Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Có : 10A = 10^15-10/10^15-11 = (10^15-11)+1/10^15-11 = 1 + 1/10^15-11
10B = 10^15+10/10^15+9 = (10^15+9)+1/10^15+9 = 1 + 1/10^15+9
Vì 10^15-11 < 10^15-9 => 1/10^15-11 > 1/10^15+9 => 10A > 10B
=> A < B
k mk nha
a, Ta có:
\(\dfrac{-13}{39}=\dfrac{-1}{3}\) và \(-\dfrac{21}{63}=\dfrac{-1}{3}\)
Vì \(\dfrac{-1}{3}=\dfrac{-1}{3}\) nên \(\dfrac{-13}{39}=-\dfrac{21}{63}\)
b, Ta có:
\(\dfrac{1}{234567}>0\) (số hữu tỉ dương) và \(-\dfrac{2}{14}< 0\) (số hữu tỉ âm)
=> \(\dfrac{1}{234567}>-\dfrac{2}{14}\)
c\(\dfrac{1}{2012}>-\dfrac{1}{14}\), Ta có:
\(\dfrac{-39}{65}=\dfrac{-3}{5}\) và \(-\dfrac{21}{35}=\dfrac{-3}{5}\)
mà \(\dfrac{-3}{5}=\dfrac{-3}{5}\) nên \(\dfrac{-39}{65}=-\dfrac{21}{35}\)
d,Ta có:
\(\dfrac{1}{2012}>0\) (số hữu tỉ dương) và \(-\dfrac{1}{14}< 0\) (số hữu tỉ âm)
Vậy suy ra: \(\dfrac{1}{2012}>-\dfrac{1}{14}\)
Lời giải:
Ta có:
\(\frac{1}{13}; \frac{1}{14}; \frac{1}{15}<\frac{1}{12}\)
\(\Rightarrow \frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{61}; \frac{1}{62};\frac{1}{63}< \frac{1}{60}\)
\(\Rightarrow \frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{3}{60}=\frac{1}{20}\)
Do đó:
\(A< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{9}{20}+\frac{1}{20}\)
\(\Leftrightarrow A< \frac{1}{2}\) (đpcm)
Đặt biểu thức bằng A:
\(\Rightarrow A=\dfrac{1}{5}\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
Ta thấy: \(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< 3.\dfrac{1}{61}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< 3.\dfrac{1}{61}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{3}{31}+\dfrac{3}{61}< \dfrac{1}{2}\left(đpcm\right)\)
10:15 ; \(\dfrac{16}{9}\):\(\dfrac{16}{24}\) ; \(\dfrac{2}{3}\):\(\dfrac{1}{4}\) ; 16:(-4) ; 14:21 ; -5:15 ; 12:(-3) ; -1,2:3,6
10:15=\(\dfrac{2}{3}\) ;\(\dfrac{16}{24}\)=\(\dfrac{2}{3}\) ;16:(-4)=-4 ;14:21=\(\dfrac{2}{3}\) :-5:15=\(\dfrac{-1}{3}\) ;12:(-3)=-4
-1,2:3,6=\(\dfrac{-1}{3}\)
Ta có các tỉ lệ thức: \(\dfrac{10}{15}\)=\(\dfrac{16}{24}\)=\(\dfrac{14}{21}\)=\(\dfrac{2}{3}\) ;\(\dfrac{16}{-4}\)=\(\dfrac{12}{-3}\)=-4 ;\(\dfrac{-5}{15}\)=\(\dfrac{-1,2}{3,6}\)=\(\dfrac{-1}{3}\)
\(a,x^2=16\)
\(x^2=4^2=\left(-4\right)^2\)
\(x=2\) hoặc \(x=-2\)
\(b,x^3=-8\)
\(x^3=\left(-2\right)^3\)
\(x=-2\)
\(c,\left(x+2\right)^2=4\)
\(\left(x+2\right)^2=2^2=\left(-2\right)^2\)
\(x+2=2\Rightarrow x=0\) hoặc \(x+2=-2\Rightarrow x=-4\)
\(d,\left(1-x\right)^3=1\)
\(1-x=1\)
\(x=0\)
e,phần này mk chưa nghĩ ra,sorry bn nha!
