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a) Ta có: \(\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-1}{5}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-5}{25}=\dfrac{2}{25}\)
hay \(x=\dfrac{6}{25}\)
Vậy: \(x=\dfrac{6}{25}\)
b) Ta có: \(\dfrac{4}{9}+\dfrac{x}{5}=\dfrac{5}{11}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{45}{99}-\dfrac{44}{99}=\dfrac{1}{99}\)
hay \(x=\dfrac{5}{99}\)
Vậy: \(x=\dfrac{5}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,2/5 = 2/5 ; 3/8=6/16 ; 1/9=3/27
b, 4/3=8/6 ; -1=-1 ; -4/-2=-8/4
tick cho mik nhé
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\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)
\(x=\dfrac{7}{35}+\dfrac{-15}{35}\)
\(x=-\dfrac{8}{35}\)
\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)
\(\dfrac{4}{7}:x=\dfrac{3}{2}\)
\(x=\dfrac{4}{7}:\dfrac{3}{2}\)
\(x=\dfrac{4}{7}\times\dfrac{2}{3}\)
\(x=\dfrac{8}{21}\)
\(c,x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)
\(x+\dfrac{3}{4}=\dfrac{-4}{6}-\dfrac{3}{6}\)
\(x+\dfrac{3}{4}=-\dfrac{7}{6}\)
\(x=-\dfrac{7}{6}-\dfrac{3}{4}\)
\(x=-\dfrac{23}{12}\)
\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)
\(\dfrac{-5}{9}-x=\dfrac{13}{18}\)
\(x=\dfrac{-5}{9}-\dfrac{13}{18}\)
\(x=\dfrac{-10}{18}-\dfrac{13}{18}\)
\(x=-\dfrac{23}{18}\)
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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: \(x\in\left\{2;-\dfrac{4}{3}\right\}\)
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a, 2/5 + 3/4 : x = -1/2
3/4 : x = -1/2 - 2/5
3/4 : x = -9/10
x = 3/4 : -9/10
x = -5/6
b, 5/7 - 2/3 . x = 4/5
2/3 . x = 4/5 + 5/7
2/3 . x = 53/35
x = 53/35 : 2/3
x = 159/70
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`5/9+(-4)/5<x/15<-4/9+3/5`
`-11/15 < x/15 <7/45`
`-33/45 <(3x)/45 <7/45`
`-33<3x<7`
`-11<x<7/3`
`=> x \in {-10;-9;....2}` (Nếu `x \in ZZ)`
`x \in {0;1;2}` (Nếu `x \in NN`).
![](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: x+2/5=1/2
=>x=1/2-2/5=5/10-4/10=1/10
b; x-2/5=2/7
=>x=2/7+2/5=10/35+14/35=24/35
c: 3/5-x=1/10
=>x=3/5-1/10=6/10-1/10=5/10=1/2
d: x*3/4=9/20
=>x=9/20:3/4=9/20*4/3=36/60=3/5
e: x:1/7=14
=>x=14*1/7=2
f: =>x+1/4=2/5:1/2=4/5
=>x=4/5-1/4=16/20-5/20=11/20
g: =>x*2/3=9/12+2/3=3/4+2/3=9/12+8/12=17/12
=>x=17/12:2/3=17/12*3/2=51/24=17/8
\(\dfrac{x-5}{-4}=\dfrac{-9}{x-5}\)
\(\Leftrightarrow\left(x-5\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)
Ta có : (x-5) . (x-5)= (-9).(-4)
⇒ (x-5) . (x-5)= 36
Mà 36 = 6.6 hoặc 36 = (-6).(-6)
⇒ x-5 = 6 hoặc x-5 = (-6)
x = 6+5 x = (-6) + 5
x = 11 x = (-1)
Vậy x = 11 hoặc x = -1
Chúc bạn học tốt ! ^^