So sánh \(3\frac{9}{10}\) và \(2\frac{9}{10}\)
SO SÁNH CÁC HỖN SỐ
A) \(3\frac{9}{10}\)VÀ \(2\frac{9}{10}\)
B)\(3\frac{4}{10}\) VÀ \(3\frac{9}{10}\)
C)\(5\frac{1}{10}\) VÀ \(2\frac{9}{10}\)
D)\(3\frac{4}{10}\) VÀ \(3\frac{2}{5}\)
so sánh $A=\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}$ và $B=\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}$
So Sánh :
a) \(\frac{9^{10}-4}{9^{10}-5}\) và \(\frac{9^{10}-2}{9^{10}-3}\)
b)\(\frac{2.7^{10}-1}{7^{10}}\) và \(\frac{2.7^{10}+1}{7^{10}+1}\)
a ) Ta có :
\(\frac{9^{10}-4}{9^{10}-5}=\frac{9^{10}-5+1}{9^{10}-5}=1+\frac{1}{9^{10}-5}\)
\(\frac{9^{10}-2}{9^{10}-3}=\frac{9^{10}-3+1}{9^{10}-3}=1+\frac{1}{9^{10}-3}\)
Do \(\frac{1}{9^{10}-5}>\frac{1}{9^{10}-3}\)
\(\Rightarrow1+\frac{1}{9^{10}-5}>1+\frac{1}{9^{10}-3}\)
\(\Rightarrow\frac{9^{10}-4}{9^{10}-5}>\frac{9^{10}-2}{9^{10}-3}\)
b ) Ta có :
\(\frac{2.7^{10}-1}{7^{10}}=2-\frac{1}{7^{10}}\)
\(\frac{2.7^{10}+1}{7^{10}+1}=\frac{2.7^{10}+2-1}{7^{10}+1}=\frac{2\left(7^{10}+1\right)-1}{7^{10}+1}=2-\frac{1}{7^{10}+1}\)
Do \(\frac{1}{7^{10}}>\frac{1}{7^{10}+1}\)
\(\Rightarrow2-\frac{1}{7^{10}}< 2-\frac{1}{7^{10}+1}\)
\(\Rightarrow\frac{2.7^{10}-1}{7^{10}}< \frac{2.7^{10}+1}{7^{10}+1}\)
So sánh hai phân số
\(A=\frac{10^9+5}{10^9-2}\)và \(B=\frac{10^9}{10^9-7}\)
So sánh hai phân số
\(A=\frac{10^9+5}{10^9-2}\) và \(B=\frac{10^9}{10^9-7}\)
\(A=\frac{10^9+5}{10^9-2}\)
\(=\frac{10^9-2}{10^9-2}+\frac{7}{10^9-2}\)
\(=1+\frac{7}{10^9-2}\)
\(B=\frac{10^9}{10^9-7}\)
\(=\frac{10^9-7}{10^9-7}+\frac{7}{10^9-7}\)
\(=1+\frac{7}{10^9-7}\)
Vì \(7\over10^9-5\)<\(7\over10^9-7\) nên A<B
So sánh :
\(\frac{10^8+1}{10^9+1}và\frac{10^9+1}{10^{10}+1}\)
\(A=\frac{10^8+1}{10^9+1}=\frac{1}{10}\left(\frac{10^9+10}{10^9+1}\right)=\frac{1}{10}\left(1+\frac{9}{10^9+1}\right)\)
\(B=\frac{10^9+1}{10^{10}+1}=\frac{1}{10}\left(\frac{10^{10}+10}{10^{10}+1}\right)=\frac{1}{10}\left(1+\frac{9}{10^{10}+1}\right)\)
\(\frac{9}{10^9+1}>\frac{9}{10^{10}+1}\)
\(\Rightarrow A>B\)
Đặt \(M=\frac{10^8+1}{10^9+1}\) và \(N=\frac{10^9+1}{10^{10}+1}\)
Có : \(M=\frac{10^8+1}{10^9+1}\)
\(\Rightarrow10M=\frac{10^9+10}{10^9+1}=\frac{10^9+1+9}{10^9+1}=1+\frac{9}{10^9+1}\)
Lại có : \(N=\frac{10^9+1}{10^{10}+1}\)
\(\Rightarrow10N=\frac{10^{10}+10}{10^{10}+1}=\frac{10^{10}+1+9}{10^{10}+1}=1+\frac{9}{10^{10}+1}\)
Vì \(\frac{9}{10^9+1}>\frac{9}{10^{10}+1}\) nên \(1+\frac{9}{10^9+1}>1+\frac{9}{10^{10}+1}\)
\(\Rightarrow10M>10N\Rightarrow M>N\)
Vậy M > N.
\(A=\frac{10^9+2}{10^9-1}vaB=\frac{10^9}{10^9-3}\)
so sánh
trả lời đầy đủ một tích(phải đúng)
ta có : \(\frac{10^9+2}{10^9-1}=\frac{10^9}{10^9-3}\)
\(\Leftrightarrow\left(10^9+2\right)\left(10^9-3\right)=\left(10^9-1\right)10^9\)
\(\Leftrightarrow10^{18}-10^9.3+2.10^9-6=10^{18}-10^9\)
\(\Rightarrow10^{18}-10^9.3+2.10^9-6=10^{18}-\left(10^9.3-2.10^9+6\right)\)
\(=10^{18}-\left(10^9+6\right)\)
vì \(-10^9>-\left(10^9+6\right)\Rightarrow10^{18}-10^9>10^{18}-\left(10^9+6\right)\)
\(\Rightarrow A>B\)
Ta có: A=\(\frac{10^9+2}{10^9-1}=\frac{10^9-1+3}{10^9-1}=1+\frac{3}{10^9-1}\)
B=\(\frac{10^9}{10^9-3}=\frac{10^9-3+3}{10^9-3}=1+\frac{3}{10^9-3}\)
Mà \(\frac{3}{10^9-1}< \frac{3}{10^9-3}\Rightarrow1+\frac{3}{10^9-1}< 1+\frac{3}{10^9-3}\Rightarrow A< B\)
Vậy A<B
Ta có :
\(A=\frac{10^9+2}{10^9-1}=\frac{10^9-1+3}{10^9-1}=1+\frac{3}{10^9-1}\)
\(B=\frac{10^9}{10^9-3}=\frac{10^9-3+3}{10^9-3}=1+\frac{3}{10^9-3}\)
Vì \(10^9-1>10^9-3\) nên \(\frac{3}{10^9-1}< \frac{3}{10^9-3}\) \(\Rightarrow1+\frac{3}{10^9-1}< 1+\frac{3}{10^9-3}\)
Do đó : \(A< B\)
So sánh A và B biết:
\(A=\frac{7^{10}}{1+7+7^2+...+7^9}\)và \(B=\frac{3^{10}}{1+3+3^2+...+3^9}\)
xét A và B có: số mũ từ 2 đến 9 giống nhau; mẫu đều cộng 1
=> Ta chỉ có thể so sánh phần cơ số
vì 7>3 => 7 mũ n>3 mũ n
=> A lớn hơn B
Cho \(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
So sánh S với 10
Ta có :
\(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
\(S=\frac{2+1}{2}+\frac{3+1}{3}+\frac{4+1}{4}+...+\frac{11+1}{11}\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{11}\right)\)
\(S=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)\)
\(S=10+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)>10\)
\(\Rightarrow\)\(S>10\)
Vậy \(S>10\)
Chúc bạn học tốt ~