\(\frac{5}{9}x\frac{9}{6}x\frac{7}{3}x\frac{6}{5}x\frac{3}{8}\)=
người nào giải nhanh và đúng nhất tớ sẻ tích
K
Khách
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Những câu hỏi liên quan
28 tháng 6 2017
= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)
= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)
= \(x^8.\frac{1}{10}.1.1.1.1\)
= \(x^8.\frac{1}{10}\)
Mk ko pik co dung ko nua
23 tháng 3 2019
\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)
\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)
2 câu sau tự làm nha
23 tháng 3 2019
\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)
\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)
=> \(x\in\left\{-2;-1;0;1;2\right\}\)
S
19 tháng 2 2017
\(a,\frac{x+5}{20}=-\frac{5}{4}\)
\(\left(x+5\right):20=\frac{-5}{4}\)
\(x+5=-\frac{5}{4}\times20\)
\(x+5=-25\)
\(x=-25-5\)
\(x=-30\)
các câu khác thì theo đó tự làm nha
19 tháng 2 2017
bn ơi các câu a, b thì mk làm đc nhưng vướng câu c đó bn ạ
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!
NT
Tìm x biết: \(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}.\)
4
13 tháng 8 2017
\(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}\)
\(\Rightarrow\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}+3=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}+3\)
\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+4}{6}+1\right)+\left(\frac{x+5}{5}+1\right)=\left(\frac{x+2}{8}+1\right)\)\(+\left(\frac{x+3}{7}+1\right)+\left(\frac{x+6}{4}\right)\)
\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}\right)=\left(x+10\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{4}\right)\)
\(\Rightarrow\left(x+10\right)\frac{43}{90}=\left(x+10\right)\frac{29}{56}\)
\(\Rightarrow x+10=0\)
\(\Rightarrow x=-10\)
S
13 tháng 8 2017
cộng 3 vào cả hai vế nên phương trình vẫn bằng nhau
Ta có \(\frac{x+1}{9}+1+\frac{x+4}{6}+1+\frac{x+5}{5}+1=\frac{x+2}{8}+1+\frac{x+3}{7}+1+\frac{x+6}{4}+1\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}-\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{4}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
mà \(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)
\(\Rightarrow x+10=0\)
\(\Leftrightarrow x=-10\)
DL
1 tháng 6 2017
\(\frac{1}{2}:\frac{3}{2}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\)
\(=\frac{1\cdot\left(2\cdot5\cdot6\cdot7\right)}{8\cdot3\cdot\left(2\cdot5\cdot6\cdot7\right)}\)
\(=\frac{1}{24}\)
DL
1 tháng 6 2017
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\cdot\frac{9}{10}\)
\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)}{\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)\cdot10}\)
\(=\frac{1}{10}\)
=21/16
k cho mình nhé