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bài 3 :
Số học sinh trung bình là :
\(1200\times\dfrac{5}{8}=750\) ( hs)
Số học sinh khá là :
\(750\times\dfrac{2}{5}=300\) (hs)
Số học sinh giỏi là :
\(1200-750-300=150\left(hs\right)\)
b) So với cả trường chứ ?
3b ) Tỉ số của hs giỏi so với toàn trường :150: 1200 = 0,125
Tỉ số phần trăm của hs giỏi so vs toàn trường là : 12,5%
Bài 1:
Ta có: \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2\ge0\\\left|y+\dfrac{1}{4}\right|\ge0\end{matrix}\right.\Rightarrow\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|\ge0\)
\(\Rightarrow A=\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|+\dfrac{13}{14}\ge\dfrac{13}{14}\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2=0\\\left|y+\dfrac{1}{4}\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{4}=0\\y+\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{-1}{4}\end{matrix}\right.\)
Vậy \(MIN_A=\dfrac{13}{14}\) khi \(x=\dfrac{1}{4};y=-\dfrac{1}{4}\)
\(a.\)
\(\dfrac{1}{3}\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)
\(\Rightarrow\dfrac{1}{6}-2+\dfrac{3}{5}=-5x+x\)
\(\Rightarrow-4x=-\dfrac{37}{30}\)
\(\Rightarrow4x=\dfrac{37}{30}\)
\(\Rightarrow x=\dfrac{37}{120}\)
\(b.\)
\(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{3}{2}=x-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{3}{2}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{2}\)
\(c.\)
\(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\)
\(\Rightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\)
\(\Rightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)
\(\Rightarrow x=\dfrac{19}{21}\)
\(d.\)
\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-\dfrac{39}{10}=x+1\)
\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{39}{10}+1\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{49}{10}\)
\(\Rightarrow x=\dfrac{49}{5}\)
\(e.\)
\(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=\dfrac{3}{5}\\4x+3=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{12}{5}\\4x=-\dfrac{18}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{9}{10}\end{matrix}\right.\)
a) \(\dfrac{1}{3}.\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\Leftrightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)
\(\Leftrightarrow5x-x=-\dfrac{3}{5}-\dfrac{1}{6}+2\Leftrightarrow4x=\dfrac{37}{30}\Leftrightarrow x=\dfrac{\dfrac{37}{30}}{4}=\dfrac{37}{120}\)
vậy \(x=\dfrac{37}{120}\)
b) \(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\Leftrightarrow\dfrac{3}{2}x-x=\dfrac{-3}{4}+1\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}x=\dfrac{-3}{4}+\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.2=\dfrac{6}{4}=\dfrac{3}{2}\) vậy \(x=\dfrac{3}{2}\)
