\(\frac{-7}{25}\)x \(\frac{39}{-14}\)x \(\frac{50}{78}\)
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\(a)5-\left(x-6\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-x+6=12-8x\)
\(\Leftrightarrow-x+8x=12-5-6\)
\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)
a) 5-(x-6)=4(3-2x)
<=>5-x-6=12-8x
<=>-x+8x=2-5-6
<=>7x=1
<=>x=1/7
a ) \(\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{-12}{19}\)
\(=\frac{7}{19}.\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{-12}{19}\)
\(=\frac{7}{19}.\frac{11}{11}+\frac{-12}{19}\)
\(=\frac{7}{19}.1+\frac{-12}{19}\)
\(=\frac{7}{19}+\frac{-12}{19}\)
\(=\frac{7+\left(-12\right)}{19}\)
\(=-\frac{5}{19}\)
b ) \(\frac{-7}{25}.\frac{39}{-14}.\frac{50}{78}=\frac{-7.39.50}{25.-14.78}=\frac{-1.1.2}{1.-2.2}=\frac{-2}{-4}=\frac{1}{2}\)
c ) \(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{3}{4}\)
a)\(\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{-12}{19}=\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{7}{19}.\frac{-12}{7}=\frac{7}{19}.\left(\frac{8}{11}+\frac{3}{11}+-\frac{12}{7}\right)=\frac{7}{19}.\left(\frac{-5}{7}\right)=-\frac{5}{19}\)
b)\(\frac{-7}{25}.\frac{39}{-14}.\frac{50}{78}=\frac{\left(-7\right).39.50}{25.\left(-14\right).78}=\frac{\left(-7\right).3.13.2.5.5}{5.5.\left(-7\right).2.2.13.3}=\frac{1}{2}\)
c)\(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}=\frac{2}{7}+\frac{1}{2}=\frac{11}{14}\)
a) Ta có: \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}=4\)
\(\Leftrightarrow\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)
\(\Leftrightarrow\frac{x-91-37}{37}+\frac{x-86-42}{42}+\frac{x-78-50}{50}+\frac{x-49-79}{79}=0\)
\(\Leftrightarrow\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)
\(\Leftrightarrow\left(x-128\right)\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)
Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)
nên x-128=0
hay x=128
Vậy: x=128
b) Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}-8=0\)
\(\Leftrightarrow\frac{x-29}{1970}-1+\frac{x-27}{1972}-1+\frac{x-25}{1974}-1+\frac{x-23}{1976}-1+\frac{x-1970}{29}-1+\frac{x-1972}{27}-1+\frac{x-1974}{25}-1+\frac{x-1976}{23}-1=0\)
\(\Leftrightarrow\frac{x-29-1970}{1970}+\frac{x-27-1972}{1972}+\frac{x-25-1974}{1974}+\frac{x-23-1976}{1976}+\frac{x-1970-29}{29}+\frac{x-1972-27}{27}+\frac{x-1974-25}{25}+\frac{x-1976-23}{23}=0\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}\right)=0\)
Vì \(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}>0\)
nên x-1999=0
hay x=1999
Vậy: x=1999
a) Ta có \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}\)=4
<=>\(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}-4=0\)
<=>\(\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)
<=>\(\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)
<=>(x-128)\(\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)
Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)=>x-128=0<=>x=128
b)Tương tự
<=>x-128=0
<=>x=128
Chú ý \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\)>0
b)tương tự
\(\frac{-7}{25}\times\frac{39}{-14}\times\frac{50}{78}\)
\(=\frac{39}{50}\times\frac{50}{78}\)
\(=\frac{1}{2}\)
\(\frac{-7}{25}\)x \(\frac{39}{-14}\)x \(\frac{50}{78}\)
= \(\frac{-7}{25}\)x \(\frac{-14}{39}\)x \(\frac{25}{39}\)
=\(\frac{-7}{25}\)x\(\frac{25}{39}\)x \(\frac{-14}{39}\)
=\(\frac{-7}{39}\)x \(\frac{-14}{39}\)
=\(\frac{98}{1521}\)
chúc bạn học tốt !!!
\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{36}+\frac{x}{45}+\frac{x}{55}+\frac{x}{66}+\frac{x}{78}+\frac{x}{78}=\frac{220}{39}\)
\(\Leftrightarrow x\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+\frac{1}{78}+\frac{1}{78}\right)=\frac{220}{39}\)
\(\Leftrightarrow x\cdot\frac{20}{39}=\frac{220}{39}\Rightarrow x=11\)
\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{36}+\frac{x}{45}+\frac{x}{55}+\frac{x}{66}+\frac{x}{78}+\frac{x}{78}=\frac{220}{39}\)
\(=>x=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+\frac{1}{78}+\frac{1}{78}=\frac{220}{39}\)
\(x\cdot\frac{20}{39}=\frac{220}{39}\)
\(x=\frac{220}{39}:\frac{20}{39}=11\)
\(\frac{-7}{25}\)x \(\frac{39}{-14}\)x\(\frac{50}{78}\)=\(\frac{\left(-7\right)x39x5x10}{5x5x\left(-7\right)x2x39x2}\)=\(\frac{10}{20}\)=\(\frac{1}{2}\)