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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
\(\dfrac{-x-27}{27}=\dfrac{2}{3}\Rightarrow-x-27=18\Leftrightarrow x=-45\)
-> chọn B
a) (-37 - 17) . (-9) + 35 . (-9 - 11)
=-54*(-9)+35*(-20)
=486+(-700)
=-214
b) (-25) . (75 - 45) - 75 . ( 45 - 25)
=(-25)*30-75*20
=-750-1500
=-2250
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
Ta có: \(\frac{x}{45}=\frac{2}{x+1}\)
\(\Leftrightarrow x.\left(x+1\right)=90\)
\(\Leftrightarrow x^2+x-90=0\)
\(\Leftrightarrow x^2-9x+10x-90=0\)
\(\Leftrightarrow x.\left(x-9\right)+10.\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+10\right).\left(x-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-10\end{cases}}\)
Vậy......
thank