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ta có 50^40>50^39
50^39>50^38=)1/50^39<1/50^38
=)50^40+1/50^39>50^39+1/50^38
=)50^40+1/50^39+1>50^39+1/50^38+1
=)A>B
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Ta có: \(\dfrac{1}{4}=\dfrac{10}{40}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}\)
Mà \(\dfrac{1}{31}>\dfrac{1}{40}\)
\(\dfrac{1}{32}>\dfrac{1}{40}\)
\(\dfrac{1}{33}>\dfrac{1}{40}\)
\(\dfrac{1}{34}>\dfrac{1}{40}\)
\(\dfrac{1}{35}>\dfrac{1}{40}\)
\(\dfrac{1}{36}>\dfrac{1}{40}\)
\(\dfrac{1}{37}>\dfrac{1}{40}\)
\(\dfrac{1}{38}>\dfrac{1}{40}\)
\(\dfrac{1}{39}>\dfrac{1}{40}\)
\(\Rightarrow\) \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{39}+\dfrac{1}{40}>\dfrac{10}{40}=\dfrac{1}{4}\)
Vậy \(S>\dfrac{1}{4}\)
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a) \(9^3\cdot3^2\)
\(=\left(3^2\right)^3\cdot3^2\)
\(=3^6\cdot3^2\)
\(=3^8\)
b) \(x^7\cdot x:x^4\)
\(=x^8:x^4\)
\(=x^4\)
c) \(7\cdot3^9+3^{10}+51\cdot3^8\)
\(=3^8\cdot\left(7\cdot3+3^2+51\right)\)
\(=3^8\cdot81\)
\(=3^8\cdot3^4\)
\(=3^{12}\)
d) \(5^{15}\cdot125^3\cdot625\)
\(=5^{15}\cdot5^9\cdot5^4\)
\(=5^{28}\)
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\(\frac{195}{1890}>\frac{39}{379}\)
\(\frac{193}{310}>\frac{173}{350}\)
\(k-mk-nha\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{7^{38}+10}{7^{38}-3}\)như thế này có đúng không bạn???
Ta có: \(A=\frac{7^{38}+10}{7^{38}-3}=\frac{7^{38}-3+13}{7^{38}-3}=1+\frac{13}{7^{38}-3}\)
Lại có: \(B=\frac{7^{39}+18}{7^{39}+5}=\frac{7^{39}+5+13}{7^{39}+5}=1+\frac{13}{7^{39}+5}\)
Vì \(\frac{13}{7^{38}-3}>\frac{13}{7^{39}+5}\) nên \(1+\frac{13}{7^{38}-3}>1+\frac{13}{7^{39}+5}\)
\(\Rightarrow A>B\)
Vậy \(A>B\).
phân số đấy ạ bà con
nhanh tay giải toán dược ngay điểm 10