Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{-x}{4}=\dfrac{-9}{x}\)
\(\Rightarrow-x^2=-36\)
\(\Rightarrow x^2=36\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
Vậy: \(x\in\left\{6;-6\right\}\)
b) \(\dfrac{5}{9}+\dfrac{x}{-1}=-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{5}{9}+\dfrac{-9x}{9}=\dfrac{-3}{9}\)
\(\Rightarrow5-9x=-3\)
\(\Rightarrow-9x=-8\)
\(\Rightarrow x=\dfrac{8}{9}\)
Vậy: \(x=\dfrac{8}{9}\)
c) \(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)
\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{3}{2}.\dfrac{16}{5}\)
\(\Rightarrow x=\dfrac{24}{5}\)
Vậy: \(x=\dfrac{24}{5}\)
d) \(\dfrac{3x-1}{-5}=\dfrac{-5}{3x-1}\)
\(\Rightarrow\left(3x-1\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-\dfrac{4}{3}\right\}\)
1.\(x\times\dfrac{2}{3}=\dfrac{1}{5}\)
\(x=\dfrac{1}{5}:\dfrac{2}{3}\)
\(x=\dfrac{3}{10}\)
2.\(\dfrac{2}{x}=\dfrac{12}{5}\)
\(12x=2.5\)
\(x=\dfrac{10}{12}=\dfrac{5}{6}\)
3.\(\dfrac{1}{2}x-\dfrac{3}{5}=-\dfrac{4}{5}\)
\(\dfrac{1}{2}x=\dfrac{-4}{5}+\dfrac{3}{5}\)
\(\dfrac{1}{2}x=-\dfrac{1}{5}\)
\(x=-\dfrac{1}{5}:\dfrac{1}{2}\)
\(x=-\dfrac{2}{5}\)
4.\(3x+2x=-5,05\)
\(5x=-5,05\)
\(x=-\dfrac{5,05}{5}\)
\(x=-1,01\)
\(x\cdot\dfrac{2}{3}=\dfrac{1}{5}\)
\(x=\dfrac{1}{5}:\dfrac{2}{3}\)
\(x=\dfrac{1}{5}\cdot\dfrac{3}{2}\)
\(x=\dfrac{3}{10}\)
\(\dfrac{2}{x}=\dfrac{12}{5}\)
\(x=\dfrac{\left(2\cdot5\right)}{12}\)
\(x=\dfrac{10}{12}\)
\(x=\dfrac{5}{6}\)
\(\dfrac{1}{2}x-\dfrac{3}{5}=\dfrac{-4}{5}\)
\(\dfrac{1}{2}x=\left(\dfrac{-4}{5}+\dfrac{3}{5}\right)\)
\(\dfrac{1}{2}x=-\dfrac{1}{5}\)
\(x=\left(-\dfrac{1}{5}\right):\dfrac{1}{2}\)
\(x=\left(-\dfrac{1}{5}\right)\cdot2\)
\(x=-\dfrac{2}{5}\)
\(3x+2x=-5,05\)
\(\left(3+2\right)x=-5,05\)
\(5x=-5,05\)
\(x=\left(-5,05\right):5\)
\(x=-1,01\)
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
\(=\left(\dfrac{x}{x+1}+\dfrac{1}{x+1}-\dfrac{x}{x-1}\right)\cdot\dfrac{3x-1-x-1}{x+1}\)
\(=\left(1-\dfrac{x}{x-1}\right)\cdot\dfrac{2x-2}{x+1}\)
\(=\dfrac{x-1-x}{x-1}\cdot\dfrac{2\left(x-1\right)}{x+1}=\dfrac{-2}{x+1}\)
`#040911`
`a)`
`3 1/3 x + 16 3/4 = -13,25`
`=> 3 1/3 x = -13,25 - 16 3/4`
`=> 3 1/3 x = -30`
`=> x = -30 \div 3 1/3`
`=> x =-9`
Vậy, `x = -9`
`b)`
`3 2/7*x - 1/8 = 2 3/4`
`=> 3 2/7x = 2 3/4 + 1/8`
`=> 3 2/7x = 23/8`
`=> x = 23/8 \div 3 2/7`
`=> x = 7/8`
Vậy, `x = 7/8`
`c)`
`x \div 4 1/3 = -2,5`
`=> x = -2,5 * 4 1/3`
`=> x = -65/6`
Vậy, `x = -65/6`
`d)`
`( (3x)/7 + 1) \div (-4) = (-1)/28`
`=> (3x)/7 +1 = (-1)/28 * (-4)`
`=> (3x)/7 + 1 = 1/7`
`=> (3x)/7 = 1/7 - 1`
`=> (3x)/7 = -6/7`
`=> 3x = -6`
`=> x = -6 \div 3`
`=> x = -2`
Vậy, `x = -2.`
a
=>10/3 . x + 16 + 3/4 = -13,25
=>10/3 x + 3/4 = -29,25
=>10/3 x = -30
=>x=-30 : 10/3
=>x=-30 . 3/10
=>x=-9
b.
=>23/7 x - 1/8 = = 11/4
=>23/7 x = 11/4 + 1/8
=>23/7 x= 22/8 + 1/8
=>23/7 x= 23/8
=>x=23/8 : 23/7
=>x=23/8 . 7/23
=>x=7/8
c.
=>x : 13/3 =-5/2
=>x=-5/2 . 13/3
=>x=-65/6
d.
=>3x/7 +1 = (-1/28) . (-4)
=>3x/7 + 1 = 1/7
=>3x/7 = -6/7
=>3x=-6
=>x=-2
a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)
\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)
\(\Leftrightarrow-7x=\dfrac{10}{3}\)
hay \(x=-\dfrac{10}{21}\)
b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)
\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)
\(\Leftrightarrow85x+85=84x+168\)
\(\Leftrightarrow x=83\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)
Vậy ...
2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)
Vậy ...
3) PT \(\Rightarrow3x-4-2x+5=3\)
\(\Rightarrow x=2\)
Vậy ...
4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy ...
3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)
\(\Leftrightarrow3x-4-2x+5=3\)
\(\Leftrightarrow x+1=3\)
hay x=2
a, dễ, tự làm
b, \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+.........+\dfrac{3x}{11.14}=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+.........+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.....+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x.\dfrac{3}{7}=\dfrac{1}{21}\)
\(\Leftrightarrow x=\dfrac{1}{9}\)
Vậy ...
a) (x-2)3 = (x-2)2
<=> (x-2)3-(x-2)2 = 0
<=> (x-2)2(x-2-1) = 0
<=> \(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\x-3=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+...+\dfrac{3x}{11.14}=\dfrac{1}{21}\)
<=> \(x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)
<=> \(x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
<=> \(x=\dfrac{1}{21}:\dfrac{3}{7}\)
<=> \(x=\dfrac{1}{9}\)
\(\dfrac{1}{2}=\dfrac{x+1}{3x}\)
\(=>3x=2(x+1)\)
`=>3x=2x+2`
`=>3x-2x=2`
`=>x=2`
\(\dfrac{1}{2}=\dfrac{x+1}{3x}\) ( ĐKXĐ : x khác 0 )
\(\Rightarrow3x=2x+2\)
\(\Leftrightarrow x=2\left(T/m\right)\)