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Lời giải:
a.
\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)
\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)
b.
\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)
$\Rightarrow 10A< 10B\Rightarrow A< B$
Lời giải:
a) Xét hiệu \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{(a+n).b-a(b+n)}{b(b+n)}=\frac{n(b-a)}{b(b+n)}\)
Nếu $b>a$ thì $\frac{a+n}{b+n}-\frac{a}{b}>0\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$
Nếu $b<a$ thì $\frac{a+n}{b+n}-\frac{a}{b}<0\Rightarrow \frac{a+n}{b+n}<\frac{a}{b}$
Nếu $b=a$ thì $\frac{a+n}{b+n}-\frac{a}{b}=0\Rightarrow \frac{a+n}{b+n}=\frac{a}{b}$
b) Rõ ràng $10^{11}-1< 10^{12}-1$.
Đặt $10^{11}-1=a; 10^{12}-1=b; 11=n$ thì: $a< b$; $A=\frac{a}{b}$ và $B=\frac{10^{11}+10}{10^{12}+10}=\frac{a+n}{b+n}$
Áp dụng kết quả phần a:
$b>a\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$ hay $B>A$
Lời giải:
$B=\frac{10^{11}+10}{10^{12}+10}$
Đặt $10^{11}-1=a; 10^{12}-1=b$ thì $0< a< b$. Khi đó:
$A-B=\frac{a}{b}-\frac{a+11}{b+11}=\frac{11(a-b)}{b(b+11)}<0$
$\Rightarrow A< B$
Giải:
a) A=1718+1/1719+1
17A=1719+17/1719+1
17A=1719+1+16/1719+1
17A=1+16/1719+1
Tương tự:
B=1717+1/1718+1
17B=1718+17/1718+1
17B=1718+1+16/1718+1
17B=1+16/1718+1
Vì 16/1719+1<16/1718+1 nên 17A<17B
⇒A<B
b) A=108-2/108+2
A=108+2-4/108+2
A=1+-4/108+2
Tương tự:
B=108/108+4
B=108+4-4/108+1
B=1+-4/108+1
Vì -4/108+2>-4/108+1 nên A>B
c)A=2010+1/2010-1
A=2010-1+2/2010-1
A=1+2/2010-1
Tương tự:
B=2010-1/2010-3
B=2010-3+2/2010-3
B=1+2/2010-3
Vì 2/2010-3>2/2010-1 nên B>A
⇒A<B
Chúc bạn học tốt!
17A=1719+1+16/1719+1
17A=1+16/1719+1
phần in nghiêng mình không hiểu lắm, bn giải thích cho mình được ko?
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
Ta có:
\(10A=\dfrac{10\left(10^{2020}+1\right)}{10^{2021}+1}=\dfrac{10^{2021}+10}{10^{2021}+1}=1+\dfrac{9}{10^{2021}+1}\)
\(10B=\dfrac{10\left(10^{2021}+1\right)}{10^{2022}+1}=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
⇒ \(10A>10B\) ( vì \(\dfrac{9}{10^{2021}+1}>\dfrac{9}{10^{2022}+1}\) )
Suy ra: \(A>B\)
2:
a: 2/9-x=-5/9
=>x=2/9+5/9=7/9
b: x-7/13=1/2
=>x=1/2+7/13=27/26
câu a
\(\dfrac{7}{4}+\dfrac{3}{2}+\dfrac{-9}{16}\\ =\dfrac{28}{16}+\dfrac{24}{16}-\dfrac{9}{16}=\dfrac{43}{16}\)
câu b
\(-\dfrac{2}{7}+\dfrac{3}{5}+\dfrac{9}{7}+\dfrac{-18}{5}\\ =-\dfrac{10}{35}+\dfrac{21}{35}+\dfrac{45}{35}-\dfrac{126}{35}\\ =-\dfrac{70}{35}=-2\)
câu c
\(-\dfrac{5}{13}+\dfrac{11}{10}-\dfrac{-9}{10}+\dfrac{-8}{13}\\ =-\dfrac{5}{13}+\dfrac{11}{10}+\dfrac{9}{10}-\dfrac{8}{13}\\ =-\dfrac{50}{130}+\dfrac{143}{130}+\dfrac{117}{130}-\dfrac{80}{130}\\ =\dfrac{130}{130}=1\)
bài 2
câu a
\(\dfrac{2}{9}-x=-\dfrac{5}{9}\\ x=\dfrac{2}{9}-\dfrac{-5}{9}\\ x=\dfrac{7}{9}\)
câu b
\(x+\dfrac{-7}{13}=\dfrac{1}{2}\\ x=\dfrac{1}{2}-\dfrac{-7}{13}\\ x=\dfrac{13}{26}+\dfrac{14}{26}\\ x=\dfrac{17}{26}\)
a, Ta có : \(10^{15}\cdot11=10^{15}\left(10+1\right)=10^{16}+10^{15}\)
Vì \(10^{16}+10^{15}>10^{16}+10\)
\(\Rightarrow\dfrac{10^{16}+10^{15}}{10^{16}+1}>\dfrac{10^{16}+10}{10^{16}+1}\)
Hay A>B
b, Ta có : \(C=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}}{10^{10}-1}+\dfrac{1}{10^{10}-1}\)
\(D=\dfrac{10^{10}-1}{10^{13}-3}=\dfrac{10^{10}}{10^{13}-3}+\dfrac{-1}{10^{13}-3}\)
Vì \(\dfrac{10^{10}}{10^{10}-1}>\dfrac{10^{10}}{10^{13}-3};\dfrac{1}{10^{10}-1}>\dfrac{-1}{10^{13}-3}\)
\(\Rightarrow\dfrac{10^{10}+1}{10^{10}-1}>\dfrac{10^{10}-1}{10^{13}-3}\)
Hay C > D