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2,
a) \(315-\left(135-x\right)=215\)
\(\Rightarrow135-x=315-215\)
\(\Rightarrow135-x=100\)
\(\Rightarrow x=135-100\)
\(\Rightarrow x=35\)
b) \(x-320:32=25\cdot16\)
\(\Rightarrow x-10=5^2\cdot4^2\)
\(\Rightarrow x-10=20^2\)
\(\Rightarrow x-10=400\)
\(\Rightarrow x=410\)
c) \(3\cdot x-2018:2=23\)
\(=3\cdot x-1009=23\)
\(\Rightarrow3\cdot x=1032\)
\(\Rightarrow x=1032:3\)
\(\Rightarrow x=344\)
d) \(280-9\cdot x-x=80\)
\(\Rightarrow280-x\cdot\left(9+1\right)=80\)
\(\Rightarrow280-10\cdot x=80\)
\(\Rightarrow10\cdot x=280-80\)
\(\Rightarrow10\cdot x=200\)
\(\Rightarrow x=20\)
e) \(38\cdot x-12\cdot x-x\cdot16=40\)
\(\Rightarrow x\cdot\left(38-12-16\right)=40\)
\(\Rightarrow x\cdot10=40\)
\(\Rightarrow x=40:10\)
\(\Rightarrow x=4\)
Câu 1 :
a, 8.( -5 ).( -4 ).2
= [ 8.2 ].[( -5 ).(-4 ]
= 16.20
= 320
b, \(1\frac{3}{7}+\frac{-1}{3}+2\frac{4}{7}\)
\(=\frac{10}{7}+\frac{-1}{3}+\frac{18}{7}\)
\(=\frac{11}{3}\)
c, \(\frac{8}{5}.\frac{2}{3}+\frac{-5.5}{3.5}\)
\(=\frac{8}{3}+\frac{-5}{3}\)
\(=\frac{3}{3}=1\)
d, \(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-2\right)^2\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{16}.4\)
\(=\frac{55}{56}-\frac{3}{4}\)
\(=\frac{13}{56}\)
Câu 2 :
a, 2x + 10 = 16
2x = 16 + 10
2x = 26
x = 26 : 2
x = 13
b, \(x-\frac{1}{3}=\frac{5}{4}\)
\(x=\frac{5}{4}+\frac{1}{3}\)
\(x=\frac{19}{12}\)
c, \(2x+3\frac{1}{3}=7\frac{1}{3}\)
\(2x+\frac{10}{3}=\frac{22}{3}\)
\(2x=\frac{22}{3}-\frac{10}{3}\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
d, \(\left(\frac{2}{11}+\frac{1}{3}\right)x=\left(\frac{1}{7}-\frac{1}{8}\right).56\)
\(\frac{17}{33}x=1\)
\(x=1-\frac{17}{33}\)
\(x=\frac{16}{33}\)
à nhầm cái này bạn chỉ cần viết ra rồi giản ước là xong :)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\)\(\left(1-\frac{1}{5}\right)\)
=\(\frac{1}{2}.\)\(\frac{2}{3}\cdot\frac{3}{4}\)\(\cdot\frac{4}{5}\)
=\(\frac{1}{5}\)
( 1 - 12 ) x ( 1 - 13 ) x ( 1 - 14 ) x ( 1 - 15 )
= \(\left(\frac{2}{2}-\frac{1}{2}\right)\times\left(\frac{3}{3}-\frac{1}{3}\right)\times\left(\frac{4}{4}-\frac{1}{4}\right)\times\left(\frac{5}{5}-\frac{1}{5}\right)\)
= \(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\)
= \(\frac{1\times2\times3\times4}{2\times3\times4\times5}\)
= \(\frac{1}{5}\)
<br class="Apple-interchange-newline"><div id="inner-editor"></div>14 + 18 +116 + 132 + 164 + \(\frac{1}{128}\) MC : 128
= \(\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\)
= \(\frac{32+16+8+4=2+1}{128}\)
= \(\frac{207}{128}\)
\(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4=\left(\dfrac{3}{2}\right)^7\)
\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8=\left(\dfrac{1}{7}\right)^7\times2^{16}\)
\(\left(-\dfrac{1}{7}\right)^4\times125\times5=\left(-\dfrac{1}{7}\right)^4\times5^3\times5=\left(-\dfrac{1}{7}\right)^4\times5^4\)
\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:2^3:2^{-4}=2^0\)
\(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^3\times7^3=1^3\)
6, \(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4\)
7,\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8\)
8,\(\left(-\dfrac{1}{7}\right) ^4\times125\times5=\left(\dfrac{1}{7}\right)^4\times5^3\times5\)
9,\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:\left[2^3\times\left(\dfrac{1}{2}\right)^4\right]\)
10, \(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times7^2\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(< =>\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)
\(< =>\frac{128+64+32+16+8+4+2+1}{256}\)
\(< =>\frac{255}{256}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(< =>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(< =>\frac{1}{1}-\frac{1}{100}\)
\(< =>\frac{99}{100}\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
\(< =>\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
\(< =>\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)
\(< =>\frac{1}{100}\)
mk chuc ban hoc tot nhe :))
a) \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)
\(=\dfrac{1}{2004}\)
b) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{3}{5}\)
