\(\dfrac{a+2}{a-2}=\dfrac{b+3}{b-3}\Rightarrow\dfrac{a+2}{b+3}=\dfrac{a-2}{b-3}=\dfrac{a}{b}=\dfrac{2}{3}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}\)
Chúc bạn học tốt nhé
\(\dfrac{a+2}{a-2}=\dfrac{b+3}{b-3}\Rightarrow\dfrac{a+2}{b+3}=\dfrac{a-2}{b-3}=\dfrac{a}{b}=\dfrac{2}{3}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}\)
Chúc bạn học tốt nhé
a) Chứng minh rằng: \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
b) Tìm số nguyên a để: \(\dfrac{2a+9}{a+3}+\dfrac{5a+17}{a+3}-\dfrac{3a}{a+3}\) là số nguyên.
Cho a,b,c là số đo 3 cạnh tam giác:
Chứng minh rằng: \(1< \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\)
a, \(P=\dfrac{1+2}{1^2\cdot2^2}+\dfrac{2+3}{2^2\cdot3^2}+...+\dfrac{9+10}{9^2\cdot10^2}\)
Chứng minh rằng: \(P< 1\)
b, \(Q=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
Chứng minh rằng: \(Q< \dfrac{1}{2}\)
__________
- Mình cần gấp. Giúp mình nhé.
Chứng minh:
a. \(A=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
b.\(B=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}< \dfrac{3}{16}\)
c. \(C=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)
Cho \(\dfrac{a}{b}=\dfrac{b}{c}\). Chứng minh rằng: \(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)
8 Chứng minh rằng :
a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\) ; b) \(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\dfrac{99.100}{100!}\)
c) \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)
d) \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}< \dfrac{1}{2}\)
Cho 3 số đôi một khác nhau. Chứng minh rằng : \(\dfrac{b-c}{\left(a-b\right)\left(a-c\right)}+\dfrac{c-a}{\left(b-c\right)\left(b-a\right)}+\dfrac{a-b}{\left(c-a\right)\left(c-b\right)}\) =\(2\left(\dfrac{1}{a-b}+\dfrac{1}{b-c}+\dfrac{1}{c-a}\right)\)
bài 3
Cho \(\dfrac{a+2}{a-2}=\dfrac{b+3}{b-3}\) Chứng minh \(\dfrac{a}{2}=\dfrac{b}{3}\)
giúp mk với
Bài 1: Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) chứng minh rằng :
a, \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b, \(\dfrac{a^2-b^{2^{ }}}{c^2-d^2}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)