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1. \(\Leftrightarrow\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{51-x}{49}+1=-5+5\)
\(\Leftrightarrow\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\frac{100-x}{49}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)
\(\Leftrightarrow x-100=0\Leftrightarrow x=100\)
2. \(\Leftrightarrow\frac{x-5}{1990}+1+\frac{x-15}{1980}+1+\frac{x-25}{1970}=\frac{x-1990}{5}+1+\frac{x-1980}{15}+1+\frac{x-1970}{25}+1\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)
\(\Leftrightarrow\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)
\(\Leftrightarrow x-1995=0\Leftrightarrow x=1995\)
\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Rightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)\(\Rightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Rightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
\(\Rightarrow100-x=0\Rightarrow x=100\)
\(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)
<=>\(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}-\frac{x-4}{56}-\frac{x-5}{55}-\frac{x-6}{54}=0\)
<=>\(\frac{x-1}{59}-1+\frac{x-2}{58}-1+\frac{x-3}{57}-1-\frac{x-4}{56}+1-\frac{x-5}{55}+1-\frac{x-6}{54}+1=0\)
<=>\(\frac{x-60}{59}+\frac{x-60}{58}+\frac{x-60}{57}-\frac{x-60}{56}-\frac{x-60}{55}-\frac{x-60}{54}=0\)
<=>\(\left(x-60\right)\left(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\right)=0\)
Do \(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\ne0\)
=>x-60=0
<=>x=60
Vậy x=60
a) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{4}{5}\right)\)
\(=\frac{47}{47}+\left(\frac{1}{5}+\frac{4}{5}\right)\)
\(=1+1=2\)
b) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
\(=12.\frac{4}{9}+\frac{4}{3}\)
\(=\frac{16}{3}+\frac{4}{3}\)
\(=\frac{20}{3}\)
c) \(12,5.\left(-\frac{5}{7}\right)+15.\left(-\frac{5}{7}\right)\)
\(=\left(-\frac{5}{7}\right).\left(12,5+15\right)\)
\(=\left(-\frac{5}{7}\right).27,5\)
\(=\left(-\frac{5}{7}\right).\frac{55}{2}\)
\(=-\frac{275}{14}\)
d) \(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{14}{4}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{15}{4}\right)^2\)
\(=\frac{4}{5}.\frac{225}{16}\)
\(=\frac{45}{4}\)
a)\(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
=\(\frac{21}{47}+\frac{1}{5}+\frac{26}{47}+\frac{4}{5}\)
=\(\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)\)
=\(\frac{47}{47}+\frac{5}{5}=1+1=2\)
b)\(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
=\(12.\frac{4}{9}+\frac{4}{3}\)
=\(\frac{12}{1}.\frac{4}{9}+\frac{4}{3}=\frac{48}{9}+\frac{4}{3}\)
=\(\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)
c)\(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)\)
=\(\left(-\frac{5}{7}\right).\left(12,5+1,5\right)\)
=\(\left(-\frac{5}{7}\right).14=\left(-\frac{5}{7}\right).\frac{14}{1}=-10\)
d)\(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
=\(\frac{4}{5}.\left(\frac{14}{4}+\frac{1}{4}\right)^2\)
=\(\frac{4}{5}.\left(\frac{15}{4}\right)^2\)
=\(\frac{4}{5}.\frac{225}{16}\)
=\(\frac{900}{80}=\frac{45}{4}\)
Nhớ tick cho mình nha!
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
a) Ta thấy:
\(\frac{x}{2}=\frac{y}{3}\)\(\Rightarrow\frac{x}{2}\cdot\frac{3}{5}=\frac{y}{3}\cdot\frac{3}{5}\)\(\Rightarrow\frac{3x}{10}=\frac{y}{5}\)
Mà \(\frac{y}{5}=\frac{z}{6}\) nên ta có biểu thức: \(\frac{3x}{10}=\frac{y}{5}=\frac{z}{6}\) ( 1 )
Biểu thức ( 1 ) tương đương với:
\(\frac{3x}{10}=\frac{3y}{15}=\frac{3z}{18}=\frac{3x+3y+3z}{10+15+18}=\frac{3\left(x+y+z\right)}{43}=\frac{3\cdot43}{43}=3\)
Khi đó:
\(\frac{3x}{10}=3\) \(\Rightarrow x=\frac{3\cdot10}{3}=10\)
\(\frac{3y}{15}=3\)\(\Rightarrow\frac{y}{5}=3\) \(\Rightarrow y=3\cdot5=15\)
\(\frac{3z}{18}=3\)\(\Rightarrow\frac{z}{6}=3\) \(\Rightarrow z=3\cdot6=18\)
a, Nhân cả hai vế cho 5, ta được: X/10 = Y/15
Tương tự ta có: Y/15 = Z/18
Do đó: X/10 = Z/18 (=Y/15)
Theo đề bài, ta có: (X+Y+Z)/(10+15+18) = 43/43 = 1
X/10=1 => X=10
Y/15=1 => Y=15
Z/18=1 => Z=18
mk lam luon nhe!
Bot vao moi ve 3 don vi, ta co
\(\left(\frac{x-7}{50}-1\right)+\left(\frac{x-6}{51}-1\right)+\left(\frac{x-5}{52}-1\right)=\left(\frac{x-52}{5}-1\right)+\left(\frac{x-51}{6}-1\right)+\left(\frac{x-50}{7}-1\right)\)
Quy dong len ,ta co
\(\frac{x-57}{50}+\frac{x-57}{51}+\frac{x-57}{52}=\frac{x-57}{5}+\frac{x-57}{6}+\frac{x-57}{7}\)
\(\frac{x-57}{50}+\frac{x-57}{51}+\frac{x-57}{52}-\frac{x-57}{5}-\frac{x-57}{6}-\frac{x-57}{7}=0\)
(x-57).\(\left(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}\right)=0\)
Ma \(\left(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}\right)\) khac 0 nen => x-57=0
x=0+57 =57
Vay x =57.
Mk chac chan 100% bai nay dung