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đề bài là gì vậy bạn uiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Thui mình giải đại nhennnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
=(x+x+x)+(1/2+2/3+3/6)
=3x+(3/6+4/6+3/6)
=3x+15
Hết ùiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Bye nhennnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
2x+1-3=13
2x+1=13+3
2x+1=16
2x+1=24
Vậy x+1=4
x=4-1=3
Vậy x=3
Ta có :
\(\left|3-x\right|\ge0\)
\(\Rightarrow x-5\ge0\)
\(\Rightarrow x\ge5>0\)
=> |3 - x| = 3 - x = x - 5
=> 3 - x - x = -5
=> 3 - 2x = -5
=> 2x = 3 - (-5)
=> 2x = 8
=> x = 4
(x+1) + (x+2) + (x+3) + (x+4) + ...+ (x+30) = 795
x.30 + (1+30).30 : 2 = 795
30x + 465 = 795
30x = 330
x = 11
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + .... + ( x + 30 ) = 795
( x + x + x + x + ...+ x ) + ( 1 + 2 + 3 + 4 + ... + 30 ) = 795
Tổng số các số hạng là : ( 1 + 30 ) . 30 : 2 = 465
=> x. 30 + 465 = 795
=> x . 30 = 330
=> x = 11
Vậy x = 11
(x+1)+(x+2)+(x+3)+.....+(x+30)=795
(x+x+x+...+x)+(1+2+3+...+30)=795
30x+465=795
30x=795-465
30x=330
x=330:30
x=11
(x+1) + (x+2)+(x+3) +...+(x+30) = 795
=> 30x + (1+2+3+...+29+30) = 795
=> 30x + 465 = 795 (1+2+3+...+30 bạn áp dụng quy tắc tính tổng dãy số cách đề là Ok)
=> 30x = 330
=> x = 11
1)\(3^{2x}=27\cdot3^x\)
\(\Leftrightarrow3^{2x}=3^3\cdot3^x\)
\(\Leftrightarrow3^{2x}=3^{3+x}\)
\(\Leftrightarrow2x=3+x\)
\(\Leftrightarrow2x-x=3\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
x.30+(1+2+3+......+29+30)=795
x.30+465=795
x.30=795-465
x.30=330
x=330:30
x=11
\(\text{(x+1)+(x+2)+(x+3)+...+(x+29)+(x+30)=795}\)
Số số hạng là:
(30-1):1+1 = 30 ( số hạng )
=> \(\text{x+1+x+2+x+3+...+x+29+x+30=795}\)
Đặt A = 1+2+3+...+30
A = \(\left(\left(30+1\right)\cdot30\right):2\)
A = 465
=> x+x+x+...+x+1+2+3+...+30=795
30x + 465 = 795
30x = 795 - 465 = 330
x = 330 : 30 = 11
Vậy x là 11