1)tìm x
a)\(\dfrac{x^2}{6}=\dfrac{24}{25}\)
b)\(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
c)\(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
2)tìm 2 số x,y biết
\(\dfrac{x}{7}=\dfrac{y}{13}\)và x+y =40
3)chứng minh rằng
tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)(b,d khác 0)
ta suy ra được \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
giúp mình nhé giải rõ giùm mình mai đi học rồi!
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 \(x=14;y=26\)
Bài 3:
Vì \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\)
Nên \(ab+ad=ab+bc\Leftrightarrow a\left(b+d\right)=b\left(a+c\right)\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
