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a: \(A=\dfrac{-13}{21}=\dfrac{-26}{42}\)
\(B=\dfrac{-9}{14}=\dfrac{-27}{42}\)
mà -26>-27
nên A>B
b: \(A=\dfrac{99}{101}=1-\dfrac{2}{101}\)
\(B=\dfrac{2011}{2013}=1-\dfrac{2}{2013}\)
mà 2/101>2/2013
nên A<B
\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{6}{12}=\dfrac{193}{1066}\)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=\left(17\dfrac{2}{31}-6\dfrac{2}{31}\right)-\dfrac{15}{17}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{1}{2}=\dfrac{193}{1066}\) (Casio :>)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}\)
\(=20+\dfrac{1}{3}=\dfrac{61}{3}\)
Lời giải:
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk, c=dk \)
Khi đó:
\(\frac{2002a+2003b}{2002a-2003b}=\frac{2002bk+2003b}{2002bk-2003b}=\frac{b(2002k+2003)}{b(2002k-2003)}=\frac{2002k+2003}{2002k-2003}(1)\)
\(\frac{2002c+2003d}{2002c-2003d}=\frac{2002dk+2002d}{2002dk-2003d}=\frac{d(2002k+2003)}{d(2002k-2003)}=\frac{2002k+2003}{2002k-2003}(2)\)
Từ \((1);(2)\Rightarrow \frac{2002a+2003b}{2002a-2003b}=\frac{2002c+2003d}{2002c-2003d}\)
Ta có đpcm.
Xét tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) . Gọi giá trị chung của các tỉ số đó là k, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=> \(a=k.b,c=k.d\)
Ta có :
( 1 )
= \(\dfrac{2002a+2003b}{2002a-2003b}=\dfrac{2002kb+2003b}{2002kb-2003b}\)
= \(\dfrac{b.\left(2002k+2003\right)}{b.\left(2002k-2003\right)}=\dfrac{2002k+2003}{2002k-2003}\)
( 2 ) \(\dfrac{2002c+2003d}{2002c-2003d}=\dfrac{2002kd+2003d}{2002kd-2003d}\)
= \(\dfrac{d.\left(2002k+2003\right)}{d.\left(2002k-2003\right)}=\dfrac{2002k+2003}{2002k-2003}\)
Từ ( 1 ) và ( 2 ) => \(\dfrac{2002a+2003b}{2002a-2003b}=\dfrac{2002c+2003d}{2002c-2003d}\)
a;\(\dfrac{17}{24}\) < \(\dfrac{17}{34}\) ⇒ \(\dfrac{-17}{24}\) > \(\dfrac{-17}{34}\) = - \(\dfrac{1}{2}\)
\(\dfrac{25}{31}\) > \(\dfrac{25}{50}\) ⇒ - \(\dfrac{25}{31}\) < \(\dfrac{-25}{50}\) = - \(\dfrac{1}{2}\)
Vậy - \(\dfrac{17}{34}\) > - \(\dfrac{25}{31}\)
b; \(\dfrac{27}{38}\) > \(\dfrac{27}{39}\) > \(\dfrac{25}{39}\)
⇒ - \(\dfrac{27}{38}\) < - \(\dfrac{25}{39}\) = \(\dfrac{-125}{195}\)
Vậy - \(\dfrac{27}{38}\) < - \(\dfrac{125}{195}\)
Bài 1:
a: \(=17+\dfrac{2}{31}-\dfrac{15}{17}-6-\dfrac{2}{31}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
b: \(=31+\dfrac{6}{13}+5+\dfrac{9}{41}-36-\dfrac{9}{41}-36-\dfrac{6}{13}\)
=36
c: \(=27+\dfrac{51}{59}-7-\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
a,\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}\)
câu b và c xem lại đề nha
Chúc bạn học tốt!!!
