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1) \(x+\dfrac{30}{100}x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}x=-\dfrac{131}{100}\)
\(\Leftrightarrow100x+30x=-131\)
\(\Leftrightarrow130x=-131\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
Vậy \(x=-\dfrac{131}{130}\)
b) \(\left(4,5-2x\right)\cdot\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow\left(\dfrac{9}{2}-2x\right)\cdot\left(-\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow-\dfrac{18}{7}+\dfrac{8}{7}x=\dfrac{11}{4}\)
\(\Leftrightarrow-72+32x=77\)
\(\Leftrightarrow32x=77+72\)
\(\Leftrightarrow32x=149\)
\(\Leftrightarrow x=\dfrac{149}{32}\)
Vậy \(x=\dfrac{149}{32}\)
A=1+2+3+4+5+...+99+100
A=\(\dfrac{100.\left(100+1\right)}{2}\)=5050
Vậy A=5050
B=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
B=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
B=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B=\(1-\dfrac{1}{100}\)=\(\dfrac{99}{100}\)
Vậy B=\(\dfrac{99}{100}\)
Chuẩn ồi còn j
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\left(n\in N\right)\)
\(B=\dfrac{10^{20}+1}{10^{21}+1}< 1\)
\(B< \dfrac{10^{20}+1+9}{10^{21}+1+9}\Rightarrow B< \dfrac{10^{20}+10}{10^{21}+10}\Rightarrow B< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\Rightarrow B< \dfrac{10^{19}+1}{10^{20}+1}=A\)\(\Rightarrow B< A\)
Áp dụng tính chất : \(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) (\(a;b,m\in N\)*)
Ta có :
\(A=\dfrac{100^{2007}+1}{100^{2008}+1}< \dfrac{100^{2007}+1+99}{100^{2008}+1+99}=\dfrac{100^{2007}+100}{100^{2008}+100}=\dfrac{100\left(100^{2006}+1\right)}{100\left(100^{2007}+1\right)}=\dfrac{100^{2006}+1}{100^{2007}+1}=B\)
\(\Rightarrow A< B\)
Ta có:
A = \(\dfrac{100^{100}+1}{100^{90}+1}\)> 1 \(\Rightarrow\) A > \(\dfrac{100^{100}+1+99}{100^{90}+1+99}\) = \(\dfrac{100^{100}+100}{100^{90}+100}\)
\(\Rightarrow\) A > \(\dfrac{100\left(100^{99}+1\right)}{100\left(100^{89}+1\right)}\) = B
Vậy A > B
\(100A=\dfrac{100^{2016}+100}{100^{2016}+1}=1+\dfrac{99}{100^{2016}+1}\)
\(100B=\dfrac{100^{2017}+100}{100^{2017}+1}=1+\dfrac{99}{100^{2017}+1}\)
mà \(100^{2016}< 100^{2017}\)
nên A>B
Bài 4:
=>(x-5)*3/10=1/5x+5
=>3/10x-3/2=1/5x+5
=>1/10x=5+3/2=6,5
=>0,1x=6,5
=>x=65
Ta thấy:
A < 1
B >1
=> A > B