cho a và b thuộc N
CMR 7*a+3*b \(⋮23\Leftrightarrow4\cdot a+5\cdot b⋮23\)
Những câu hỏi liên quan
3,Tính hợp lý
a,\(\frac{3}{7}\cdot\frac{4}{9}+\frac{3}{7}\cdot\frac{5}{9}+\frac{5}{14}\)
b,\(\frac{-11}{23}\cdot\frac{6}{7}+\frac{8}{9}\cdot\frac{-11}{23}-\frac{1}{23}\)
Giúp mình nhé!!!
a,\(\frac{3}{7}.\frac{4}{9}+\frac{3}{7}.\frac{5}{9}+\frac{5}{14}\)
\(=\frac{3}{7}.\left(\frac{4}{9}+\frac{5}{9}\right)+\frac{5}{14}\)
\(=\frac{3}{7}.1+\frac{5}{14}\)
\(=\frac{3}{7}+\frac{5}{14}=\frac{6}{14}+\frac{5}{14}=\frac{11}{14}\)
b,\(\frac{-11}{23}.\frac{6}{7}+\frac{8}{9}.\frac{-11}{23}-\frac{1}{23}\)
\(=\)\(\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{9}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.\frac{110}{63}-\frac{1}{23}\)
=\(\frac{-1210}{1449}\)-\(\frac{1}{23}\)
\(=\frac{-1273}{1449}\)
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tính nhanh
A= \(\frac{19}{23}\cdot\frac{-4}{7}-\frac{4}{23}\cdot\frac{2}{7}\)
B= \(\frac{3}{5}+\frac{2}{5}\cdot\frac{-11}{3}+\frac{2}{3}\cdot\frac{-2}{5}+\frac{14}{15}\)
a) A = \(\frac{19}{23}.\frac{-4}{27}-\frac{4}{23}.\frac{2}{7}\)
= \(\frac{19}{7}.\frac{-4}{23}+\frac{-4}{23}.\frac{2}{7}\)
= \(\frac{-4}{23}.\left(\frac{19}{7}+\frac{2}{7}\right)\)
= \(\frac{-4}{23}.3\)
= \(\frac{-12}{23}\)
b) B = \(\frac{3}{5}+\frac{2}{5}.\frac{-11}{3}+\frac{2}{3}.\frac{-2}{5}+\frac{14}{15}\)
= \(\frac{9+14}{15}+\frac{2}{5}.\frac{-11}{3}+\frac{-2}{3}.\frac{2}{5}\)
= \(\frac{23}{15}+\frac{2}{5}\left(\frac{-11}{3}+\frac{-2}{3}\right)\)
= \(\frac{23}{15}+\frac{2}{5}.\frac{-13}{3}\)
= \(\frac{23}{15}+\frac{-26}{15}\)
= \(\frac{-3}{15}=\frac{-1}{5}\)
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a) Chứng minh biểu thức sau không phụ thuộc vào x:
\(\left(\frac{5\cdot a+b}{5\cdot a^2-a\cdot b}+\frac{5\cdot a-b}{5\cdot a^2-a\cdot b}\right)\div\frac{100\cdot a^2+4\cdot b^2}{25\cdot a^3-a\cdot b^2}\)
b) Tìm x; y sao cho \(x^3+y^3=3\cdot x\cdot y-1\)
a)Tìm các số tự nhiên a,b sao cho: (\(\left(2008\cdot a+3\cdot b+1\right)\cdot\left(2008^a+2008\cdot a+b\right)=225\)
b)Tìm x thỏa mãn: \(11\frac{1}{2}\cdot\frac{1}{|3x-1|}=\frac{23}{28}\)
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chúc bạn học tốt !