c) \(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\Leftrightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\Leftrightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)
\(\Leftrightarrow x=\dfrac{\dfrac{19}{12}}{\dfrac{7}{4}}=\dfrac{19}{21}\) vậy \(x=\dfrac{19}{21}\)
d) \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\Leftrightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)
\(\Leftrightarrow\dfrac{3}{2}x-x=1+\dfrac{5}{2}+\dfrac{7}{5}\Leftrightarrow\dfrac{1}{2}x=\dfrac{49}{10}\Leftrightarrow x=\dfrac{49}{10}.2=\dfrac{49}{5}\)
vậy \(x=\dfrac{49}{5}\)
e) \(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\Leftrightarrow\left|4x+3\right|=\dfrac{\dfrac{6}{15}}{\dfrac{2}{3}}=\dfrac{3}{5}\)
th1 : \(4x+3\ge0\Leftrightarrow4x\ge-3\Leftrightarrow x\ge\dfrac{-3}{4}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow4x+3=\dfrac{3}{5}\Leftrightarrow4x=\dfrac{3}{5}-3=\dfrac{-12}{5}\)
\(\Leftrightarrow x=\dfrac{\dfrac{-12}{5}}{4}=\dfrac{-3}{5}\left(tmđk\right)\)
th2: \(4x+3< 0\Leftrightarrow4x< -3\Leftrightarrow x< \dfrac{-3}{4}\)
\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow-\left(4x+3\right)=\dfrac{3}{5}\Leftrightarrow-4x-3=\dfrac{3}{5}\)
\(\Leftrightarrow4x=-3-\dfrac{3}{5}=\dfrac{-18}{5}\Leftrightarrow x=\dfrac{\dfrac{-18}{5}}{4}=\dfrac{-9}{10}\left(tmđk\right)\)
vậy \(x=\dfrac{-3}{5};x=\dfrac{-9}{10}\)
1,=0 . [2017/2018+2018/2019]
=>0
2,TH1 x-3=0=>x=3
TH2 y-4=0=>y=4
3, -2/4 = -x/10 = 16/y
=>-1/2 = -x/10 = 16/y
=>-1/2 = -x/10 => -5/10 = -x/10 => x=5
-1/2 = 16/y => 16/-32 = 16/y => y = -32
a) Ta có :
\(x:4\dfrac{1}{3}=-2,5\)
\(x=-2,5\times4\dfrac{1}{3}\)
\(x=-\dfrac{65}{6}\)
b) Ta có :
\(x:\dfrac{-3}{5}=\dfrac{-10}{21}\)
\(x=\dfrac{-10}{21}\times\dfrac{-3}{5}\)
\(x=\dfrac{2}{7}\)
c)\(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}\)
\(\dfrac{2}{3}x=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}\)
\(x=\dfrac{3}{5}\)
d) \(\dfrac{1}{2}x+\dfrac{1}{2}=\dfrac{5}{2}\)
\(\dfrac{1}{2}x=2\)
\(x=2:\dfrac{1}{2}\)
\(x=4\)
~ Chúc bn học tốt ~
a, \(x:4\dfrac{1}{3}=-2,5.\)
\(x:\dfrac{13}{3}=\dfrac{-5}{2}.\)
\(x=\dfrac{-5}{2}.\dfrac{13}{3}.\)
\(x=\dfrac{-65}{6}.\)
b, \(x:\dfrac{-3}{5}=\dfrac{-10}{21}.\)
\(x=\dfrac{-10}{21}.\dfrac{-3}{5}.\)
\(x=\dfrac{30}{105}=\dfrac{2}{7}.\)
c, \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}.\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}.\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{5}{10}.\)
\(\dfrac{2}{3}x=\dfrac{3}{5}.\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}.\)
\(x=\dfrac{3}{5}.\dfrac{3}{2}.\)
\(x=\dfrac{9}{10}.\)
d, \(\dfrac{1}{2}x+\dfrac{1}{2}=\dfrac{5}{2}.\)
\(\dfrac{1}{2}x=\dfrac{5}{2}-\dfrac{1}{2}.\)
\(\dfrac{1}{2}x=2.\)
\(x=2:\dfrac{1}{2}.\)
\(x=2.2.\)
\(x=4.\)
~ Chúc bn học tốt!!! ~
Bài mik đúng thì nhớ tick mik nha!!!
