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\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{6}{6}=1\)
\(\frac{13}{14}+\frac{14}{8}=\frac{13.4}{14.4}+\frac{14.7}{8.7}=\frac{52}{56}+\frac{98}{56}=\frac{150}{56}\simeq2,68\)
Như vậy: \(1\le x\le2,68\)
Mà x thuộc N => x=1 và x=2
Đáp số: x=1 và x=2
Bài 3
\(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow\left(x-1\right).3=8.9\)
\(\Rightarrow\left(x-1\right).3=72\)
\(\Rightarrow x-1=24\)
\(\Rightarrow x=25\)
\(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow\left(-x\right).x=\left(-9\right).4\)
\(\Rightarrow-x=-36\)
\(\Rightarrow x=36\)
\(\frac{x}{4}=\frac{18}{x+1}\)
\(\Rightarrow x.\left(x+1\right)=4.18\)
\(\Rightarrow x.\left(x+1\right)=72\)
Vì x và x + 1 là 2 số tự nhiên liên tiếp
\(\Rightarrow x\left(x+1\right)=8.9\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=8\end{cases}}\)
Bài 4
\(\frac{x-4}{y-3}=\frac{4}{3},x-y=5\)
Ta có :
\(x-y=5\)
\(\Rightarrow x=5+y\)
\(\Rightarrow\frac{y+5-4}{y-3}=\frac{4}{3}\)
\(\Rightarrow\frac{y+1}{y-3}=\frac{4}{3}\)\(\)
\(\Rightarrow\left(y+1\right).3=\left(y-3\right).4\)
\(\Rightarrow y.3+1.3=y.4-3.4\)
\(\Rightarrow y.3+3=y.4-12\)
\(\Rightarrow y.3-y.4=-12-3\)
\(\Rightarrow-1y=-15\)
\(\Rightarrow y=\left(-15\right):\left(-1\right)\)
\(\Rightarrow y=15\)
Vì x = y + 5
\(\Rightarrow x=15+4\)
\(\Rightarrow x=19\)
Vậy x = 19 , y = 15
\(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow\left(-x\right).x=4.\left(-9\right)\)
\(\Rightarrow-x=-9;x=4\)
\(\Rightarrow x=9;x=4\)
\(\frac{x}{5}=\frac23\)
\(x\) = \(\frac23\times5\)
\(x=\frac{10}{3}\)
Vậy \(x=\frac{10}{3}\)
\(\frac{x}{3}-\frac12=\frac15\)
\(\frac{x}{3}\) = \(\frac15\) + \(\frac12\)
\(\frac{x}{3}\) = \(\frac{2}{10}+\frac{5}{10}\)
\(\frac{x}{3}=\frac{7}{10}\)
\(x=\frac{7}{10}\times3\)
\(x=\frac{21}{10}\)
Vậy \(x=\frac{21}{10}\)
\(\frac{x}{5}+\frac12=\frac{6}{10}\)
\(\frac{x}{5}=\frac{6}{10}-\frac12\)
\(\frac{x}{5}=\frac{6}{10}-\frac{5}{10}\)
\(\frac{x}{5}=\frac{1}{10}\)
\(x=\frac{1}{10}\times5\)
\(x=\frac12\)
Vậy \(x=\frac12\)
\(\frac{x+3}{15}\) = \(\frac13\)
\(x+3=\frac13\times15\)
\(x+3=5\)
\(x=5-3\)
\(x=2\)
Vậy \(x=2\)
a)\(\left(4\frac{5}{37}-3\frac45+8\frac{15}{29}\right)-\left(3\frac{5}{57}-6\frac{14}{29}\right)\)
=\(4\frac{5}{37}-3\frac45+8\frac{15}{29}-3\frac{5}{37}+6\frac{14}{29}\)
=\(\left(4\frac{5}{37}-3\frac{5}{37}\right)+\left(8\frac{15}{29}+6\frac{14}{29}\right)-3\frac45\)
=\(\left\lbrack\left(4-3\right)+\left(\frac{5}{37}-\frac{5}{37}\right)\right\rbrack+\left\lbrack\left(8+6\right)+\left(\frac{15}{29}\right.\right.\)+\(\frac{14}{29})\) -\(\frac{19}{5}\)
=\(1+0+14+1-\frac{19}{5}\)
=\(15+1-\frac{19}{5}\)
=\(16-\frac{19}{5}\)
=\(\frac{80}{5}-\frac{19}{5}\)
=\(\frac{61}{5}\)
a; A = 1 + 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 2
1 = 1 = 1
1/2^2 < 1/1.2 = 1/1 - 1/2
1/3^2 < 1/2.3 = 1/2 - 1/3
.......................
1/100^2 < 1/99.100 = 1/99 - 1/100
Cộng vế với vế ta có:
A = 1 + 1 - 1/100
A = 2 - 1/100 < 2 (đpcm)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
Bài giải
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{13}{4}+\frac{14}{8}\)
\(1\le x\le5\text{ }\Rightarrow\text{ }x\in\left\{1\text{ ; }2\text{ ; }3\text{ ; }4\text{ ; }5\right\}\)