tim x:
1+5+9+13+...+x=91
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a: =>4/9*a/b=5/27+3/27=8/27
=>a/bb=8/27:4/9=8/27*9/4=72/108=2/3
b: =>9/13*a/b=40/91+1/7=53/91
=>a/b=53/63
`a)4/9xxa/b-1/9=5/27`
`=>4/9xxa/b=5/27+1/9`
`=>4/9xxa/b=5/27+3/27`
`=>4/9xxa/b=8/27`
`=>a/b=8/27:4/9`
`=>a/b=8/27xx9/4`
`=>a/b=2/3`
Vậy `a=2` và `b=3`
__
`b)9/13xxa/b+1/7=40/91`
`=>9/13xxa/b=40/91-1/7`
`=>9/13xxa/b=40/91-13/91`
`=>9/13xxa/b=27/91`
`=>a/b=27/91:9/13`
`=>a/b=27/91xx13/9`
`=>a/b=39/91`
Vậy `a=39` và `b=91`
Ta có
1 = 0 + 1;
5 = 2 + 3;
9 = 4 + 5;
....
x = a + ( a + 1 )
=> 1 + 5 + 9 + 13 + 17 + ... + x = 501
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + ... + a + ( a + 1 ) = 506
Hay [ ( a - 1 ) - 1 ] x ( a - 1) 2 [ ( a - 1 ) - 1 ] x ( a - 1 ) 2 = 506
=> ( a + 1 ) x ( a + 2 ) = 506
=> a = 21
Do đó x = 21 + 21 + 1 = 43
Ta có
1 = 0 + 1;
5 = 2 + 3;
9 = 4 + 5;
....
x = a + ( a + 1 )
=> 1 + 5 + 9 + 13 + 17 + ... + x = 501
= 1 + 2 + 3 + 4 + 5 + 6 + 7 + ... + a + ( a + 1 ) = 506
Hay [ ( a - 1 ) - 1 ] x ( a - 1) 2 [ ( a - 1 ) - 1 ] x ( a - 1 ) 2 = 506
=> ( a + 1 ) x ( a + 2 ) = 506
=> a = 21
Do đó x = 21 + 21 + 1 = 43
Ta co :
x+1:4+1=501501
x+1:4 = 501501-1
x+1:4 =501500
x+1 = 501500:4
x+1 = 125375
x = 125375 -1
x =125374
Vay x=125735
**** nhe
\(1+5+9+...+x=501501\)
\(\Leftrightarrow\frac{\left(1+x\right).\left[\left(x-1\right)\div4+1\right]}{2}=501501\)
\(\Leftrightarrow\left(1+x\right).\left[\left(x-1\right)\div4+1\right]=501501.2\)
\(\Leftrightarrow\left(1+x\right).\left[\frac{x-1}{4}+1\right]=1003002\)
\(\Leftrightarrow\frac{x-1}{4}\left(1+x\right)+x+1=1003002\)
\(\Leftrightarrow\frac{x-1}{4}+\frac{x-1}{4}x+x=1003001\)
\(\Leftrightarrow\frac{x-1+x\left(x-1\right)+4x}{4}=1003001\)
Tự giải tiếp
Tìm x thuoc z:
1) \(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)
Vậy: \(x\in\left\{30;-48\right\}\)
2) \(\left|x+7\right|-13=25\)
\(\Leftrightarrow\left|x+7\right|=25+13=38\)
\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)
\(\Leftrightarrow x\in\left\{31;-45\right\}\)
Vậy:.................
tim x biet
\(1)123-3.\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=123-23\)
\(\Leftrightarrow3\left(x+4\right)=100\)
\(\Leftrightarrow x+4=\frac{100}{3}\)
\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)
Vậy:................
2) Tương tự
a) 3x = 32 . 9
3x = 32 . 32
3x = 34
=> x = 4
b) ( 21 + x ) : 9 = 95 : 94
(21 + x ) : 9 = 9
21 + x = 9 . 9
21 + x = 81
x = 81 - 21
x = 60
c) 5x + 1 - 13 = 613
5x = 613 + 13 - 1
5x = 625
5x = 54
=> x = 4
x= 25