Giúp v ạ m.ơn mn trc:3
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Do \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Rightarrow1-\dfrac{b}{a}=1-\dfrac{d}{c}\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\) (đpcm)
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5:
a: =>x+1/5-1/9+1/9-1/13+...+1/41-1/45=-37/45
=>x+9/45-1/45=-37/45
=>x+8/45=-37/45
=>x=-1
b: 3/x+1 nguyên
=>x+1 thuộc {1;-1;3;-3}
=>x thuộc {0;-2;2;-4}
c; -5/2x+1 nguyên
=>2x+1 thuộc {1;-1;5;-5}
=>x thuộc {0;-1;2;-3}
![](https://rs.olm.vn/images/avt/0.png?1311)
c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\)
Ở nơi x=9/4-1/2 là x-9/4-1/2 nha
a. -1,5 + 2x = 2,5
<=> 2x = 2,5 + 1,5
<=> 2x = 4
<=> x = 2
b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)
<=> 9x + 45 - 3 = 8
<=> 9x = 8 + 3 - 45
<=> 9x = -34
<=> x = \(\dfrac{-34}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(54\left(\dfrac{km}{h}\right)=15\left(\dfrac{m}{s}\right);9\left(\dfrac{m}{s}\right)=32,4\left(\dfrac{km}{h}\right)\)
Baì 2:
\(t'=s':v'=5:\left(5.3,6\right)=\dfrac{5}{18}h\)
\(\Rightarrow v_{tb}=\dfrac{s'+s''}{t'+t''}=\dfrac{5+3,8}{\dfrac{5}{18}+\left(\dfrac{15}{60}\right)}\simeq16,67\left(\dfrac{km}{h}\right)\)
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Câu 2:
\(\Leftrightarrow\left(x+2\right)\left(10x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{3}{10}\end{matrix}\right.\)
Câu 45 : | -1,5 | = 1,5 => Chọn A
Câu 46 : \(\left|\frac{-6}{10}\right|=\frac{6}{10}=\frac{3}{5}\)=> Chọn A
Câu 47 : | -0,4 | = 0,4 => đúng => A đúng
| -0,4 | = -0,4 => Sai
| -0,4 | = ± 0,4 => Sai
| -0,4 | = 0 => Sai
=> Chọn A
Câu 48 : Ta có : | -1,5 | = 1,5 => đúng
| 0 | = 0 => đúng
| 1,8 | = ± 1,8 => Sai
| 0,2 | = 0,2 > 0 => đúng
=> Chọn C
Câu 49 :
Vì \(\left|x\right|\)\(=\frac{1}{2}\Rightarrow x=\pm\frac{1}{2}\)=> Chọn B ( do bài bảo tìm TẤT CẢ giá trị x thỏa mãn )