Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
M=\(\dfrac{10^{100^{ }}+1}{10^{101}+1}\)
M=\(\dfrac{10^{99+1}+1}{10^{100+1}+1}\)
M=\(\dfrac{10^{99}.10+1}{10^{100}.10+1}\)
N=\(\dfrac{10^{99^{ }}+1}{10^{100}+1}\)
=>M lớn hơn N
M>N,vì:\(\dfrac{10^{100}+1}{10^{101}+1}=\dfrac{10^{100}}{10^{101}}\)
\(\dfrac{10^{99}+1}{10^{100}+1}=\dfrac{10^{99}}{10^{100}}\)
\(\dfrac{10^{100}}{10^{101}}>\dfrac{10^{99}}{10^{100}}\)
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\left(n\in N\right)\)
\(B=\dfrac{10^{20}+1}{10^{21}+1}< 1\)
\(B< \dfrac{10^{20}+1+9}{10^{21}+1+9}\Rightarrow B< \dfrac{10^{20}+10}{10^{21}+10}\Rightarrow B< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\Rightarrow B< \dfrac{10^{19}+1}{10^{20}+1}=A\)\(\Rightarrow B< A\)
Trước hết ta hãy so sánh :
\(\dfrac{10^{100}+1}{10^{101}+1}\)với \(\dfrac{10^{100}+1}{10^{102}+1}\)
Ta có: Cả hai phân số trên cùng tử.
\(\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{102}+1}\)
Tiếp đó so sánh : \(\dfrac{10^{101}+1}{10^{102}+1}\)với \(1\)
Ta được: \(\dfrac{10^{101}+1}{10^{102}+1}< 1\)
Ta lại so sánh được:\(\dfrac{10^{100}+1}{10^{102}+1}< 1\) (*)
Từ (*) suy ra \(\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+2}< \dfrac{10^{101}+1}{10^{102}+1}< 1\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+1}\)
Ngoài ra còn một cách như sau:
\(\dfrac{10^{101}+1}{10^{102}+1}=\dfrac{10^{\left(100+1\right)}+1}{10^{\left(101+1\right)}+1}=\dfrac{10}{10}.\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{101}+1}\) hay B > A hay A < B
Bài 1:
d)
\(\dfrac{x+5}{95}+\dfrac{x+10}{90}+\dfrac{x+15}{85}+\dfrac{x+20}{80}=-4\)
\(\Leftrightarrow\dfrac{x+5}{95}+1+\dfrac{x+10}{90}+1+\dfrac{x+15}{85}+1+\dfrac{x+20}{80}+1=-4+1+1+1+1\)
\(\Leftrightarrow\dfrac{x+100}{95}+\dfrac{x+100}{90}+\dfrac{x+100}{85}+\dfrac{x+100}{80}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\right)=0\)
\(\Leftrightarrow x+100=0\) ( vì: \(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\ne0\))
\(\Leftrightarrow x=-100\)
\(B=\dfrac{9}{10!}+\dfrac{10}{11!}+...........+\dfrac{99}{100!}\)
Ta thấy :
\(\dfrac{9}{10!}=\dfrac{10-1}{10!}=\dfrac{1}{9!}-\dfrac{1}{10!}\)
\(\dfrac{10}{11!}< \dfrac{11-1}{11!}=\dfrac{1}{10!}-\dfrac{1}{11!}\)
..........................
