Cho A=5^1990+1/5^1991+1 và B=5^1991+1/5^1992+1.
So sánh A và B
So sánh:
A=\(\dfrac{10^{1990}+1}{10^{1991}+1}\) và B=\(\dfrac{10^{1991}+1}{10^{1992}+1}\)
đáng ra là toán lớp 6 đó nhưng mik thích đặt toán lớp 5 :)
A = \(\dfrac{10^{1990}+1}{10^{1991}+1}\) ⇒ 10A = \(\dfrac{10^{1991}+10}{10^{1991}+1}\) = \(1+\dfrac{9}{10^{1991}+1}\)
B = \(\dfrac{10^{1991}+10}{10^{1992}+1}\) ⇒ 10B = \(\dfrac{10^{1992}+10}{10^{1992}+1}\) = 1 + \(\dfrac{9}{10^{1992}+1}\)
Vì \(\dfrac{9}{10^{1991}+1}\) > \(\dfrac{9}{10^{1992}+1}\)
10A > 10B => A > B
So sánh: A=10^1990+1/10^1991+1 và B=10^1991+1/10^1992+1
so sánh: A=(101990+1)/(101991+1) VÀ B=(101991+1)/(101992+1)
so sánh A=10 mũ 1990+1/10 mũ 1991+1 và B=10 mũ 1991+1/10 mũ 1992+1 lời giải chi tiết
so sánh A=10 mũ 1990+1/10 mũ 1991+1 và B=10 mũ 1991+1/10 mũ 1992+1 lời giải chi tiết nha
A=\(\frac{10^{1990+1}}{10^{1991+1}}\);; B=\(\frac{10^{1991+1}}{10^{1992+1}}\)
Hãy so sánh A và B
Ta có : \(A=\frac{10^{1990}+1}{10^{1991}+1}=>10A=\frac{10.\left(10^{1990}+1\right)}{10^{1991}+1}\)
\(=>10A=\frac{10^{1991}+10}{10^{1991}+1}=\frac{\left(10^{1991}+1\right)+9}{10^{1991}+1}\)
\(=>10A=1+\frac{9}{10^{1991}+1}\)
Ta lại có : \(B=\frac{10^{1991}+1}{10^{1992}+1}=>10B=\frac{10.\left(10^{1991}+1\right)}{10^{1992}+1}\)
Tương tự như A => \(10B=1+\frac{9}{10^{1992}+1}\)
Vì \(\frac{9}{10^{1991}+1}>\frac{9}{10^{1992}+1}=>10A>10B\)
\(=>A>B\)
đăt 10A=\(\frac{10^{1991}+1}{10^{1991}+1}\)=1+\(\frac{9}{10^{1991}}\)
Câu B tương tự
ta có:\(\frac{9}{10^{1991}+1}\)>\(\frac{9}{10^{1992}}\)
nên 10A>10B
=>A>b
so sánh:
\(A=\frac{10^{1990}+1}{10^{1991}+1}\)và\(B=\frac{10^{1991}+1}{10^{1992}+1}\)
Áp dụng a/b < 1 => a/b < a+m/b+m (a;b;m thuộc N*)
=> \(B=\frac{10^{1991}+1}{10^{1992}+1}< \frac{10^{1991}+1+9}{10^{1992}+1+9}\)
=> \(B< \frac{10^{1991}+10}{10^{1992}+10}\)
=> \(B< \frac{10.\left(10^{1990}+1\right)}{10.\left(10^{1991}+1\right)}\)
=> \(B< \frac{10^{1990}+1}{10^{1991}+1}=A\)
=> B < A
Bài này mình biết làm nè , nhưng ... dài dòng lắm
So Sánh:
A=101990+1 phần 101991+1 và B= 101991+1 phần 101992+ 1
A=10^1990+1/10^1991
A=10.(10^1990+1 / 10^1991+1)
10A=10^1991+10 / 10^1991+1
10A=10^1991+1 / 10^1991+1 +9/10^1991+1
10A=1 + 9/10^1991
B=10^1991+1 / 10^1992+1
