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c) {[(20 - 2.3).5] - 2.5} : 2 + (4.5)²
= {(20 - 6).5] - 10} : 2 + 20²
= (14.5 - 10) : 2 + 400
= (70 - 10) : 2 + 400
= 60 : 2 + 400
= 30 + 400
= 430
d) (1² + 2² + 3² + ... + 100²) . (2⁴ - 4²)
= (1² + 2² + 3² + ... + 100²) . (16 - 16)
= (1² + 2² + 3² + ... + 100²) . 0
= 0
Bài 2
a) -2x < 5
2x > 5
x > 5/2
b) [31 - (x + 5)].11 = 121
31 - (x + 5) = 121 : 11
31 - (x + 5) = 11
x + 5 = 31 - 11
x + 5 = 20
x = 20 - 5
x = 15
c) (x + 1)³ = 27
(x + 1)³ = 3³
x + 1 = 3
x = 3 - 1
x = 2
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a) 100 - 7 (x - 5) = 31 +33
100 - 7 (x - 5) = 31 + 27
100 - 7 (x - 5) = 58
7 (x - 5) = 100 - 58
7 (x - 5) = 42
x - 5 = 42 : 7
x - 5 = 6
x = 6 + 5
x = 11
b) 24 + 5x = 75 : 73
24 + 5x = 75 - 3
24 + 5x = 72 = 49
5x = 49 - 24
5x = 25
x = 25 : 5
x = 5
mik làm từng bước một nên hơi dài, mong bạn thông cảm nha !!!!
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Giải:
a) \(100-7\left(x-5\right)=31+3^3\)
\(\Leftrightarrow100-7\left(x-5\right)=31+27\)
\(\Leftrightarrow100-7\left(x-5\right)=58\)
\(\Leftrightarrow7\left(x-5\right)=42\)
\(\Leftrightarrow x-5=6\)
\(\Leftrightarrow x=11\)
Vậy ...
b) \(12\left(x-1\right):3=4^3+2^3\)
\(\Leftrightarrow4\left(x-1\right)=64+8\)
\(\Leftrightarrow4\left(x-1\right)=72\)
\(\Leftrightarrow x-1=18\)
\(\Leftrightarrow x=19\)
Vậy ...
c) \(24+5x=7^5:7^3\)
\(\Leftrightarrow24+5x=7^2\)
\(\Leftrightarrow24+5x=49\)
\(\Leftrightarrow5x=25\)
\(\Leftrightarrow x=5\)
Vậy ...
d) \(5x-206=2^4.4\)
\(\Leftrightarrow5x-206=2^6\)
\(\Leftrightarrow5x-206=64\)
\(\Leftrightarrow5x=270\)
\(\Leftrightarrow x=54\)
Vậy ...
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a) \(5+5^2+5^3+....+5^{100}\)
đặt \(A=5+5^2+5^3+....+5^{100}\) ( \(A\) có \(100\) số hạng )
\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+....+\left(5^{99}+5^{100}\right)\) ( có \(100\div2=50\) nhóm )
\(A=5\left(1+5\right)+5^3\left(1+5\right)+....+5^{99}\left(1+5\right)\)
\(A=5.6+5^3.6+....+5^{99}.6\)
\(A=6\left(5+5^3+....+5^{99}\right)\)
vì \(6⋮6\Rightarrow6\left(5+5^3+....+5^{99}\right)⋮6\Rightarrow A⋮6\)
b) \(2+2^2+2^3+....+2^{100}\)
đặt \(B=2+2^2+2^3+....+2^{100}\) ( \(B\) có \(100\) số hạng )
\(B=\left(2+2^2+2^3+2^4+2^5\right)+.....+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\) ( có \(100\div5=20\) nhóm )
\(B=2\left(1+2+2^2+2^3+2^4\right)+....+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(B=2.31+....+2^{96}.31\)
\(B=31\left(2+...+2^{96}\right)\)
vì \(31⋮31\Rightarrow31\left(2+...+2^{96}\right)\Rightarrow B⋮31\)
a) 5+5^2+5^3..+5^100
=(5+5^2)+(5^3+5^4)+....+(5^99+5^100)
=5.(1+5)+5^3.(1+5)+....+5^99.(1+5)
=5.6+5^3.6+.....+5^99.6
=6.(5+5^3+.....+5^99):6
(5x-3)5-1=31
(5x-3)5=32
(5x-3)5=25
5x-3=2
x=1