\(\dfrac{x}{3}=\dfrac{-2}{9}\)
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30 tháng 12 2021
\(\text{a)}\dfrac{-3}{5}-x=\dfrac{21}{10}\)
\(x=\dfrac{-3}{5}-\dfrac{21}{10}=\dfrac{-27}{10}\)
\(\text{b)}x:\dfrac{2}{9}=\dfrac{9}{2}\)
\(x\) \(=\dfrac{9}{2}.\dfrac{2}{9}=1\)
\(\text{c) }\dfrac{x}{9}=\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{9.5}{3}=15\)
\(\text{d)}x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)
\(x:\dfrac{8}{125}=\dfrac{125}{8}\)
\(x\) \(=\dfrac{125}{8}.\dfrac{8}{125}=1\)
27 tháng 8 2021
a: Ta có: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{x-4}\)
\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+22}{x-4}\)
d: Ta có: \(D=\dfrac{1}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-12}{x-9}\)
\(=\dfrac{\sqrt{x}-3+x+3\sqrt{x}+2\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+6\sqrt{x}-15}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
H
30 tháng 12 2021
\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)
\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)
\(x=\)-\(\dfrac{27}{10}\)
\(x:\dfrac{2}{9}=\dfrac{9}{2}\)
\(x.\dfrac{9}{2}=\dfrac{9}{2}\)
\(x=\dfrac{9}{2}:\dfrac{9}{2}\)
\(x=1\)
\(\dfrac{x}{9}=\dfrac{5}{3}\)
\(x.3=5.9\)
\(x.3=45\)
\(x=45:3=15\)
\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)
\(x:\dfrac{8}{125}=\dfrac{125}{8}\)
\(x.\dfrac{125}{8}=\dfrac{125}{8}\)
\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)
28 tháng 7 2021
a) Ta có: \(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(=\dfrac{x^2}{x\left(x-3\right)}-\dfrac{6\left(x-3\right)}{x\left(x-3\right)}-\dfrac{9}{x\left(x-3\right)}\)
\(=\dfrac{x^2-6x+18-9}{x\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
b) Ta có: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(=\dfrac{7\left(x+6\right)-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-7x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-13x+6x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left[x\left(x-13\right)+6\left(x-13\right)\right]}{x\left(x+6\right)}\)
\(=\dfrac{13-x}{x}\)
c) Ta có: \(\dfrac{6}{x-3}-\dfrac{2x-6}{x^2-9}-\dfrac{4}{x+3}\)
\(=\dfrac{6\left(x+3\right)-2x+6-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{6x+18-2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
17 tháng 7 2023
a) \(\dfrac{8}{9}x=\dfrac{2}{7}-\dfrac{2}{3}=-\dfrac{8}{21}\)
\(x=-\dfrac{8}{21}:\dfrac{8}{9}=-\dfrac{3}{7}\)
b) \(\dfrac{2}{5}x=\dfrac{2}{5}-\dfrac{2}{5}=0\)
\(x=0:\dfrac{2}{5}=0\)
c)\(\dfrac{7}{8}x=\dfrac{2}{9}-\dfrac{1}{3}=-\dfrac{1}{9}\)
\(x=-\dfrac{1}{9}:\dfrac{7}{8}=-\dfrac{8}{63}\)
NM
17 tháng 7 2023
a) 2/7 - 8/9 . x = 2/3
⇒ 8/9 . x = 2/7 - 2/3
⇒ 8/9 .x = -8/21
⇒ x = -8/21 : 8/9
⇒ x = -3/7.
Vậy...
20 tháng 12 2022
a: \(=\dfrac{x^2-x+x+1+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\)
b: \(=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
c: \(=\dfrac{2x^2-3x-9-x^2+3x+x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2+6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x}{x-3}\)
NA
Ngoc Anh Thai
Giáo viên
8 tháng 5 2021
a)
\(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{x+3}=\dfrac{9}{38}\\\\ \dfrac{1}{x+3}=\dfrac{1}{4}-\dfrac{9}{38}\\ \dfrac{1}{x+3}=\dfrac{1}{76}\\ x+3=76\\ x=73.\)
NA
Ngoc Anh Thai
Giáo viên
8 tháng 5 2021
b)
\(\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ \dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ 2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ 2.\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}=\dfrac{1}{18}\\ x+1=18\\ x=17.\)
26 tháng 4 2021
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}
26 tháng 4 2021
cảm ơn bạn Hoàng Bình Bảo nha nhưng mà đây là toán lớp 6 mà bạn
\(\dfrac{x}{3}=\dfrac{-2}{9}\)
\(\Leftrightarrow=\dfrac{-2}{9}\times3\)
\(\Leftrightarrow x=\dfrac{-2}{3}\)
`x/3=(-2)/9`
`=>x*9=-2*3`
`=>9x=-6`
`=>x=-6/9=-2/3`