Question

[Trần Thanh Tùng]

1/ So sánh A=\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)và B=\(\frac{7}{12}\)

Vậy A...B

2/ Giá trị của x thỏa mãn:

Ix²+Ix-1II=x²

^{3/ \(\frac{\left(-\frac{1}{2}\right)^n}{\left(-\frac{1}{2}\right)^{n+2}}\)}. Tìm giá trị n=?

Answer 1

[Mai Ngọc]

2)lx^2+lx+1ll=x^2

=>x^2+lx+1l=x^2=>lx+1l=0=>x=-1

3)\(\frac{\left(-\frac{1}{2}\right)^n}{\left(-\frac{1}{2}\right)^{n-2}}=\left(-\frac{1}{2}\right)^{n-n-2}=\left(-\frac{1}{2}\right)^{-2}=4\)

Answer 2

[Mai Ngọc]

1)\(A=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)

\(\Rightarrow A=\left(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{80}\right)\)

\(\Rightarrow A=C+D\)

Ta có:\(\frac{1}{41}>\frac{1}{60};>\frac{1}{60}:\frac{1}{43}>\frac{1}{60};...;\frac{1}{59}>\frac{1}{60};\frac{1}{60}=\frac{1}{60}\)

\(\Rightarrow C=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)

Ta thấy C có 20 số hạng

\(\Rightarrow C>\frac{1}{60}.20=\frac{1}{3}\)

Ta có:\(\frac{1}{61}>\frac{1}{80};\frac{1}{62}>\frac{1}{80};\frac{1}{63}>\frac{1}{80};...;\frac{1}{79}>\frac{1}{80};\frac{1}{80}=\frac{1}{80}\)

\(\Rightarrow D=\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}\)

Ta thấy D có 20 số hạng.

\(\Rightarrow D>\frac{1}{80}.20=\frac{1}{4}\)

\(\Rightarrow A=C+D>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

\(\Rightarrow A>B\)