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a) \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
b) b = a - c => b + c = a
\(\left\{{}\begin{matrix}\frac{a}{b}\cdot\frac{a}{c}=\frac{a^2}{bc}\\\frac{a}{b}+\frac{a}{c}=\frac{ac+ab}{bc}=\frac{a\left(b+c\right)}{bc}=\frac{a^2}{bc}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{b}\cdot\frac{a}{c}=\frac{a}{b}+\frac{a}{c}\)
Bước 2 bạn sai rồi. Vd: \(\frac{1}{3x3}\) đâu bằng hay nhỏ hơn \(\frac{1}{2x3}\)
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\Rightarrow\frac{bc+ac}{abc}=\frac{ab}{abc}\Rightarrow bc+ac=ab\)
\(\Rightarrow ab-ac-bc=0\Rightarrow a\left(b-c\right)-c\left(b-c\right)=c^2\)
\(\Rightarrow\left(b-c\right)\left(a-c\right)=c^2\Rightarrow\frac{a-c}{c}=\frac{c}{b-c}\)
1.a.ta có:\(\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
mà \(\frac{2017}{2018}>\frac{2017}{2018+2019};\frac{2018}{2019}>\frac{2018}{2018+2019}\)
\(\Rightarrow M>N\)
b.ta thấy:
\(\frac{n+1}{n+2}>\frac{n+1}{n+3}>\frac{n}{n+3}\Rightarrow\frac{n+1}{n+2}>\frac{n}{n+3}\)
=> A>B
a) Ta có: \(\frac{a}{b}+\frac{a}{c}=\frac{ac+ab}{bc}=\frac{a\left(b+c\right)}{bc}=\frac{a.a}{bc}\) (thay b+c = a) (1)
\(\frac{a}{b}\times\frac{a}{c}=\frac{a.a}{bc}\) (2)
Từ (1) và (2) suy ra: \(\frac{a}{b}+\frac{a}{c}=\frac{a}{b}\times\frac{a}{c}\) (đpcm)
b) \(c=a+b\)\(\Rightarrow\)\(a=c-b\)
Ta có: \(\frac{a}{b}-\frac{a}{c}=\frac{ac-ab}{bc}=\frac{a\left(c-b\right)}{bc}=\frac{a^2}{bc}\) (thay c-b = a) (3)
\(\frac{a}{b}\times\frac{a}{c}=\frac{a^2}{bc}\) (4)
Từ (3) và (4) suy ra: \(\frac{a}{b}-\frac{a}{c}=\frac{a}{b}\times\frac{a}{c}\) (đpcm)
Bài 1:
a) \(\frac{a}{5}=\frac{-3}{b}\)
\(\Rightarrow ab=-15\)
Ta có bảng sau:
| a | 1 | -1 | 15 | -15 |
| b | -15 | 15 | -1 | 1 |
Vậy cặp số \(\left(a;b\right)\) là \(\left(1;-15\right);\left(-1;15\right);\left(15;-1\right);\left(-15;1\right)\)
b) @Nguyễn Huy Thắng
Bài 2:
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)
\(\left\{\begin{matrix}\frac{a}{b}=1\\\frac{b}{c}=1\\\frac{c}{a}=1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c\left(đpcm\right)\)
Vậy a = b = c
Sửa đề: chứng minh \(S\ge6\)
Ta có:
\(S=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\left(\frac{a}{b}-2+\frac{b}{a}\right)+\left(\frac{b}{c}-2+\frac{c}{b}\right)+\left(\frac{a}{c}-2+\frac{c}{a}\right)+6\)
\(=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2+\left(\sqrt{\frac{b}{c}}-\sqrt{\frac{c}{a}}\right)^2+\left(\sqrt{\frac{a}{c}}-\sqrt{\frac{c}{a}}\right)^2+6\ge6\)
\(\Rightarrow\)ĐPCM
Đây nè k cho mình nha:
Ta có \(\frac{a+b}{c}>\frac{a+b}{a+b+c}\)
\(\frac{b+c}{a}>\frac{b+c}{a+b+c}\)
\(\frac{a+c}{b}>\frac{a+c}{a+b+c}\)
Suy ra \(S>\frac{a+b}{a+b+c}+\frac{b+c}{a+b+c}+\frac{a+c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Vậy S > 2
Vì \(\frac{a}{b}< \frac{c}{d}\)
⇒ \(ad< bc\)
⇒ \(2018ad< 2018bc\)
⇒ \(2018ad+cd< 2018bc+cd\)
⇒ \(\left(2018a+c\right)d< \left(2018b+d\right)c\)
⇒ \(\frac{2018a+c}{2018b+d}< \frac{c}{d}\)
Vậy \(\frac{2018a+c}{2018b+d}< \frac{c}{d}\) (ĐPCM)