = 2 . 74
( 3x - 16 ) . 343 = 2 . 2401
( 3x - 16 ) . 343 = 4802
3x - 16 = 4802 : 343
3x - 16 = 14
3x = 14 + 16
3x = 30
x = 30 : 3
x = 10
123 - 5( x + 4 ) = 38
=> 5( x + 4 ) = 123 - 38
=> x + 4 = 85 ÷ 5
=> x = 17 - 4
=> x = 13
(3x - 24 ) · 73 = 2·74
=> ( 3x - 16 ) = 2 . 74 ÷ 73
=> 3x - 16 = 14
=> 3x = 16 + 14
=> x = 30 ÷ 3
=> x = 10
\(\left[8x-120:4\right]\times3^3=3^6.\)
\(8x-120:4=3^3\)
\(8x-30=27\)
\(8x=27+30\)
\(8x=57\)
\(x=19\)
123 - 5.(x + 4) = 38
123 - 5x - 20 = 38
-5x = 38 - 123 + 20
-5x = -65
x = -65 : -5
x = 13
Vậy x = 13
\(A=\frac{\frac{4}{111}+\frac{4}{123}-\frac{4}{999}}{\frac{2}{111}+\frac{2}{123}+\frac{2}{999}}\)
\(=\frac{4.\left(\frac{1}{111}+\frac{1}{123}+\frac{1}{999}\right)}{2.\left(\frac{1}{111}+\frac{1}{123}+\frac{1}{999}\right)}\)
\(=\frac{4}{2}=2\)
1) 5^2 . x + 2^4 . x = 123
=> 25 .x + 16 .x = 123
=> x ( 25 + 16 ) = 123
=> x . 41 = 123
=> x = 123 : 41
=> x = 3
b, 3^6 : 3 ^ 3 + 5.x = 57
=> 3^3 + 5.x = 57
=> 27 + 5x = 57
=> 5x = 57 - 27
=> 5x = 30
=> x = 30 : 5
=> x = 6
\(\frac{\frac{4}{111}+\frac{4}{123}+\frac{4}{999}}{\frac{2}{111}+\frac{2}{123}+\frac{2}{999}}=\frac{4.\left(\frac{1}{111}+\frac{1}{123}+\frac{1}{999}\right)}{2.\left(\frac{1}{111}+\frac{1}{123}+\frac{1}{999}\right)}=\frac{4}{2}=2\)
\(=\frac{2\left(\frac{2}{111}+\frac{2}{123}+\frac{2}{999}\right)}{\frac{2}{111}+\frac{2}{123}+\frac{2}{999}}\)
\(=2\)
a)\(-7-25+27+5=\left(27-7\right)+\left(5-25\right)\)
\(=20-20=0\)
b)\(-8+16-\left(-8\right)=-8+16+8=\left(8-8\right)+16=16\)
c)\(1+1-\left(-8\right)=1+1+8=10\)(Vì số nào mũ 0 cũng bằng 1(ngoại trừ số 0))
d)\(65.23+65.35-35.65+35.23\)
\(=\left(65.35-65.35\right)+23.\left(65+35\right)\)
\(=0+23.100=2300\)
f) \(8+\left[1800-\left(64-54\right)^3\right].8-5\)
\(=8+\left(1800-10^3\right).8-5\)
\(=8+800.8-5=6403\)
h)\(-123-46+46+123=\left(123-123\right)+\left(46-46\right)\)
\(=0+0=0\) (phá ngoặc ra thôi nhé)
Đặt \(A=2^2+2^3+2^4+...+2^{123}\)
\(2A=2^3+2^4+2^5+...+2^{124}\)
\(2A-A=\left(2^3+2^4+2^5+...+2^{124}\right)-\left(2^2+2^3+2^4+...+2^{123}\right)\)
\(A=2^{124}-4\)
\(A=2^2+2^3+2^4+...+2^{123}\)
\(2\cdot A=2^3+2^4+2^5+...+2^{124}\)
\(A=2^{124}-2^2=2^{124}-4\)