\(x^2+\dfrac{1}{9}=\dfrac{5}{3}:3\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a/`
` 7/5 + 3 2/5 - 1 1/2 `
`= (7/5 + 17/5) - 3/2`
`= 24/5 - 3/2 `
`= 48/10 - 15/10 `
`= 33/10 `
`b/`
` 3 xx 2 4/9 xx 3/2 `
` = 3 xx 22/9 xx 3/2 `
` = 22/3 xx 3/2`
`= 11.`
`c/`
` 5/9 xx ( 2 1/6 - 1 2/3 ) `
`= 5/9 xx ( 13/6 -5/3 )`
`= 5/9 xx ( 13/6 - 10/6 ) `
`= 5/9 xx 3/6 `
`= 5/9 xx 1/2 `
`= 5/18`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)
\(x=\dfrac{7}{35}+\dfrac{-15}{35}\)
\(x=-\dfrac{8}{35}\)
\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{2}\)
\(x=\dfrac{4}{7}:\dfrac{3}{2}\)
\(x=\dfrac{4}{7}\times\dfrac{2}{3}\)
\(x=\dfrac{8}{21}\)
\(c,x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)
\(x+\dfrac{3}{4}=\dfrac{-4}{6}-\dfrac{3}{6}\)
\(x+\dfrac{3}{4}=-\dfrac{7}{6}\)
\(x=-\dfrac{7}{6}-\dfrac{3}{4}\)
\(x=-\dfrac{23}{12}\)
\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{13}{18}\)
\(x=\dfrac{-5}{9}-\dfrac{13}{18}\)
\(x=\dfrac{-10}{18}-\dfrac{13}{18}\)
\(x=-\dfrac{23}{18}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
\(\Rightarrow \left[\begin{array}{} x+\frac{1}{2}=0\\ \frac{2}{3}-2x=0 \end{array} \right.\)
\(\Leftrightarrow \left[\begin{array}{} x=\frac{-1}{2}\\ x=\frac{1}{3} \end{array} \right.\)
Vậy phương trình có tập nghiệm S={\(\frac{-1}{2};\frac{1}{3}\)}
Câu 2:
\(\Rightarrow \left[\begin{array}{} 3x-10=0\\ 5-\frac{1}{2}x=0 \end{array} \right.\)
\(\Leftrightarrow \left[\begin{array}{} x-=\frac{10}{3}\\ x=10 \end{array} \right.\)
Vậy phương trình có tập nghiệm S={\(10;\frac{10}{3}\)}
Câu 3:
\(\Leftrightarrow \frac{1}{3}x=\frac{65}{4}-\frac{53}{4}\)
\( \Leftrightarrow \frac{1}{3}x=\frac{12}{4}\)
\(\Leftrightarrow x=9\)
Vậy phương trình có tập nghiệm S={9}
Câu 4:
\(\Leftrightarrow \frac{2}{3}x=\frac{2}{3}\)
\(\Leftrightarrow x=1\)
Vậy phương trình có tập nghiệm S={1}
Câu 5:
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x(x+1)}=\frac{2010}{2011}\)
\(\Leftrightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)
\(\Leftrightarrow 1-\frac{1}{x+1}=\frac{2010}{2011}\)
\(\Leftrightarrow \frac{x}{x+1}=\frac{2010}{2011}\)
\(\Rightarrow 2010x+2010=2011x\)
\(\Leftrightarrow x=2010\)
Vậy phương trình có tập nghiệm S={2010}
cảm ơn bạn Hoàng Bình Bảo nha nhưng mà đây là toán lớp 6 mà bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
`2/3+5/2-3/4`
`=10/4-3/4+2/3`
`=7/4+2/3`
`=21/12+8/12`
`=29/12`
b)
`2/5xx1/2:1/3`
`=2/10xx3/1`
`=6/10=3/5`
c)
`2/9:2/9xx1/3`
`=2/9xx9/2xx1/3`
`=1xx1/3`
`=1/3`
a, \(\dfrac{2}{3}\) + \(\dfrac{5}{2}\) - \(\dfrac{3}{4}\)
= \(\dfrac{8}{12}\) + \(\dfrac{30}{12}\) - \(\dfrac{9}{12}\)
= \(\dfrac{38-9}{12}\)
= \(\dfrac{29}{12}\)
b, \(\dfrac{2}{5}\) x \(\dfrac{1}{2}\) : \(\dfrac{1}{3}\)
= \(\dfrac{1}{5}\) x \(\dfrac{3}{1}\)
= \(\dfrac{3}{5}\)
c, \(\dfrac{2}{9}\) : \(\dfrac{2}{9}\) x \(\dfrac{1}{3}\)
= 1 x \(\dfrac{1}{3}\)
= \(\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a, 3-(x+5/7 )=9/21`
`=>x+5/7= 3-9/21`
`=>x+5/7= 63/21-9/21`
`=>x+5/7= 54/21`
`=>x= 54/21-5/7`
`=>x= 54/21 - 15/21`
`=>x= 39/21`
`=>x= 13/7`
`b, x/2+ x/5 = 17/10`
`=> (5x)/10 + (2x)/10=17/10`
`=> 7x/10=17/10`
`=> 7x.10=10.17`
`=>7x.10=170`
`=>7x=170:10`
`=>7x=17`
`=>x=17/7`
`c, 1/2x + 1/3 -1= 3 1/3`
`=> 1/2x + 1/3 -1= 10/3`
`=> 1/2x + 1/3=10/3+1`