Để a nguyên \(\Leftrightarrow\) 3x + 7 \(⋮\) x - 1
\(\Rightarrow\) 3x + 7 \(⋮\) x - 1
\(\Rightarrow\) 3x - 3 + 10 \(⋮\) x - 1
\(\Rightarrow\) 3(x-1)+10\(⋮\) x-1
\(\Rightarrow\) 10 \(⋮\) x-1
| x-1 | -10 | 10 | -1 | 1 |
| x | -9 | 11 | 0 | 2 |
Vậy x \(\in\){-9 ; 11 ; 0 ; 2 }
a, Giải:
Ta có: \(\dfrac{x}{5}=\dfrac{y}{7}\Rightarrow\dfrac{7}{5}x=y\)
\(x-y=-10\)
\(\Rightarrow x-\dfrac{7}{5}x=-10\)
\(\Rightarrow\dfrac{-2}{5}x=-10\)
\(\Rightarrow x=25\Rightarrow y=35\)
Vậy x = 25, y = 35
b, Giải:
Ta có: \(3x=4y\Rightarrow\dfrac{3}{4}x=y\)
\(y+x=14\)
\(\Rightarrow\dfrac{3}{4}x+x=14\)
\(\Rightarrow\dfrac{7}{4}x=14\)
\(\Rightarrow x=8\Rightarrow y=6\)
Vậy x = 8, y = 6
c, Ta có: \(\dfrac{4}{x}=\dfrac{2}{y}\Rightarrow\dfrac{x}{4}=\dfrac{y}{2}\Rightarrow x=2y\)
\(2x-y=12\)
\(\Rightarrow4y-y=12\)
\(\Rightarrow3y=12\)
\(\Rightarrow y=4\)
\(\Rightarrow x=8\)
Vậy x = 8, y = 4
a) Ta có: \(\dfrac{x}{5}=\dfrac{y}{7}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=7k\end{matrix}\right.\)
Do \(x-y=-10\Leftrightarrow5k-7k=-10\)
\(\Leftrightarrow\left(5-7\right)k=-10\)
\(\Leftrightarrow\left(-2\right)k=-10\)
\(\Leftrightarrow k=\left(-10\right):\left(-2\right)=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k=5.5=25\\y=7k=7.5=35\end{matrix}\right.\)
b) Xét \(3x=4y\Leftrightarrow\dfrac{x}{4}=\dfrac{y}{3}=m\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4m\\y=3m\end{matrix}\right.\)
Do \(y+x=14\Leftrightarrow3m+4m=14\)
\(\Leftrightarrow\left(3+4\right)m=14\)
\(\Leftrightarrow7m=14\)
\(\Leftrightarrow m=14:7=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=4m=4.2=8\\y=3m=3.2=6\end{matrix}\right.\)
c) Ta có \(\dfrac{4}{x}=\dfrac{2}{y}=n\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{n}\\y=\dfrac{2}{n}\end{matrix}\right.\)
Do \(2x-y=12\Leftrightarrow2.\dfrac{4}{n}-\dfrac{2}{n}=12\)
\(\Leftrightarrow\dfrac{8}{n}-\dfrac{2}{n}=12\)
\(\Leftrightarrow\dfrac{6}{n}=12\)
\(\Leftrightarrow n=\dfrac{6}{12}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{n}=\dfrac{4}{\dfrac{1}{2}}=8\\y=\dfrac{2}{n}=\dfrac{2}{\dfrac{1}{2}}=4\end{matrix}\right.\)
Bài 1:
a) \(\dfrac{x^2}{6}=\dfrac{24}{25}\)
\(\Leftrightarrow x^2.25=6.24\)
\(\Leftrightarrow x^2.25=144\)
\(\Leftrightarrow x^2=144:25\)
\(\Leftrightarrow x^2=5,76\)
\(\Leftrightarrow x=2,4\)
b) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
\(\Leftrightarrow7\left(x-1\right)=6\left(x+5\right)\)
\(\Leftrightarrow7x-7=6x+30\)
\(\Leftrightarrow7x=6x+30+7\)
\(\Leftrightarrow7x=6x+37\)
\(\Leftrightarrow7x-6x=37\)
\(\Leftrightarrow x=37\)
c) \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x-2\right).x+\left(x-2\right).7=\left(x+4\right).x-\left(x+4\right)\)
\(\Leftrightarrow x^2-2x+7x-14=x^2+4x-x-4\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow x^2+5x-14+4-3x-x^2=0\)
\(\Leftrightarrow\left(x^2-x^2\right)+\left(5x-3x\right)-\left(14-4\right)=0\)
\(\Leftrightarrow2x-10=0\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=10:2=5\)
Bài 2:
\(\dfrac{x}{7}=\dfrac{y}{13}\) và \(x+y=40\)
Ta có: \(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)
Do đó \(\left\{{}\begin{matrix}\dfrac{x}{7}=2\Rightarrow x=14\\\dfrac{y}{13}=2\Rightarrow y=26\end{matrix}\right.\)
Vậy \(x=14;y=26\)