\(\color{red}{\text{Tìm X }}\):
\(a) |x|=\)\(\dfrac{1}{5}\)\(+\)\(\dfrac{3}{4}\)
\(b)|x-5|=-\)\(\dfrac{5}{3}+\dfrac{1}{4}\)
\(\color{red}{\text{Giúp em với ak}}\)
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câu c) mang tính mua vui hay gì hả bn
mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)
\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)
1. Tính
\(a,5\times\dfrac{7}{3}=\dfrac{35}{3}\)
\(b,\dfrac{13}{4}:7=\dfrac{13}{4}\times\dfrac{1}{7}=\dfrac{13}{28}\)
2. Tính
\(a,\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}\)
\(=\dfrac{15}{35}+\dfrac{14}{35}+\dfrac{3}{4}\)
\(=\dfrac{29}{35}+\dfrac{3}{4}\)
\(=\dfrac{116}{140}+\dfrac{105}{140}\)
\(=\dfrac{221}{140}\)
\(b,\dfrac{9}{7}-\dfrac{5}{11}\times\dfrac{11}{7}\)
\(=\dfrac{9}{7}-\dfrac{55}{77}\)
\(=\dfrac{99}{77}-\dfrac{55}{77}\)
\(=\dfrac{44}{77}=\dfrac{4}{7}\)
\(c,\dfrac{3}{5}\times\dfrac{5}{7}+\dfrac{4}{7}\)
\(=\dfrac{3}{5}\times\left(\dfrac{5}{7}+\dfrac{4}{7}\right)\)
\(=\dfrac{3}{5}\times\dfrac{9}{7}\)
\(=\dfrac{27}{35}\)
\(d,\dfrac{7}{9}\times\dfrac{2}{5}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}\times\dfrac{11}{3}\)
\(=\dfrac{154}{135}\)
\(e,\dfrac{9}{7}+\dfrac{2}{3}-\dfrac{1}{4}\)
\(=\dfrac{27}{21}+\dfrac{14}{21}-\dfrac{1}{4}\)
\(=\dfrac{41}{21}-\dfrac{1}{4}\)
\(=\dfrac{164}{84}-\dfrac{21}{84}\)
\(=\dfrac{143}{84}\)
\(g,\dfrac{4}{9}:\dfrac{3}{5}\times\dfrac{2}{11}\)
\(=\dfrac{4}{9}\times\dfrac{5}{3}\times\dfrac{2}{11}\)
\(=\dfrac{20}{27}\times\dfrac{2}{11}\)
\(=\dfrac{40}{297}\)
\(h,\dfrac{7}{2}-\dfrac{3}{10}:\dfrac{2}{5}\)
\(=\left(\dfrac{7}{2}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\left(\dfrac{35}{10}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\dfrac{32}{10}:\dfrac{2}{5}\)
\(=\dfrac{16}{5}\times\dfrac{5}{2}\)
\(=\dfrac{80}{10}=8\)
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)
Do đó: x=18; y=12; z=9
a) Thay x + 3y - 2z vào biểu thức ta có:
\(\dfrac{x - 1}{3} = \dfrac{3(y + 2)}{3 . 4} = \dfrac{2(z - 2)}{2 . 3}\) = \(\dfrac{x - 1}{3} = \dfrac{3x + 6}{12} = \dfrac{2z - 4}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhua ta có:
\(\dfrac{x - 1}{3} = \dfrac{3y + 6}{12} = \dfrac{2z - 4}{6} = \dfrac{x - 1}{3}+ \dfrac{3y + 6}{12} -\dfrac{2z - 4}{6}\)
=\(\dfrac{x - 1 + 3y + 6 - 2z + 4}{3 + 12 -6} \) = \(\dfrac{(x + 3y - 2z) + ( -1 + 6 +4)}{3 + 12 - 6} \)
=\(\dfrac{36 + 9}{9}\) = 5
=> \(\dfrac{x - 1}{3} =\) 5 => x - 1 = 5.3 =15 => x = 5+1 = 6
=>
=>
Vậy ...
(Bạn dựa theo cách này và lm những bài tiếp nhé!)
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)
Do đó: x=18; y=12; z=9
\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\left(4-x\right)^2=25=5^2=\left(-5\right)^2\)
4-x=5 hoặc 4-x=-5
x=-1 hoặc x=9
c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{138}{30}< X< \dfrac{200}{3}\)
\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)
Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)
\(\Rightarrow X=\dfrac{105381}{52}\)
\(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{38\times39}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...\dfrac{1}{38}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{13}{39}-\dfrac{1}{39}\)
\(=\dfrac{12}{39}=\dfrac{4}{13}\)
`1/(3xx4)+1/(4xx5)+1/(5xx6)+...+1/(38xx39)`
`=1/3-1/4+1/4-1/5+1/5-1/6+...+1/38-1/39`
`=1/3-1/39`
`=4/13`
a \(\Rightarrow\left|x\right|=\dfrac{19}{20}\Rightarrow\left[{}\begin{matrix}x=\dfrac{19}{20}\\x=-\dfrac{19}{20}\end{matrix}\right.\)
b \(\Rightarrow\left|x-5\right|=-\dfrac{17}{12}\) vô lí vì\(VT=\left|x-5\right|\ge0\) mà \(VP=-\dfrac{17}{20}< 0\)
\(\Rightarrow\) ko có x
a/ /x/= \(\dfrac{1}{5}+\dfrac{3}{4}\)
/x/= 0.95
=> x= \(\pm0.95\)
Vạy....
b/ /x-5/ =\(-\dfrac{5}{3}+\dfrac{1}{4}\)
/x-5/ =\(-\dfrac{17}{12}\) (vô lý)
Vậy...