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HT
25 tháng 7 2017
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
TG
30 tháng 3 2020
a/ 2x = 5y và x - 2y = -12
Ta có: 2x = 5y => \(\frac{x}{5}=\frac{y}{2}\)
Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{5}=\frac{y}{2}=\frac{x-y}{5+2}=\frac{x-2y}{5+2.2}=\frac{-12}{9}=-\frac{4}{3}\)
\(\frac{x}{5}=-\frac{4}{3}\Rightarrow x=\frac{-4}{3}.5=-\frac{20}{3}\)
\(\frac{y}{2}=-\frac{4}{3}\Rightarrow y=-\frac{4}{3}.2=-\frac{8}{3}\)
Vậy:.................
b/ 2x = 3y = 4z và x + y + z =21
Ta có: 2x = 3y = 4z
=> \(\frac{2x}{12}=\frac{3y}{12}=\frac{4z}{12}\)
=> \(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}\)
Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{6}=\frac{y}{4}=\frac{z}{3}=\frac{x+y+z}{6+4+3}=\frac{21}{13}\)
\(\frac{x}{6}=\frac{21}{13}\Rightarrow x=\frac{21}{13}.6=\frac{126}{13}\)
\(\frac{y}{4}=\frac{21}{13}\Rightarrow y=\frac{21}{13}.4=\frac{84}{13}\)
\(\frac{z}{3}=\frac{21}{13}\Rightarrow z=\frac{21}{13}.3=\frac{63}{13}\)
Vậy:...............
c/Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{32}{8}=4\)
\(\frac{x}{3}=4\Rightarrow x=4.3=12\)
\(\frac{y}{5}=4\Rightarrow y=4.5=20\)
Vậy:................
d/ Ta có: 7x = 3y
=> \(\frac{7x}{21}=\frac{3y}{21}\)
=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng: tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{16}{-4}=-4\)
\(\frac{x}{4}=-4\Rightarrow x=\left(-4\right).4=-16\)
\(\frac{y}{7}=-4\Rightarrow y=\left(-4\right).7=-28\)
Vậy:................
HN
1
22 tháng 12 2018
\(\left|\frac{1}{15}-x\right|+\left|\frac{2}{25}-y\right|+\left|z-\frac{16}{5}\right|=0\)
Ta có: \(\hept{\begin{cases}\left|\frac{1}{15}-x\right|\ge0\forall x\\\left|\frac{2}{25}-y\right|\ge0\forall y\\\left|z-\frac{16}{5}\right|\ge0\forall z\end{cases}}\)\(\Rightarrow\left|\frac{1}{15}-x\right|+\left|\frac{2}{25}-y\right|+\left|z-\frac{16}{5}\right|\ge0\forall x;y;z\)
Mà \(\left|\frac{1}{15}-x\right|+\left|\frac{2}{25}-y\right|+\left|z-\frac{16}{5}\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|\frac{1}{15}-x\right|=0\\\left|\frac{2}{25}-y\right|=0\\\left|z-\frac{16}{5}\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{15}\\y=\frac{2}{25}\\z=\frac{16}{5}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=\frac{1}{15}\\y=\frac{2}{25}\\z=\frac{16}{5}\end{cases}}\)
TL
30 tháng 6 2016
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. \(\Leftrightarrow\frac{x-5}{1990}+1+\frac{x-15}{1980}+1+\frac{x-25}{1970}=\frac{x-1990}{5}+1+\frac{x-1980}{15}+1+\frac{x-1970}{25}+1\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)
\(\Leftrightarrow\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)
\(\Leftrightarrow x-1995=0\Leftrightarrow x=1995\)
21 tháng 7 2019
\(\frac{2^{4-x}}{16^5}=32^6\)
=> \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
=> \(\frac{2^{4-x}}{2^{20}}=2^{30}\)
=> \(2^{4-x}=2^{30}.2^{20}\)
=> \(2^{4-x}=2^{50}\)
=> 4 - x = 50
=> x = 4 - 50 = -46
\(\frac{3^{2x+3}}{9^3}=9^{14}\)
=> \(\frac{3^{2x+3}}{\left(3^2\right)^3}=\left(3^2\right)^{14}\)
=> \(\frac{3^{2x+3}}{3^6}=3^{28}\)
=> \(3^{2x+3}=3^{28}.3^6\)
=> \(3^{2x+3}=3^{34}\)
=> 2x + 3 = 34
=> 2x = 34 - 3
=> 2x = 31
=> x = 31/2
28 tháng 8 2019
a,\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\) (1)
<=> \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
<=> \(\left(x+1\right)\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
=> x+1=0 (vì \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\ne0\))
<=> x=-1
Vậy pt (1) có tập nghiệm S\(=\left\{-1\right\}\)
b, \(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)(2)
<=> \(\frac{x+6}{2015}+1+\frac{x+5}{2016}+1+\frac{x+4}{2017}+1=\frac{x+3}{2018}+1+\frac{x+2}{2019}+1+\frac{x+1}{2020}+1\)
<=> \(\frac{x+2021}{2015}+\frac{x+2021}{2016}+\frac{x+2021}{2017}-\frac{x+2021}{2018}-\frac{x+2021}{2019}-\frac{x+2021}{2020}=0\)
<=> \(\left(x+2021\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x+2021=0(vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=-2021
Vậy pt (2) có tập nghiệm S=\(\left\{-2021\right\}\)
c,\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\) (3)
<=> \(\frac{x+6}{2016}-1+\frac{x+7}{2017}-1+\frac{x+8}{2018}-1=\frac{x+9}{2019}-1+\frac{x+10}{2020}-1+1-1\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}=\frac{x-2010}{2019}+\frac{x-2010}{2020}\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}-\frac{x-2010}{2019}-\frac{x-2010}{2020}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x-2010=0 (vì \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=2010
Vậy pt (3) có tập nghiệm S=\(\left\{2010\right\}\)
d, \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\) (4)
<=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=15-1-2-3-4-5\)
<=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
<=> (x-100)(\(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\))=0
=> x -100=0(vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\))
<=> x=100
Vậy pt (4) có tập nghiệm S=\(\left\{100\right\}\)
VM
28 tháng 8 2019
a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!
x - x/2 = 32/15 : -16/5
=> x(1 - 1/2) = -2/3
=> x.1/2 = -2/3
=> x = -4/3
vậy_
\(x-\frac{x}{2}=\frac{32}{15}:-\frac{16}{5}\)
\(x\cdot\left(1-\frac{1}{2}\right)=-\frac{2}{3}\)
\(x\cdot\frac{1}{2}=-\frac{2}{3}\)
\(x=-\frac{2}{3}:\frac{1}{2}\)
\(x=-\frac{4}{3}\)