1+2+...+x=45
tìm x
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x(x-y)=\(\frac{2}{45}\) y(x-y)=\(\frac{1}{45}\)
\(\frac{x\left(x-y\right)}{y\left(x-y\right)}\)=\(\frac{2}{45}\):\(\frac{1}{45}\)
=> \(\frac{x}{y}\)=2
=> x=2y
thay vào biểu thức 2 ta có
y(x-y)=\(\frac{1}{45}\)<=> y(2y-y)=\(\frac{1}{45}\)=> \(y^2\)=\(\frac{1}{45}\) => y=\(\sqrt{\frac{1}{45}}\)hoặc y= -\(\sqrt{\frac{1}{45}}\)
với y=\(\sqrt{\frac{1}{45}}\) =>x=\(\frac{1}{2}\)\(\sqrt{\frac{1}{45}}\)
với y=-\(\sqrt{\frac{1}{45}}\)=> x=-\(\frac{1}{2}\)\(\sqrt{\frac{1}{45}}\)
a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)
=> \(1-\frac{1}{x+1}=\frac{44}{45}\)
=> \(\frac{x}{x+1}=\frac{44}{45}\)
=> x = 44
b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
.................
\(\frac{1}{45^2}< \frac{1}{44.45}=\frac{1}{44}-\frac{1}{45}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}=1-\frac{1}{45}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)
a) 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/(x+1)=1-1/(x+1)=x/(x+1)=44/45
=> x=44
b/ 1/22 < 1/1.2; 1/32 < 1/2.3; ....; 1/452 < 1/44.45
=> A < 1/1.2+1/2.3+...+1/44.45=1-1/45=44/45 < 1
=> A < 1
1+2+3+4+5+...+x= 45
1+2+3+4+5+6+7+8+x=45
x=45-8-7-6-5-4-3-2-1
x=9
\(5x+\left(1+2+3+4\right)=45\)
\(5x+10=45\)
\(5x=35\)
\(x=7\)
Vậy \(x=7\)
x+1 + x+2 + x+3 + x +4 + x+5 = 45
x + x + x + x + x = 45 -1 -2 -3 -4 -5
5x = 30
x =6
x=9
nha
Ta có:
\(1+2+...+x=45\)
\(\Rightarrow\frac{x.\left(x+1\right)}{2}=45\)
\(\Rightarrow x.\left(x+1\right)=90\Leftrightarrow x=9\)