`=> 1/2x + 1/3=10/3 + 3/3`
`=> 1/2x + 1/3=13/3`
`=>1/2 x= 13/3 -1/3`
`=> 1/2x= 12/3`
`=> 1/2x= 4`
`=>x= 4 :1/2`
`=>x= 4 xx 2`
`=>x=8`
\(a,3-\left(x+\dfrac{5}{7}\right)=\dfrac{9}{21}\\ x+\dfrac{5}{7}=3-\dfrac{9}{21}\\ x+\dfrac{5}{7}=\dfrac{18}{7}\\ x=\dfrac{18}{7}-\dfrac{5}{7}\\ x=\dfrac{13}{7}\\ b,\dfrac{x}{2}+\dfrac{x}{5}=\dfrac{17}{10}\\ \dfrac{5x}{10}+\dfrac{2x}{10}=\dfrac{17}{10}\\ \dfrac{7x}{10}=\dfrac{17}{10}\\ 7x=17\\ x=\dfrac{17}{7}\\ c,\dfrac{1}{2}x+\dfrac{1}{3}-1=3\dfrac{1}{3}\\ \dfrac{1}{2}x+\dfrac{1}{3}-1=\dfrac{10}{3}\\ \dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{10}{3}+1\\ \dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{13}{3}\\ \dfrac{1}{2}x=\dfrac{13}{3}-\dfrac{1}{3}\\ \dfrac{1}{2}x=4\\ x=4:\dfrac{1}{2}\\ x=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{8}{15}\\ =\dfrac{3\times6}{5\times6}+\dfrac{1\times15}{2\times15}+\dfrac{8\times2}{15\times2}\\ =\dfrac{18}{30}+\dfrac{15}{30}+\dfrac{16}{30}\\ =\dfrac{49}{30}\\ \dfrac{6}{9}+\dfrac{14}{18}-\dfrac{5}{6}\\ =\dfrac{6\times2}{9\times2}+\dfrac{14}{18}-\dfrac{5\times3}{6\times3}\\ =\dfrac{12}{18}+\dfrac{14}{18}-\dfrac{15}{18}\\ =\dfrac{11}{18}\)
\(\dfrac{9}{20}-\dfrac{3}{5}:\dfrac{4}{1}\\ =\dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}\\ =\dfrac{9}{20}-\dfrac{3}{20}\\ =\dfrac{6}{20}\\ =\dfrac{3}{10}\)
\(\dfrac{1}{6}+\dfrac{2}{3}\times\dfrac{8}{9}\\=\dfrac{1}{6}+\dfrac{16}{27}\\ =\dfrac{1\times9}{6\times9}+\dfrac{16\times2}{27\times2}\\ =\dfrac{9}{54}+\dfrac{32}{54}\\ =\dfrac{41}{54}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2}{5}\times\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{1}{5}\times\dfrac{3}{2}-\dfrac{1}{8}=\dfrac{3}{10}-\dfrac{1}{8}=\dfrac{24}{80}-\dfrac{10}{80}=\dfrac{14}{80}=\dfrac{7}{40}\\ \dfrac{4}{3}+\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{3}-\dfrac{1}{5}=\dfrac{25}{15}-\dfrac{3}{15}=\dfrac{22}{15}\\ \dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\\ \dfrac{2}{8}+\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{2}{8}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{2}{8}+\dfrac{1}{3}\times\dfrac{5}{2}=\dfrac{2}{8}+\dfrac{5}{6}=\dfrac{1}{4}+\dfrac{5}{6}=\dfrac{6}{24}+\dfrac{20}{24}=\dfrac{26}{24}=\dfrac{13}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
1: Ta có: \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
\(\Leftrightarrow2x-8+12x=4x-2\)
\(\Leftrightarrow10x=6\)
hay \(x=\dfrac{3}{5}\)
2: Ta có: \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
\(\Leftrightarrow15x-6-30=10-20x\)
\(\Leftrightarrow35x=46\)
hay \(x=\dfrac{46}{35}\)
3: Ta có: \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
\(\Leftrightarrow3x-6-4=6x-6\)
\(\Leftrightarrow-3x=4\)
hay \(x=-\dfrac{4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\Leftrightarrow\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=9+\dfrac{5}{7}-\dfrac{5}{7}=9\)
=>x-1/2=27
hay x=55/2
c: =>1/2x-3/4=42/63=2/3
=>1/2x=17/12
hay x=17/6
\(x^2+\dfrac{1}{9}=\dfrac{5}{3}:\dfrac{1}{3}\)
\(x^2+\dfrac{1}{9}=\dfrac{5}{9}\)
\(x^2=\dfrac{5}{9}+\dfrac{1}{9}\)
\(x^2=\dfrac{4}{9}\)
\(x=\dfrac{2}{3};\dfrac{-2}{3}\)
x2 + \(\dfrac{1}{9}\) = \(\dfrac{5}{9}\)
x2 = \(\dfrac{5}{9}\)- \(\dfrac{1}{9}\)
x2 = \(\dfrac{4}{9}\)
x2 = 2/32
=> x = 2/3
vậy x = 2/3