mk có câu này làm ko ra mong các bạn giúp các bạn có ra kết quả nhớ thử lại nhé
\(5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x-3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)
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(5\(\dfrac{8}{17}\)+\(\dfrac{-4}{17}\)): x +\(\dfrac{22}{7}\):\(\dfrac{52}{3}\)=\(\dfrac{4}{11}\)
\(^{\dfrac{89}{17}}\):x+\(\dfrac{22}{7}\)x\(\dfrac{3}{52}\)=\(\dfrac{4}{11}\)
\(\dfrac{89}{17}\):x+\(\dfrac{33}{182}\)=\(\dfrac{4}{11}\)
\(\dfrac{89}{17}\):x=\(\dfrac{385}{2002}\)
x=\(\dfrac{2314}{85}\)
7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)
\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)
\(\Rightarrow-2\le x\le2\)
\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)
\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)
\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)
\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)
\(\Rightarrow x\in\left\{11;12;13;14\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)
tìm x:
\(5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)
Cho link đây, hồi chiều mới làm: Câu hỏi của nguyen thu trang - Toán lớp 6 | Học trực tuyến
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b:
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
`a)4/5+5 1/2 xx (4,5-2)+7/10`
`=4/5+11/2*2,5+7/10`
`=0,8+2,2+0,7`
`=3+0,7=3,7`
`b)125%xx 17/4:(1 5/16-0,5)+2008`
`=1,25xx4,25:13/16+2008`
`=85/13+2008`
`=2014 7/13`
`c)5/11+(16/11+1)`
`=5/11+1+5/11+1`
`=2+10/11=32/11`
`d)3/17+11/4+5/8+14/17+3/8`
`=3/17+14/17+5/8+3/8+11/4`
`=1+1+11/4`
`=19/4`
a)
\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)
<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)
<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)
b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)
c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)
= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)
d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)
= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)
Bài 2:
1: \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\)
=>\(\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{5}{6}=\dfrac{1}{12}+\dfrac{10}{12}=\dfrac{11}{12}\)
=>x=11
2: \(\dfrac{2}{3}-1\dfrac{4}{15}x=-\dfrac{3}{5}\)
=>\(\dfrac{2}{3}-\dfrac{19}{15}x=-\dfrac{3}{5}\)
=>\(\dfrac{19}{15}x=\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10+9}{15}=\dfrac{19}{15}\)
=>\(x=\dfrac{19}{15}:\dfrac{19}{15}=1\)
3: \(\dfrac{\left(-3\right)^x}{81}=-27\)
=>\(\left(-3\right)^x=\left(-3\right)^3\cdot\left(-3\right)^4=\left(-3\right)^7\)
=>x=7
4: \(\left|x+0,237\right|=0\)
=>x+0,237=0
=>x=-0,237
5: \(\left(x-1\right)^2=25\)
=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
6: \(\left|2x-1\right|=5\)
=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
7: \(\left(x-1\right)^3=-\dfrac{8}{27}\)
=>\(\left(x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)
=>\(x-1=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}+1=\dfrac{1}{3}\)
8: \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)
=>\(\dfrac{5}{3}:\dfrac{x}{4}=20\)
=>\(\dfrac{20}{3x}=20\)
=>3x=20/20=1
=>\(x=\dfrac{1}{3}\)
9: \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:2\dfrac{2}{3}\)
=>\(\dfrac{\dfrac{8}{3}}{x}=\dfrac{\dfrac{16}{9}}{\dfrac{8}{3}}\)
=>\(\dfrac{16}{9}\cdot x=\dfrac{8}{3}\cdot\dfrac{8}{3}=\dfrac{64}{9}\)
=>16x=64
=>x=64/16=4
Bài 3:
1: Ta có: x-24=y
=>x-y=24
mà \(\dfrac{x}{7}=\dfrac{y}{3}\)
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{24}{4}=6\)
=>\(x=6\cdot7=42;y=6\cdot3=18\)
2: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}\)
mà x-y=48
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}=\dfrac{x-y}{5-7}=\dfrac{48}{-2}=-24\)
=>\(x=-24\cdot5=-120;y=-24\cdot7=-168;z=-24\cdot2=-48\)
3: \(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}\)
mà x-y=4009
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}=\dfrac{x-1+3-y}{2005+2006}=\dfrac{4009+2}{4011}=1\)
=>\(x-1=2005;3-y=2006\)
=>x=2005+1=2006; y=3-2006=-2003
5: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)
mà 2x+3y-z=-14
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{2\cdot3+3\cdot5-7}=\dfrac{-14}{14}=-1\)
=>\(x=-3;y=-5;z=-7\)
Bạn tách ra từng CH khác nhau đi nhé. Gộp 1 trong tất cả rất khó nhìn và lâu.
\(5\dfrac{8}{17}:x+\left(\dfrac{-4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)
\(\Leftrightarrow\left(5\dfrac{8}{17}+\dfrac{-4}{17}\right):x+\dfrac{22}{7}:\dfrac{52}{3}=\dfrac{4}{11}\)
\(\Leftrightarrow\left(\dfrac{43}{17}+\dfrac{-4}{17}\right):x+\dfrac{22}{7}.\dfrac{3}{52}=\dfrac{4}{11}\)
\(\Leftrightarrow\dfrac{39}{17}:x+\dfrac{33}{182}=\dfrac{4}{11}\)
\(\Leftrightarrow\dfrac{39}{17}:x=\dfrac{4}{11}-\dfrac{33}{182}\)
(cứ thế mà làm tiếp nhé, mình thấy số xấu quá)
\(\Leftrightarrow\left(5+\dfrac{8}{17}-\dfrac{4}{17}\right):x=\dfrac{4}{11}+\dfrac{22}{7}:\dfrac{52}{3}\)
\(\Leftrightarrow\dfrac{89}{17}:x=\dfrac{1091}{2002}\)
hay \(x\simeq9,607\)