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Tính nhanh:
a,\(\frac{7}{13}\cdot\frac{7}{15}-\frac{5}{12}\cdot\frac{21}{39}+\frac{49}{91}\cdot\frac{8}{15}\)
b,\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
tính bằng cách thuận tiện nhất
a,\(\frac{6}{7}\cdot\frac{16}{15}\cdot\frac{7}{6}\cdot\frac{21}{32}\)b, \(\frac{21}{17}\cdot\frac{13}{14}\cdot56\cdot\frac{3}{42}\) c,\(\frac{7}{4}\cdot\frac{11}{21}+\frac{11}{21}\cdot\frac{5}{4}\) d,\(\frac{23}{14}\cdot\frac{6}{14}-\frac{9}{14}\cdot\frac{6}{13}\)
a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)
Phần b T2
c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)
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Bài 1: cho a,b,cge0 và a+b+c1. Chứng minh rằng :a,left(1-aright)cdotleft(1-bright)cdotleft(1-cright)ge8cdot acdot bcdot cb,16cdot acdot bcdot cge a+bc,frac{a}{1+a}+frac{2cdot b}{2+b}+frac{3cdot c}{3+c}lefrac{6}{7}Bài 2: cho a,b,c0 và a.b.c0 chứng minh rằng:frac{bcdot c}{a^2cdot b+a^2cdot c}+frac{acdot c}{b^2cdot c+b^2cdot a}+frac{acdot b}{c^2cdot a+c^2cdot b}gefrac{3}{2}
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Bài 1: cho \(a,b,c\ge0\) và a+b+c=1. Chứng minh rằng :
a,\(\left(1-a\right)\cdot\left(1-b\right)\cdot\left(1-c\right)\ge8\cdot a\cdot b\cdot c\)
b,\(16\cdot a\cdot b\cdot c\ge a+b\)
c,\(\frac{a}{1+a}+\frac{2\cdot b}{2+b}+\frac{3\cdot c}{3+c}\le\frac{6}{7}\)
Bài 2: cho a,b,c>0 và a.b.c=0 chứng minh rằng:
\(\frac{b\cdot c}{a^2\cdot b+a^2\cdot c}+\frac{a\cdot c}{b^2\cdot c+b^2\cdot a}+\frac{a\cdot b}{c^2\cdot a+c^2\cdot b}\ge\frac{3}{2}\)
Bài 1 :
a) Ta có : \(\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Áp dụng bđt Cauchy : \(a+b\ge2\sqrt{ab}\) , \(b+c\ge2\sqrt{bc}\) , \(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) hay \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge8abc\)
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Tính giá trị biểu thứca,Afrac{24cdot47-23}{24+47-23}cdotfrac{3+frac{3}{7}-frac{3}{11}+frac{3}{1001}-frac{3}{13}}{frac{9}{1001}-frac{9}{13}+frac{9}{7}-frac{9}{11}+9}b,Mfrac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}c,A81cdotleft[frac{12-frac{12}{7}-frac{12}{289}-frac{12}{85}}{4-frac{4}{7}-frac{4}{289}-frac{4}{85}}:frac{5+frac{5}{13}+frac{5}{169}+frac{5}{91}}{6+frac{6}{13}+frac{6}{169}+frac{6}{91}}right]:frac{158158158}{711711711}d,Afrac{5cdotleft(2^2.3^2right)^9cdotleft(2^2right)^6-2cdotleft(2^2cdot3r...
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Tính giá trị biểu thức
a,\(A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
b,\(M=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)
c,\(A=81\cdot\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right]:\frac{158158158}{711711711}\)
d,\(A=\frac{5\cdot\left(2^2.3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)
2,Thực hiện phép tính(tính nhanh nếu có thể)
a,frac{2}{3}+frac{1}{3}cdotleft(-frac{2}{5}right)
b,0,75cdot1frac{7}{9}-1frac{2}{5}:frac{-21}{20}
c,frac{-2}{17}+frac{15}{23}+frac{15}{-17}-frac{-4}{19}+frac{8}{23}
