Ta có:

\(1-\frac{-2015}{-2016}=1-\frac{2015}{2016}=\frac{1}{2016}\)

\(1-\frac{-2016}{-2017}=1-\frac{2016}{2017}=\frac{1}{2017}\)

Vì \(\frac{1}{2016}>\frac{1}{2017}\Rightarrow\frac{-2015}{-2016}< \frac{-2016}{-2017}\)

Đây là cách so sánh phần bù, bạn có thể lên mạng tham khảo thêm nhé :)

a ĐK \(a>0\)và \(a\ne1\)

. \(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right).\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

\(=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}\)

b. Ta có \(M-1=\frac{\sqrt{a}-1}{\sqrt{a}}-1=\frac{\sqrt{a}-1-\sqrt{a}}{\sqrt{a}}=\frac{-1}{\sqrt{a}}< 0\)

Vậy \(M< 1\)

tick jum mình sec cố gắng giải

chtt nghia la zj zax pan

a,Với \(a>0;a\ne1\)

 \(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

\(=\left(\frac{\sqrt{a}-1+a-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)^2}\right).\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{a-1}{a+\sqrt{a}}\)

b, Ta có : \(1=\frac{a+\sqrt{a}}{a+\sqrt{a}}\)mà \(a-1=\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\)

\(a+\sqrt{a}=\sqrt{a}\left(\sqrt{a}+1\right)\)vì \(\sqrt{a}-1< \sqrt{a}\)

Vậy \(\frac{a-1}{a+\sqrt{a}}< 1\)hay \(M< 1\)

Bài 1: \(\left(\frac{-1}{16}\right)^{100}=\frac{1}{\left(2^4\right)^{100}}=\frac{1}{2^{400}}>\frac{1}{2^{500}}=\left(\frac{-1}{2}\right)^{500}.\)

Bài 2: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)

\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)

\(\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)

\(\Leftrightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)

\(\Rightarrow a=b=c\)

\(\Rightarrow\frac{a}{b}=1;\frac{b}{c}=1;\frac{c}{a}=1\)

\(\Rightarrow M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2.2.2=8\)

Đáp số là M > 1. Bạn cần cách giải không ?

Co minh biet ket qua roi ban HiHI

mk chi cac bạn tuyet chieu;

nhung bai toan dang nay mk thuong lay so cu the nhu 1;2;3 .... thay vao se doan dc kq vi violympic thoi gian thi co hạn

cac ban co dong y k

; 

TH1 : a<b

\(\Rightarrow am< bm\)

\(\Rightarrow ab+am< ab+bm\Rightarrow a\left(b+m\right)< b\left(a+m\right)\)

\(\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\)

TH2 : a=b

\(\Rightarrow am=bm\)

\(\Rightarrow ab+am=ab+bm\Rightarrow a\left(b+m\right)=b\left(a+m\right)\)

\(\Rightarrow\frac{a}{b}=\frac{a+m}{b+m}\)

TH1 : a>b

\(\Rightarrow am>bm\)

\(\Rightarrow ab+am>ab+bm\Rightarrow a\left(b+m\right)>b\left(a+m\right)\)

\(\Rightarrow\frac{a}{b}>\frac{a+m}{b+m}\)

Vậy ... ( có 3 trường hợp )