So sanh A = 3^123+1/3^125+1 va B= 3^122+1/3^124+1
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18 tháng 8 2023 lúc 19:50
SO SÁNH : A = 3^123 +1 / 3^125 + 1 và B = 3^122/ 3^124 + 1
A = \(\dfrac{3^{123}+1}{3^{125}+1}\) Vì 3123 + 1 < 2125 + 1 Nên A = \(\dfrac{3^{123}+1}{3^{125}+1}\)< \(\dfrac{3^{123}+1+2}{3^{125}+1+2}\)
A < \(\dfrac{3^{123}+3}{3^{125}+3}\) = \(\dfrac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\) = \(\dfrac{3^{122}+1}{3^{124}+1}\) = B
Vậy A < B
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so sanh
\(A=\frac{3^{123}+1}{3^{125}+1}\)va \(b=\frac{3^{122}+1}{3^{134}+1}\)
Đề đúng là \(B=\frac{3^{122}+1}{3^{124}+1}\)nhé .
Ta có :
\(9A=9.\left(\frac{3^{123}+1}{3^{125}+1}\right)=\frac{3^{125}+9}{3^{125}+1}\)
\(=1+\frac{8}{3^{125}+1}\)
\(9B=9.\left(\frac{3^{122}+1}{3^{124}+1}\right)=\frac{3^{124}+9}{3^{124}+1}\)
\(=1+\frac{8}{3^{124}+1}\)
Dễ thấy \(3^{124}+1< 3^{125}+1\)
\(\Leftrightarrow\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)
\(\Leftrightarrow\frac{8}{3^{125}+1}+1< \frac{8}{3^{124}+1}+1\)
\(\Leftrightarrow A< B\)
Vậy....
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SO SÁNH A = 3^123 + 1 / 3^125 + 1 và B = 3^122+1 / 3^124+1
SO SÁNH : A = 3^123 +1 / 3^125 + 1 VÀ B = 3^122 + 1 / 3^124 + 1
so sánh 3^123+1/ 3^125+1 và 3^122+1 / 3^124+1
\(A=\dfrac{3^{123}+1}{3^{125}+1}\Leftrightarrow3^2A=\dfrac{3^{125}+9}{3^{125}+1}\)
\(9A=\dfrac{3^{125}+1}{3^{125}+1}+\dfrac{8}{3^{125}+1}=1+\dfrac{8}{3^{125}+1}\)
\(B=\dfrac{3^{122}+1}{3^{124}+1}\Leftrightarrow3^2B=\dfrac{3^{124}+9}{3^{124}+1}\)
\(9B=\dfrac{3^{124}+1+8}{3^{124}+1}+\dfrac{3^{124}+1}{3^{124}+1}+\dfrac{8}{3^{124}+1}=1+\dfrac{8}{3^{124}+1}\)
\(9A< 9B\Leftrightarrow A< B\)
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so sánh A và B:
A=\(\frac{3^{123}+1}{3^{125}+1}\)
B=\(\frac{3^{122}+1}{3^{124}+1}\)
Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\) (a;b;m \(\in\)N*)
Ta có:
\(A=\frac{3^{123}+1}{3^{125}+1}< \frac{3^{123}+1+2}{3^{125}+1+2}\)
\(A< \frac{3^{123}+3}{3^{125}+3}\)
\(A< \frac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\)
\(A< \frac{3^{122}+1}{3^{124}+1}=B\)
=> A < B
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\(9A=\frac{3^{125}+9}{3^{125}+1}=1+\frac{8}{3^{125}+1}\)
\(9B=\frac{3^{124}+9}{3^{124}+1}=1+\frac{8}{3^{124}+1}\)
Mà 3^125+1>3^124+1 =>\(\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)
Nên A<B
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9A=\(\frac{3^{125}+9}{3^{125}+1}\)=\(1+\frac{8}{3^{125}+1}\)
9B=\(\frac{3^{124}+9}{3^{124}+1}\)=\(1+\frac{8}{3^{124}+1}\)
Vì \(\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)\(\Rightarrow9B>9A\)\(\Rightarrow B>A\)
Vậy B>A
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So sánh :
\(\frac{3^{123}+1}{3^{125}+1}vs\frac{3^{122}}{3^{124}+1}\)
https://i.imgur.com/SBC97Yo.jpg
so sanh 230+320+430 và 3.2410
b)\(\frac{3^{123}+1}{3^{125}+1}\) và \(\frac{3^{122}}{3^{124}+1}\)
so sánh: A= \(\frac{3^{123}+1}{3^{125}+1}\)và B= \(\frac{3^{122}}{3^{124}+1}\)
các bn lm nhanh hộ mik, mik đang cần gấp
\(B=\frac{3^{122}}{3^{124}+1}=\frac{3^{123}}{3^{125}+3}< \frac{3^{123}+1}{3^{125}+3}< \frac{3^{123}+1}{3^{125}+1}=A\)
Do đó \(A>B\).