d,2019^0cdotleft(6-2frac{4}{5}right)cdot3frac{1}{8}-1frac{3}{5}:25phần trăm
e,left(frac{7}{8}-frac{1}{2}right)cdot2frac{2}{3}-frac{3}{7}cdotleft(2,5^2right)
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2,Thực hiện phép tính(tính nhanh nếu có thể)
a,\(\frac{2}{3}+\frac{1}{3}\cdot\left(-\frac{2}{5}\right)\)
b,\(0,75\cdot1\frac{7}{9}-1\frac{2}{5}:\frac{-21}{20}\)
c,\(\frac{-2}{17}+\frac{15}{23}+\frac{15}{-17}-\frac{-4}{19}+\frac{8}{23}\)
d,\(2019^0\cdot\left(6-2\frac{4}{5}\right)\cdot3\frac{1}{8}-1\frac{3}{5}:25\)phần trăm
e,\(\left(\frac{7}{8}-\frac{1}{2}\right)\cdot2\frac{2}{3}-\frac{3}{7}\cdot\left(2,5^2\right)\)
a) \(\frac{2}{3}+\frac{1}{3}\cdot\left(-\frac{2}{5}\right)\\ =\frac{2}{3}+\frac{-2}{15}\\ =\frac{10}{15}+\frac{-2}{15}\\ =\frac{8}{15}\)
b) \(0,75\cdot1\frac{7}{9}-1\frac{2}{5}:\frac{-21}{20}\\ =\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}\cdot\frac{-20}{21}\\ =\frac{4}{3}-\frac{-4}{3}\\ =\frac{4}{3}+\frac{4}{3}\\ =\frac{4}{3}\cdot2\\ =\frac{8}{3}\)
c) \(\frac{-2}{17}+\frac{15}{23}+\frac{15}{-17}-\frac{-4}{19}+\frac{8}{23}\\ =\frac{-2}{17}+\frac{15}{23}+\frac{-15}{17}+\frac{4}{19}+\frac{8}{23}\\ =\left(\frac{-2}{17}+\frac{-15}{17}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\\ =\left(-1\right)+1+\frac{4}{19}\\ =0+\frac{4}{19}\\ =\frac{4}{19}\)
d) \(2019^0\cdot\left(6-2\frac{4}{5}\right)\cdot3\frac{1}{8}-1\frac{3}{5}:25\%\\ =1\cdot\left(\frac{30}{5}-\frac{14}{5}\right)\cdot\frac{25}{8}-\frac{8}{5}:\frac{1}{4}\\ =1\cdot\frac{16}{5}\cdot\frac{25}{8}-\frac{8}{5}\cdot4\\ =\frac{16}{5}\cdot\frac{25}{8}-\frac{32}{5}\\ =\frac{50}{5}-\frac{32}{5}\\ =\frac{18}{5}\)
e) \(\left(\frac{7}{8}-\frac{1}{2}\right)\cdot2\frac{2}{3}-\frac{3}{7}\cdot\left(2,5^2\right)\\ =\left(\frac{7}{8}-\frac{4}{8}\right)\cdot\frac{8}{3}-\frac{3}{7}\cdot6,25\\ =\frac{3}{8}\cdot\frac{8}{3}-\frac{3}{7}\cdot\frac{25}{4}\\ =1-\frac{75}{28}\\ =\frac{28}{28}-\frac{75}{28}\\ =\frac{-47}{28}\)
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a, \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-2}{5}\right)\)
= \(\frac{2}{3}+\frac{-2}{15}=\frac{8}{15}\)
b, \(0,75.1\frac{7}{9}-1\frac{2}{5}:\frac{-21}{20}\)
= \(\frac{3}{4}.\frac{16}{9}-\frac{7}{5}.\frac{-20}{21}\)
= \(\frac{4}{3}-\left(\frac{-4}{3}\right)=\frac{8}{3}\)
c, \(\frac{-2}{17}+\frac{15}{23}+\frac{15}{-17}+\frac{4}{19}+\frac{8}{23}\)
= \(\left(\frac{-2}{17}+\frac{-15}{17}\right)+\left(\frac{15}{23}+\frac{8}{23}\right)+\frac{4}{19}\)
= \(\left(-1\right)+1+\frac{4}{19}=0+\frac{4}{19}=\frac{4}{19}\)
d, \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}-1\frac{3}{5}:25\%\)
=> \(\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}:25\%\)
= \(\frac{16}{5}.\frac{25}{8}-\frac{8}{5}.25:100\)
= 10 - 0,4 = 9,6
e, \(\left(\frac{7}{8}-\frac{1}{2}\right).2\frac{2}{3}-\frac{3}{7}.\left(2,5^2\right)\)
=> \(\frac{3}{8}.\frac{8}{3}-\frac{3}{7}.6,25\)
= \(1-\frac{75}{28}=\frac{-47}{28}\)
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