B=10.(10^1991+1 / 10^1992+1)
10B=10^1992+10 / 10^1992+1
10B=10^1992+1 / 10^1992+1 + 9/10^1992+1
10B= 1+9/10^1992+1
Ta có 9/10^1991 > 9/10^1992
10A > 10B
A > B
Vì \(\frac{10^{1994}+1}{10^{1992}+1}\)<1
=> \(\frac{10^{1994}+1}{10^{1992}+1}\)<\(\frac{10^{1994}+1+9}{10^{1992}+1+9}\)
Ta có \(\frac{10^{1994}+1+9}{10^{1992}+1+9}\)=\(\frac{10\left(10^{1990}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\frac{10^{1990}+1}{10^{1991}+2}\)
=>\(\frac{10^{1994}+1}{10^{1992}+1}\)<\(\frac{10^{1990}+1}{10^{1991}+2}\)
Vậy B < A
So sánh :
\(A=\frac{10^{1990}+1}{10^{1991}+1}\) và \(B=\frac{10^{1991}+1}{10^{1992}+1}\)
\(A=\frac{10^{1990}+1}{10^{1991}+1}\Rightarrow10A=\frac{10^{1991}+10}{10^{1991}+1}=1+\frac{9}{10^{1991}+1}\)
\(B=\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow10B=\frac{10^{1992}+10}{10^{1992}+1}=1+\frac{9}{10^{1992}+1}\)
Vì \(10^{1991}< 10^{1992}\Rightarrow1+\frac{9}{10^{1991}+1}>1+\frac{9}{10^{1992}+1}\)
\(\Rightarrow\frac{10^{1990}+1}{10^{1991}+1}>\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow A>B\)
Ta có : \(B=\frac{10^{1991}+1}{10^{1992}+1}< \frac{10^{1991}+1+9}{10^{1992}+1+9}\)
Mà : \(\frac{10^{1991}+1+9}{10^{1992}+1+9}=\frac{10^{1991}+10}{10^{1992}+10}\)
\(=\frac{10\left(10^{1990}+1\right)}{10\left(10^{1991}+1\right)}\)
\(=\frac{10^{1990}+1}{10^{1991}+1}\)
\(\Rightarrow B< A\)
Giải
+) Ta có \(A=\frac{10^{1990}+1}{10^{1991}+1}\)
\(10A=\frac{10\left(10^{1990}+1\right)}{10^{1991}+1}\)
\(=\frac{10.10^{1990}+10.1}{10^{1991}+1}\)
\(=\frac{10^{1991}+10}{10^{1991}+1}\)
\(=\frac{10^{1991}+1+9}{10^{1991}+1}\)
\(=\frac{10^{1991}+1}{10^{1991}+1}+\frac{9}{10^{1991}+1}\)
\(=1+\frac{9}{10^{1991}+1}\)
+) Ta có \(B=\frac{10^{1991}+1}{10^{1992}+1}\)
\(10B=\frac{10\left(10^{1991}+1\right)}{10^{1992}+1}\)
\(=\frac{10.10^{1991}+10.1}{10^{1992}+1}\)
\(=\frac{10^{1992}+10}{10^{1992}+1}\)
\(=\frac{10^{1992}+1+9}{10^{1992}+1}\)
\(=\frac{10^{1992}+1}{10^{1992}+1}+\frac{9}{10^{1992}+1}\)
\(=1+\frac{9}{10^{1992}+1}\)
+) Vì \(10^{1991}+1< 10^{1992}+1\)
\(\Rightarrow\frac{9}{10^{1991}+1}>\frac{9}{10^{1992}+1}\)
\(\Rightarrow1+\frac{9}{10^{1991}+1}>\text{}1+\frac{9}{10^{1992}+1}\text{}\)
Hay \(10A>10B\)
\(\Rightarrow A>B\)