a) \(\dfrac{12}{47}\) và \(\dfrac{11}{53}\)
Ta có: \(\dfrac{11}{47}>\dfrac{11}{53}\) mà \(\dfrac{12}{47}>\dfrac{11}{47}\)
\(\Rightarrow\dfrac{12}{47}>\dfrac{11}{53}\)
a) Ta có :\(\dfrac{12}{47}>\dfrac{12}{48}=\dfrac{1}{4}=\dfrac{11}{44}>\dfrac{11}{53}\)
\(\Rightarrow\dfrac{12}{47}>\dfrac{11}{53}\)
b) Ta có : \(\dfrac{456}{461}=1-\dfrac{5}{461}\)
\(\dfrac{123}{128}=1-\dfrac{5}{128}\)
Vì \(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow1-\dfrac{5}{461}>1-\dfrac{5}{128}\)
\(\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
c) Ta có :\(\dfrac{12}{47}>\dfrac{12}{48}=\dfrac{1}{4}=\dfrac{19}{76}>\dfrac{19}{77}\)
=> \(\dfrac{12}{47}>\dfrac{19}{77}\)
d) Ta có : \(13A=13.\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{16}+13}{13^{16}+1}=\dfrac{13^{16}+1+12}{13^{16}+1}=1+\dfrac{12}{13^{16}+1}\)
\(13B=13.\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{17}+13}{13^{17}+1}=\dfrac{13^{17}+1+12}{13^{17}+1}=1+\dfrac{12}{13^{17}+1}\)
Ta thấy : \(\dfrac{12}{13^{16}+1}>\dfrac{12}{13^{17}+1}\Rightarrow1+\dfrac{12}{13^{16}+1}>1+\dfrac{12}{13^{17}+1}\Rightarrow\dfrac{13^{15}+1}{13^{16}+1}>\dfrac{13^{16}+1}{13^{17}+1}\)
1. Tìm n, biết:
a) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)
(-2)n + 2 = (-2)5
n + 2 = 5
n = 5 - 2
n = 3.
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow\dfrac{2^3}{2^n}=2\)
\(\Rightarrow\) 2n . 2 = 23
n + 1 = 3
n = 3 - 1
n = 2.
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
2n - 1 = 3
2n = 3 + 1
2n = 4
n = 4 : 2
n = 2.
2. Tính:
a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{2}\right)^4\)
\(=\left(\dfrac{1}{2}\right)^7\)
\(=\dfrac{1}{128}\)
b) 273 : 93
= (33)3 : (32)3
= 39 : 36
= 33
= 27
c) 1252 : 253
= (53)2 : (52)3
= 56 : 56
= 1
d) \(\dfrac{27^2.8^5}{6^6.32^3}\)
\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)
\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)
\(=\dfrac{3^6}{6^6}\)
\(=\dfrac{1}{64}.\)
B2 :
b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)
c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 1
a,
\(\dfrac{89}{-13}< 0< \dfrac{1}{123}\\ \Rightarrow\dfrac{89}{-13}< \dfrac{1}{123}\)
Vậy \(\dfrac{89}{-13}< \dfrac{1}{123}\)
b,
\(\dfrac{-13}{15}>\dfrac{-15}{15}=-1=\dfrac{-30}{30}>\dfrac{-31}{30}\)
Vậy \(\dfrac{-13}{15}>\dfrac{-31}{30}\)
c,
\(\dfrac{125}{123}=\dfrac{123}{123}+\dfrac{2}{123}=1+\dfrac{2}{123}\\ \dfrac{99}{97}=\dfrac{97}{97}+\dfrac{2}{97}=1+\dfrac{2}{97}\)
Vì \(\dfrac{2}{97}>\dfrac{2}{123}\Rightarrow1+\dfrac{2}{97}>1+\dfrac{2}{123}\Leftrightarrow\dfrac{99}{97}>\dfrac{125}{123}\)
Vậy \(\dfrac{99}{97}>\dfrac{125}{123}\)
d,
\(\dfrac{125}{126}< \dfrac{126}{126}=1=\dfrac{986}{986}< \dfrac{987}{986}\)
Vậy \(\dfrac{125}{126}< \dfrac{987}{986}\)