Câu 10: Tìm x, biết
a)\(\left(-1\dfrac{2}{5}+x\right)\div\dfrac{14}{15}=2\dfrac{1}{3}\)
\(\Leftrightarrow-1\dfrac{2}{5}+x=\dfrac{98}{45}\)
\(\Leftrightarrow x=\dfrac{161}{45}\)
Bài 1 :
Ta có :
\(a+c=2b\left(1\right)\)
\(2bd=c\left(b+d\right)\left(2\right)\)
Thay \(\left(1\right)\) vào \(\left(2\right)\) ta được :
\(\left(a+c\right)d=c\left(b+d\right)\)
\(\Leftrightarrow ad+cd=cb+cd\)
\(\Leftrightarrow ad=cb\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\rightarrowđpcm\)
Bài 2 :
\(a,\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\)
\(=\dfrac{5^{21}\left(2.5-9\right)}{5^{20}}\)
\(=5\left(10-9\right)\)
\(=5\)
b, \(\dfrac{5\left(3.7^{15}-19.17^{14}\right)}{7^{14}+3.7^{15}}\)
\(=\dfrac{5.2.7^{14}}{10.7^{15}}\)
\(=\dfrac{1}{7}\)
Câu 2:
\(B=\dfrac{5^{21}\cdot\left(2\cdot5-9\right)}{5^{20}}\cdot\dfrac{7^{15}\left(7+3\right)}{15\cdot7^{15}-95\cdot7^{14}}\)
\(=\dfrac{5\cdot1}{1}\cdot\dfrac{7^{15}\cdot10}{7^{14}\cdot\left(15\cdot7-95\right)}\)
\(=5\cdot\dfrac{7\cdot10}{105-95}=5\cdot7=35\)
a)= \(\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)+\dfrac{11}{125}\)
= \(\dfrac{-1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{11}{125}\)
= 0 + \(\dfrac{11}{125}\)
= \(\dfrac{11}{125}\)
b) \(=\left(1-1\right)+\left(\dfrac{-1}{2}-\dfrac{1}{2}\right)+\left(2-2\right)\) +
\(\left(\dfrac{-2}{3}-\dfrac{1}{3}\right)+\left(3-3\right)+\left(\dfrac{-3}{4}-\dfrac{1}{4}\right)\) + 4
= 0 + (-1) + 0 + (-1) + 0 + (-1) + 4
= -1
c) = \(\dfrac{1}{3}.\dfrac{14}{25}-\dfrac{1}{2}.\dfrac{14}{25}\)
= \(\dfrac{14}{25}.\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\)
= \(\dfrac{14}{25}.\left(\dfrac{-1}{6}\right)\)
= \(\dfrac{-7}{75}\)
d) = \(\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\)
= 1 + (-1)
= 0
\(A=\dfrac{14^{14}+1}{14^{15}+1}\)
\(\Rightarrow14.A=\dfrac{14^{15}+14}{14^{15}+1}\)
\(\Rightarrow14.A=\dfrac{14^{15}+1}{14^{15}+1}+\dfrac{13}{14^{15}+1}\)
\(\Rightarrow14.A=1+\dfrac{13}{14^{15}+1}\)
\(B=\dfrac{14^{15}+1}{14^{16}+1}\)
\(\Rightarrow14.B=\dfrac{14^{16}+14}{14^{16}+1}\)
\(\Rightarrow14.B=\dfrac{14^{16}+1}{14^{16}+1}+\dfrac{13}{14^{16}+1}\)
\(\Rightarrow14.B=1+\dfrac{13}{14^{16}+1}\)
Nhận xét: \(\dfrac{13}{14^{15}+1}>\dfrac{13}{14^{16}+1}\) (cùng tử, xét mẫu)
\(\Rightarrow A>B\)
Vậy \(A>B\)