So sánh :
a) \(\left(\frac{1}{243}\right)^9\) và \(\left(\frac{1}{83}\right)^{13}\)
b)1990^10+1990^9 và 1991^10
So sánh:
\(^{ }\left(\frac{1}{243}\right)^9\) và \(\left(\frac{1}{83}\right)^{13}\)
tham khảo nha
http://olm.vn/hoi-dap/question/166511.html
ta có: \(\left(\frac{1}{243}\right)^9\)>\(\left(\frac{1}{83}\right)^{13}\)
So sánh
\(\left(\frac{1}{243}\right)^9\) và \(\left(\frac{1}{83}\right)^{13}\)
ta co( \(\frac{1}{243}\))9=(\(\frac{1}{3}\))45=(\(\frac{1}{81}\))11,25<(\(\frac{1}{83}\))13
ta co( \(\frac{1}{243}\))9=(\(\frac{1}{3}\))45=(\(\frac{1}{81}\))11,25<(\(\frac{1}{83}\))13
So sánh hai phân số :
\(\left(\frac{1}{243}\right)^9\)và \(\left(\frac{1}{83}\right)^{13}\)
Bài 3: Chứng tỏ rằng:
a, Nếu A= \(\dfrac{\left(10^{1990}+1\right)}{10^{1991}+1}\)và B = \(\dfrac{\left(10^{1991}+1\right)}{10^{1992}+1}\)thì A > B
Giúp mik vs! Thanks nha!
Giải:
a) \(A=\dfrac{10^{1990}+1}{10^{1991}+1}\) và \(B=\dfrac{10^{1991}+1}{10^{1992}+1}\)
Ta có:
\(A=\dfrac{10^{1990}+1}{10^{1991}+1}\)
\(10A=\dfrac{10^{1991}+10}{10^{1991}+1}\)
\(10A=\dfrac{10^{1991}+1+9}{10^{1991}+1}\)
\(10A=1+\dfrac{9}{10^{1991}+1}\)
Tương tự :
\(B=\dfrac{10^{1991}+1}{10^{1992}+1}\)
\(10B=\dfrac{10^{1992}+10}{10^{1992}+1}\)
\(10B=\dfrac{10^{1992}+1+9}{10^{1992}+1}\)
\(10B=1+\dfrac{9}{10^{1992}+1}\)
Vì \(\dfrac{9}{10^{1991}+1}>\dfrac{9}{10^{1992}+1}\) nên \(10A>10B\)
\(\Rightarrow A>B\left(đpcm\right)\)
Chúc bạn học tốt!
Cho A= 1990^10+ 1990^9, B= 1991^10. So sánh A và B
Lời giải:
$A=1990^{10}+1990^9=1990^9(1990+1)=1990^9.1991< 1991^9.1991=1991^{10}$
Hay $A< B$
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
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 \(B=\frac{10^{1991}+1}{10^{1992}+1}\)
\(A=\frac{10^{1990}+1}{10^{1991}+1}\Rightarrow10A=\frac{10^{1991}+10}{10^{1991}+1}=1+\frac{9}{10^{1991}+1}\)
\(B=\frac{10^{1991}+1}{10^{1992}+1}\Rightarrow10B=\frac{10^{1992}+10}{10^{1992}+1}=1+\frac{9}{10^{1992}+1}\)
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 \(10A>10B\)
\(\Rightarrow A>B\)
1. so sánh a,10 mũ 10 và 48. 50 mũ 5 b,1990 mũ 10 + 1990 mũ 9 và 1991 mũ 10 c,107 mũ 50 và 73 mũ 75 d,2 mũ 91 và 5 mũ 35 e, A = 72 mũ 45 - 72 mũ 44 và 72 mũ 44 - 72 mũ 43 2 tìm x a, x-2023 /4 = 1 phần x - 2023 b, (2x + 1) mũ 4= (2x + 1) mũ 6 c,(3x-1) mũ 10 = (3x - 1) mũ 20 d, 2 mũ x+1 . 3y = 12x