a ) 1 + 2 + 3 + 4 + ... + x = 45
b ) 2 + 4 + ...... + 2 . x = 110
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b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)
Ta có: \(x^2-y^2+2z^2=108\)
\(\Leftrightarrow\left(2k\right)^2-\left(3k\right)^2+2\cdot\left(4k\right)^2=108\)
\(\Leftrightarrow4k^2-9k^2+2\cdot16k^2=108\)
\(\Leftrightarrow k^2=4\)
Trường hợp 1: k=2
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\cdot2=4\\y=3k=3\cdot2=6\\z=4k=4\cdot2=8\end{matrix}\right.\)
Trường hợp 2: k=-2
\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\cdot\left(-2\right)=-4\\y=3k=3\cdot\left(-2\right)=-6\\z=4k=4\cdot\left(-2\right)=-8\end{matrix}\right.\)
Tách ?
`a, 70 -5.(x-3) =45`
`=> 5.(x-3)= 70-45`
`=> 5.(x-3)=25`
`=>x-3=25:5`
`=>x-3=5`
`=>x= 5+3`
`=>x=8`
______
`b,10 + 2.x = 4^5:4^3`
`=> 10 + 2.x = 4^(5-3)`
`=> 10 + 2.x =4^2=16`
`=> 2.x=16-10`
`=>2.x=6`
`=>x=6:2`
`=>x=3`
_____
`c,60-3.x-2=51`
`=> 60-3.x= 51+2`
`=> 60-3.x=53`
`=>3.x=60-53`
`=> 3.x= 7`
`=>x= 7/3`
____
`d, 4.x-20=2^5:2^3`
`=> 4.x-20=2^(5-3)`
`=> 4.x-20=2^2`
`=> 4.x= 4+20`
`=>4.x=24`
`=>x=24:4`
`=>x=6`
____
`2^x . 4=16`
`=> 2^x=16:4`
`=>2^x= 4`
`=>2^x=2^2`
`=>x=2`
____
`f, 3^x . 3=243`
`=>3^x=243:3`
`=> 3^x=81`
`=> 3^x=3^3`
`=>x=3`
_____
`g, 64. 4^x =16^8`
`=> 4^3 . 4^x=(4^2)^8`
`=> 4^3 . 4^x = 4^(16)`
`=> 4^x= 4^(16-3)`
`=>4^x=4^(13)`
`=>x=13`
_____
`2^x . 16^2 =1024`
`=> 2^x= 1024 : 16^2`
`=>2^x=4`
`=>2^x=2^2`
`=>x=2`
a: =>5(x-3)=25
=>x-3=5
=>x=8
b: =>2x=16-10=6
=>x=3
c: =>58-3x=51
=>3x=7
=>x=7/3
d: =>4x-20=4
=>4x=24
=>x=6
e: =>2^x=4
=>2^x=2^2
=>x=2
f: =>3^x=81
=>3^x=3^4
=>x=4
g: =>4^x*4^3=4^16
=>x+3=16
=>x=13
h: =>2^x=1024/256=4=2^2
=>x=2
\(2x+\left(1+2+3+...+100\right)=15150\)
\(2x+\left[\left(1+100\right)+\left(2+99\right)+...+\left(50+51\right)\right]=15150\)
\(2x+\left[101+101+...+101\right]=15150\)CÓ 50 SỐ 101
\(2x+\left[101\times50\right]=15150\)
\(2x=15150:5050\)
\(2x=3\)
\(x=3:2\)
\(x=1.5\)
a, 2x + (1+2+3+4+...+100) = 15150
=> 2x + \(\frac{\left(1+100\right).\left[\left(100-1\right)+1\right]}{2}\)= 15150
=> 2x + \(\frac{101.100}{2}\)= 15150
=> 2x + 5050 = 15150
=> 2x = 15150 - 5050
=> 2x = 10100
=> x = 10100 : 2
=> x = 5050
Vậy x = 5050
b, .(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)=36
=> (x + x + x + x +x + x +x +x ) + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36
=> 8x + 36 = 36
=> 8x = 0
=> x = 0
Vậy x = 0
c, 0+0+4+6+8+...+2x=110
Sửa đề :0 + 2 + 4 + 6 + 8 + ... + 2x = 110 = 2 + 4 + 6 + 8 + ... + 2x = 110
SSH : \(\frac{\left(2\text{x}-2\right)}{2}+1=x-1+1=x\)
Tổng : \(\frac{\left(2\text{x}+2\right).x}{2}=110\Leftrightarrow\frac{2.\left(x+1\right).x}{2}=110\)
\(\Leftrightarrow\left(x+1\right)x=110\)
\(\Leftrightarrow\left(10+1\right).10=110\)
=> x = 10
Vậy x = 10
`x+1+x+2+x+3+x+4=110`
`=>(x+x+x+x)+(1+2+3+4)=110`
`=> 4x+ 10 =110`
`=>4x=110-10`
`=> 4x=100`
`=>x=100:4`
`=>x=25`
\(x+1+x+2+x+3+x+4=110\\ \Leftrightarrow4x=100\\ \Leftrightarrow x=25\)
Vậy x = 25.
a) \(1+2+3+...+x=55\)
\(\Rightarrow\frac{x\left(x+1\right)}{2}=55\)
\(x\left(x+1\right)=55.2\)
\(x\left(x+1\right)=110=10.11\)
\(\Rightarrow x=10.\)
b) \(2+4+6+...+x=110\)
\(\frac{\left(2+x\right)\left(\frac{x-2}{2}+1\right)}{2}=110\)
\(\left(x+2\right)\left(\frac{x}{2}-1+1\right)=110.2\)
\(\left(x+2\right)\frac{x}{2}=220\)
\(x\left(x+2\right)=220.2\)
\(x\left(x+2\right)=440=20.22\)
\(\Rightarrow x=20.\)
a)VT=\(\frac{x\left(x+1\right)}{2}\)=55
=> x(x+1)=110
=. x=10
b) VT=2(1+2+3+.....+\(\frac{x}{2}\))=110
=> 1+2+3+...+\(\frac{x}{2}\)=55
Công thức tính dãy số theo quy luật:
Số số hạng=(số cuối-số đầu):khoảng cách+1
Tổng = số số hạng.(số đầu +số cuối) tất cả chia 2
Theo câu a=> \(\frac{x}{2}\)=10 => x=20
x * 4 + 1 + 2 + 3 + 4 = 110
x * 4 = 110 -1 - 2 - 3 - 4
x * 4 = 100
x = 100 : 4
x = 25
Dấu * là nhân nhé . k mình mình k lại 3 lần ở câu khác
a/ x=9 - vì:1+9=2+8=3+7=4+6=10 ;(4*10=40)+5 = 45
b/ x=10 - vì: 2(1+2+...+x) =2X55 ;(45+10=55)*2= 110