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\(A=\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)
\)
\(\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Không có gtri A=? ak bạn??
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)
\(\Leftrightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{\left(x+3\right)}=\frac{100}{101}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{100}{101}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{100}{101}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{101}\)
\(\Rightarrow x+3=101\)
\(=>x=98\)
\(D=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)
\(D=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)
\(D=1-\frac{1}{x+3}=\frac{100}{101}\)
\(D=\frac{1}{x+3}=1-\frac{100}{101}\)
\(D=\frac{1}{x+3}=\frac{1}{101}\)
\(\Rightarrow x+3=101\Rightarrow x=98\)
Ủng hộ mk nha ^_^
Câu 1:
\(A=\frac{\left(1+2+3+...+100\right)x\left(101x102-101x101-51-50\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x\left(101x\left(102-101\right)-\left(50+51\right)\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x\left(101-101\right)}{2+4+6+8+...+2048}\)
\(A=\frac{\left(1+2+3+...+100\right)x0}{2+4+6+8+...+2048}\)
\(A=0\)
Ta có:Số số hạng từ 2 đến 101 là:
(101-2):1+1=100(số hạng)
Do đó từ 2 đến 101 có số cặp là:
100:2=50(cặp)
\(B=\frac{101+100+99+...+3+2+1}{101-100+99-98+3-2+1}\)
\(B=\frac{5151}{51}\)
\(B=101\)
Câu 2:
a)697:\(\frac{15x+364}{x}\)=17
\(\frac{15x+364}{x}\)=697:17
\(\frac{15x+364}{x}\)=41
15x+364=41x
41x-15x=364
26x=364
x=14
Vậy x=14
b)92.4-27=\(\frac{x+350}{x}+315\)
\(\frac{x+350}{x}+315\)=341
\(\frac{x+350}{x}\)=26
x+350=26
x=26-350
x=-324
Vậy x=-324
c, 720 : [ 41 - ( 2x -5)] = 40
[ 41 - ( 2x -5)] =720:40
[ 41 - ( 2x -5)] =18
2x-5=41-18
2x-5=23
2x=28
x=14
Vậy x=14
d, Số số hạng từ 1 đến 100 là:
(100-1):1+1=100(số hạng)
Tổng dãy số là:
(100+1)x100:2=5050
Mà cứ 1 số hạng lại có 1x suy ra có 100x
Ta có:(x+1) + (x+2) +...+ (x+100) = 5750
(x+x+...+x)+(1+2+...+100)=5750
100x+5050=5750
100x=700
x=7
Vậy x=7
\(A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x\left(x+3\right)}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
\(=1-\frac{1}{x+3}\)
\(=\frac{x+2}{x+3}=\frac{100}{101}\)
\(\Rightarrow x=98\)
\(A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{x.\left(x+3\right)}=\frac{100}{101}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{100}{101}\)
\(A=1-\frac{1}{x+3}=\frac{100}{101}\)
\(\frac{1}{x+3}=1-\frac{100}{101}=\frac{1}{101}\)
=> x + 3 = 101
=> x = 101 - 3
=> x = 98
Vậy x = 98
Ủng hộ mk nha ^_-
ĐKXĐ : 101x \(\ge\)0 nên x \(\ge\)0
khi đó : \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
\(\Leftrightarrow\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
\(\Leftrightarrow100x+\frac{5050}{101}=101x\Leftrightarrow x=50\)
*ĐK : 101x\(\ge\) 0 => x\(\ge\)0
=> \(x+\frac{1}{101}\ge\frac{1}{101}>0\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\)
\(x+\frac{2}{101}\ge\frac{2}{101}>0\Rightarrow\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\)
...
\(x+\frac{100}{101}\ge\frac{100}{101}>0\Rightarrow\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)
Ta có :
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
<=> \(100x+\left(\frac{1+2+...+100}{101}\right)=101x\)
<=> \(100x+\frac{5050}{101}=101x\)
<=> \(100x-101x=\frac{-5050}{101}\)
<=> -x = -50
=> x = 50 ( t/m x >/ 0)