Tìm x để \(\frac{37}{42}=\frac{41-x}{32}\)
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\(\left(1-\frac{1}{35}\right)\left(1-\frac{1}{36}\right)\left(1-\frac{1}{37}\right)...\left(1-\frac{1}{2010}\right)\left(1-\frac{1}{2011}\right)\)
\(=\frac{34}{35}.\frac{35}{36}.\frac{36}{37}.....\frac{2009}{2010}.\frac{2010}{2011}\)
\(=\frac{34}{2011}\)
\(\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}+\frac{109}{110}+\frac{131}{132}+\frac{155}{156}\)
\(=1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}+1-\frac{1}{110}+1-\frac{1}{132}+1-\frac{1}{156}\)
\(=7-\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}\right)\)
\(=7-\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}\right)\)
\(=7-\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{12}-\frac{1}{13}\right)\)
\(7-\left(\frac{1}{6}-\frac{1}{13}\right)=6\frac{71}{78}\)
\(\frac{15}{37}\)x \(\left(\frac{38}{41}-\frac{74}{45}\right)-\frac{38}{41}\)x \(\left(\frac{15}{37}+\frac{82}{76}\right)\)
\(=\frac{15}{37}\)x \(\frac{38}{41}-\frac{15}{37}\)x \(\frac{74}{45}-\frac{38}{41}\)x \(\frac{15}{37}-\frac{38}{41}\)x \(\frac{82}{76}\)
\(=\frac{15}{37}\)x \(\frac{38}{41}-\frac{2}{3}-\frac{38}{41}\)x \(\frac{15}{37}-1\)
\(=\frac{15}{37}\)x \(\frac{38}{41}\)\(-\frac{38}{41}\)x \(\frac{15}{37}\)\(-\frac{2}{3}-1\)
\(=0-\frac{2}{3}-1=\frac{-5}{3}\)
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
gọi 1/41+1/42+1/43+...+1/79+1/80 là A
ta có:1/41>1/60,1/42>1/60,1/43>1/60,...,1/60=1/60
=>1/41+1/42+1/43+...+1/60>1/60
1/61>1/80,..................................,1/80=1/80
=>1/61+1/62+............+1/80>1/80
=>1/41+1/42+1/43+...+1/79+1/80>1/60+1/80
lại có 7/12=1/4+1/3
1/60.20=1/3 và 1/80.20=1/4
=>1/41+1/42+1/43+...+1/79+1/80>1/3+1/4
=>1/41+1/42+1/43+...+1/79+1/80>7/12
\(\frac{32-x}{7}=\frac{x-42}{9}\)
=\(\frac{\left(32-x\right)9}{63}=\frac{\left(x-42\right)7}{63}\)
\(\Rightarrow\)\(\left(32-x\right)9=\left(x-42\right)7\)
=\(288-x9=x7-294\)
=\(288+294=x9+x7\)
=\(x=-36\frac{6}{16}\)
=\(x\times16=-582\)
\(x=-582\div16\)
a,\(\frac{32-x}{7}=\frac{x-42}{9}\)
\(\Leftrightarrow9\left(32-x\right)=7\left(x-42\right)\)
\(\Leftrightarrow288-9x-7x-294=0\)
\(\Leftrightarrow9x+7x=288-294\)
\(\Leftrightarrow2x=-6\)
\(\Leftrightarrow x=-3\)
b. \(\left(2x-1\right)^2+\left|x+3\right|=0\)
\(\Leftrightarrow\left|x+3\right|=-4x^2+4x-1\)
\(\left|x+3\right|=x+3\)khi \(x+3\ge0\)hay \(x\ge-3\)
\(\left|x+3\right|=-\left(x+3\right)\)khi \(x+3< 0\)hay \(x< -3\)
với \(x\ge-3\Rightarrow x+3=-4x^2+4x-1\)
\(\Leftrightarrow4x^2-4x+1+x+3=0\)
\(\Leftrightarrow4x^2-3x+4=0\)\(\Leftrightarrow\)vô nghiệm
với \(x< -3\)\(\Rightarrow-x-3=-4x+4-1\)
\(\Leftrightarrow4x^2-4x+1-x-3=0\)
\(\Leftrightarrow4x^2-5x-2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{57}}{8}\left(tm\right)\\x=\frac{5-\sqrt{57}}{8}\left(L\right)\end{cases}}\)
Quy đồng \(\frac{37}{42}=\frac{37\times32}{42\times32}=\frac{1184}{1344};\frac{41-x}{32}=\frac{42\times\left(41-x\right)}{42\times32}=\frac{42\times\left(41-x\right)}{1344}\)
\(\frac{37}{42}=\frac{41-x}{32}\) nên \(\frac{1184}{1344}=\frac{42\times\left(41-x\right)}{1344}\)
=> 42 x (41 - \(x\)) = 1184
42 x 41 - 42 x \(x\) = 1184
1722 - 42 x \(x\) = 1184
42 x \(x\) = 1722 - 1184
42 x \(x\) = 538
\(x\) = 538 / 42 = 269 / 21