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Xét tử:
tử = 1/18 + 2/17 + 3/16 + ... + 18/1 + (1+1+1+...+1)(18 số 1)
=(1/18 + 1)+(2/17 + 1)+...+(18/1 + 1)
=19/18 + 19/17 + ... + 19/1
=19(1/18 + 1/17 + ... + 1/1)
Nên tử/ mẫu =19
Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
b bạn gõ lại đề giúp mình nhé
mh biết làm bài này rùi bn có cần mi2h đang cho bn ko?
\(E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)-\left(\frac{12}{17}-\frac{5}{18}+\frac{7}{5}\right)\)
\(=\frac{1}{13}+\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}-\frac{12}{17}+\frac{5}{18}-\frac{7}{5}\)
\(=\left(\frac{1}{13}-\frac{1}{13}\right)+\left(\frac{-5}{18}+\frac{5}{18}\right)+\left(\frac{12}{17}-\frac{12}{17}\right)-\frac{7}{5}\)
\(=0+0+0-\frac{7}{5}\)
\(=-\frac{7}{5}\)
\(E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)-\left(\frac{12}{17}-\frac{5}{18}+\frac{7}{5}\right)\)
\(=\frac{1}{13}+\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}-\frac{-12}{17}+\frac{5}{18}-\frac{7}{5}\)
\(=\left(\frac{1}{13}-\frac{1}{13}\right)+\left(\frac{-5}{18}+\frac{5}{18}\right)+\left(\frac{12}{17}-\frac{-12}{17}\right)-\frac{7}{5}\)
\(=0+0+\frac{24}{17}-\frac{7}{5}\)
\(=\frac{24}{17}-\frac{7}{5}\)
\(=\frac{120}{85}-\frac{119}{85}=\frac{1}{85}\)
\(\frac{3}{17}+\frac{-5}{13}+\frac{14}{17}+\frac{-18}{35}+\frac{17}{-35}+\frac{-8}{13}\)
\(=\left(\frac{3}{17}+\frac{14}{17}\right)-\left(\frac{5}{13}+\frac{8}{13}\right)-\left(\frac{18}{35}+\frac{17}{35}\right)\)
\(=1-1-1\)
\(=-1\)
2. Tìm ba số nguyên dương đôi một khác nhau:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Không mất tính tổng quát: G/s: a>b>c>0
=> \(\frac{1}{a}< \frac{1}{b}< \frac{1}{c}\)
Vì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\); a,b,c là số nguyên dương
=> \(\frac{1}{a}< \frac{1}{b}< \frac{1}{c}< 1\)
=> a>b>c>1 , với a, b, c là số nguyên dương (1)
=> \(1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}< \frac{1}{c}+\frac{1}{c}+\frac{1}{c}=\frac{3}{c}\)
=> \(1< \frac{3}{c}\Rightarrow c< 3\)
Từ (1) => c=2
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{2}=1\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{2}\)
Do đó: \(\frac{1}{2}=\frac{1}{a}+\frac{1}{b}< \frac{1}{b}+\frac{1}{b}=\frac{2}{b}\)=> b<4 => b=3
Khi đó ta có:
\(\frac{1}{a}+\frac{1}{2}+\frac{1}{3}=1\Rightarrow\frac{1}{a}=\frac{1}{6}\Rightarrow a=6\)
Vậy (a;b;c)=(6;3;2) và các hoán vị của nó
e, \(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}+\frac{28}{42}:\frac{1}{28}-8\)
\(=\left(\frac{3}{14}-\frac{13}{21}+\frac{2}{3}\right):\frac{1}{28}-8\)
\(=\frac{11}{42}:\frac{1}{28}-8\)
\(=\frac{22}{3}-8\)
\(=-\frac{2}{3}\)
e, \(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}+\frac{28}{42}:\frac{1}{28}-8\)
\(=\left(\frac{3}{14}-\frac{13}{21}+\frac{2}{3}\right):\frac{1}{28}-8\)
\(=\frac{11}{42}:\frac{1}{28}-8\)
\(=\frac{22}{3}-8\)
\(=-\frac{2}{3}\)
a, \(\frac{x}{y+z+1}=\frac{y}{x+z+3}=\frac{z}{x+y-4}=\frac{x+y+z}{y+z+1+x+z+3+x+y-4}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)
=>\(x+y+z=\frac{1}{2};\frac{x}{y+z+1}=\frac{1}{2};\frac{y}{x+z+3}=\frac{1}{2};\frac{z}{x+y-4}=\frac{1}{2}\)
=>\(\hept{\begin{cases}y+z+1=2x\\x+z+3=2y\\x+y-4=2z\end{cases}}\Rightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+3=3y\\x+y+z-4=3z\end{cases}\Rightarrow\hept{\begin{cases}3x=\frac{1}{2}+1\\3y=\frac{1}{2}+3\\3z=\frac{1}{2}-4\end{cases}}}\Rightarrow\hept{\begin{cases}3x=\frac{3}{2}\\3y=\frac{7}{2}\\3z=\frac{-7}{2}\end{cases}}\)
đến đây dễ rồi
b, =>(x-18)(x+16)=(x+4)(x-17)
=>x2+16x-18x-288=x2-17x+4x-68
=>x2-2x-288-x2+13x+68=0
=>11x-220=0
=>11x=220
=>x=20