\(\dfrac{99}{100!}< \dfrac{100-1}{100!}=\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{10!}+\dfrac{1}{10!}-\dfrac{1}{11!}+...........+\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}-\dfrac{1}{100!}\)
\(\Leftrightarrow B< \dfrac{1}{9!}\rightarrowđpcm\)
Ta có :
\(A=\dfrac{100^{10}+1}{100^{10}-1}=\dfrac{100^{10}-1+2}{100^{10}-1}=\dfrac{100^{10}-1}{100^{10}-1}+\dfrac{2}{100^{10}-1}=1+\dfrac{2}{100^{10}-1}\)
\(B=\dfrac{100^{10}-1}{100^{10}-3}=\dfrac{100^{10}-3+2}{100^{10}-3}=\dfrac{100^{10}-3}{100^{10}-3}+\dfrac{2}{100^{10}-3}=1+\dfrac{2}{100^{10}-3}\)
\(\) Vì \(1+\dfrac{2}{100^{10}-1}< 1+\dfrac{2}{100^{10}-3}\Rightarrow A< B\)
9: \(=1-\dfrac{1}{99}+1-\dfrac{1}{100}+\dfrac{100}{101}\cdot\dfrac{1-4+3}{12}=2-\dfrac{199}{9900}=\dfrac{19601}{9900}\)
10: \(=\left(\dfrac{78}{79}+\dfrac{79}{80}+\dfrac{80}{81}\right)\cdot\dfrac{6+5+9-20}{30}=0\)
Ta có A = \(\frac{10^{100}-1}{10^{98}-1}=\frac{10^{98}.10^2-10^2+99}{10^{98}-1}\)
\(=\frac{10^2\left(10^{98}-1\right)+99}{10^{98-1}}\)
\(=10^2+\frac{99}{10^{98}-1}\)
B= \(\frac{10^{101}-1}{10^{99}-1}=\frac{10^{99}.10^2-10^2+99}{10^{99}-1}\)
\(=\frac{10^2\left(10^{99}-1\right)+99}{10^{99}-1}\)
\(=10^2+\frac{99}{10^{99}-1}\)
Vì \(\frac{99}{10^{98}-1}>\frac{99}{10^{99}-1}\)nên \(10^2+\frac{99}{10^{98}-1}>10^2+\frac{99}{10^{99}-1}\)=> A > B
Vậy A > B
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
=\(\dfrac{3}{2}.\dfrac{56}{305}\)
= \(\dfrac{78}{305}\)
\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) hoặc \(6-2x=0\)
*Nếu \(x^2-4=0\)
⇒ x2 = 4
⇒ x ∈ {2 ; -2}
*Nếu \(6-2x=0\)
⇒2x = 6
⇒ x = 6 : 2 = 3
Vậy x ∈ { -2 ; 2 ; 3 }
c) E = \(\dfrac{4116-14}{10290-35}\) và K = \(\dfrac{2929-101}{2.1919+404}\)
E = \(\dfrac{4116-14}{10290-35}\)
E = \(\dfrac{14.\left(294-1\right)}{35.\left(294-1\right)}\)
E = \(\dfrac{14}{35}\)
K = \(\dfrac{2929-101}{2.1919+404}\)
K = \(\dfrac{101.\left(29-1\right)}{101.\left(38+4\right)}\)
K = \(\dfrac{29-1}{34+8}\)
K = \(\dfrac{28}{42}\) = \(\dfrac{2}{3}\)
Ta có : E = \(\dfrac{14}{35}\) và K = \(\dfrac{2}{3}\)
\(\dfrac{14}{35}\) = \(\dfrac{42}{105}\)
\(\dfrac{2}{3}\) = \(\dfrac{70}{105}\)
Vậy E < K
Các câu còn lại tương tự
a) Giải
Đặt \(M=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\)
\(\Rightarrow A< A.M\)
hay \(A< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\right)\)
\(\Rightarrow A< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{98}{99}.\dfrac{99}{100}\)
\(\Leftrightarrow A< \dfrac{1.2.3.4.5.6...98.99}{2.3.4.5.6.7...99.100}\)
\(\Rightarrow A< \dfrac{1}{100}< \dfrac{1}{10}\)
Vậy \(A< \dfrac{1}{10}\)
ta có:
1/10.A=10100+1/10(1099+1)
1/10.A=10100+1/10100+10
1/10.A=1-(9/10100+10)
1/10.B=10101+1/10(10100+1)
1/10.B=10101+1/10101+10
1/10.B=1-(9/10101+10)
vì(10101+10)>(10100+1)=> 9/10101+10 < 9/10100+10 => 1-(9/10101+10) > 1-(9/10100+10)
hay 1/10.A>1/10.B
=>A>B
ta có:
1/10.A=10100+1/10(1099+1)
1/10.A=10100+1/10100+10
1/10.A=1-(9/10100+10)
1/10.B=10101+1/10(10100+1)
1/10.B=10101+1/10101+10
1/10.B=1-(9/10101+10)
vì(10101+10)>(10100+1)=> 9/10101+10 < 9/10100+10 => 1-(9/10101+10) < 1-(9/10100+10)
hay 1/10.A<1/10.B
=>A<B