Cho x-y-z = 0
Tính C = (1-z/x) . (1-x/y) . ( 1+y/z)
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\(\dfrac{x}{3}=\dfrac{y}{5}\) \(\Leftrightarrow x=\dfrac{3}{5}y\)
Thay vào B có
\(B=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5\cdot\left(\dfrac{3}{5}y\right)^2+3y^2}{10\cdot\left(\dfrac{3}{5}y\right)^2-3y^2}=\dfrac{5\cdot\dfrac{9}{25}y^2+3y^2}{10\cdot\dfrac{9}{25}y^2-3y^2}\)
\(=\dfrac{\dfrac{9}{5}y^2+3y^2}{\dfrac{18}{5}y^2-3y^2}=\dfrac{y^2\left(\dfrac{9}{5}+3\right)}{y^2\left(\dfrac{18}{5}-3\right)}=\dfrac{\dfrac{9}{5}+3}{\dfrac{18}{5}-3}=\dfrac{\dfrac{24}{5}}{\dfrac{3}{5}}\)
\(=\dfrac{24}{5}\cdot\dfrac{5}{3}=8\)
a, \(\dfrac{7}{20}\) = \(\dfrac{8}{20}\) + (\(\dfrac{-1}{20}\))
b, \(\dfrac{7}{20}\) = \(\dfrac{1}{4}\) + \(\dfrac{1}{10}\)
a, \(\dfrac{13}{18}\) = \(\dfrac{5}{18}\) + \(\dfrac{4}{9}\)
b, \(\dfrac{13}{18}\) = \(\dfrac{14}{18}\) + ( \(\dfrac{-1}{18}\))
c, \(\dfrac{13}{18}\) = \(\dfrac{7}{18}\) + \(\dfrac{1}{3}\)
Góc xMz = tMz-xMt => xMz= 90-70 = 20 ( độ)
Góc tMy = xMy-xMt => 90-70 = 20 ( độ)
Góc zMy = xMy - zMx => 90-20 = 70 ( độ)
a,
Theo đề ra, ta có:
\(\widehat{xOt}=35^o;\widehat{xOy}=70^o\Rightarrow\widehat{xOt}< \widehat{xOy}\)
\(\Rightarrow\) Tia Ot nằm giữa tia Ox và Oy (*)
Ta có: \(\widehat{xOt}+\widehat{tOy}=\widehat{xOy}\)
\(\Rightarrow35^o+\widehat{tOy}=70^o\)
\(\Rightarrow\widehat{tOy}=35^o\) (**)
b,
Từ (*)(**) \(\Rightarrow\) Ot là tia phân giác \(\widehat{xOy}\)
c,
Theo đề ra: Ot' là tia đối của tia Ot
\(\Rightarrow Ot',Ot\) tạo thành \(\widehat{tOt'}\) (Góc bẹt)
Mà \(\widehat{tOt'}=180^o;\widehat{tOy}=35^o\Rightarrow\widehat{tOt'}>\widehat{tOy}\)
\(\Rightarrow\) Tia Oy nằm giữa tia Ot' và tia Ot
Ta có: \(\widehat{t'Oy}+\widehat{tOy}=\widehat{t'Ot}\)
\(\Rightarrow\widehat{t'Oy}=180^o-35^o=145^o\)
\(\dfrac{-3}{7}\) + \(\dfrac{15}{4}\) - \(\dfrac{2}{13}\) - \(\dfrac{3}{7}\)
= \(\dfrac{-3}{7}\)- \(\dfrac{3}{7}\) + \(\dfrac{15}{4}\) - \(\dfrac{2}{13}\)
= \(\dfrac{-6}{7}\) + \(\dfrac{15}{4}\) - \(\dfrac{2}{13}\)
= \(\dfrac{-312}{364}\) + \(\dfrac{1365}{364}\) - \(\dfrac{56}{364}\)
= \(\dfrac{997}{364}\)
= \(\dfrac{19}{7}\)
\(a)x-\dfrac{3}{15}=\dfrac{2}{5}\)
\(\Rightarrow x=\dfrac{3}{5}\)
\(b)\dfrac{3}{5}-\dfrac{1}{2}\left(x+\dfrac{3}{4}\right)=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}\left(x+\dfrac{3}{4}\right)=\dfrac{1}{10}\)
\(\Rightarrow x+\dfrac{3}{4}=\dfrac{1}{5}\)
\(\Rightarrow x=-\dfrac{11}{20}\)
\(c)\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)+\left(\dfrac{3}{2}x+\dfrac{5}{3}\right)=1\)
\(\Rightarrow\left(\dfrac{1}{2}x+\dfrac{3}{2}x\right)-\dfrac{1}{3}+\dfrac{5}{3}=1\)
\(\Rightarrow2x+\dfrac{4}{3}=1\)
\(\Rightarrow2x=-\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{1}{6}\)
\(d)\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)-\left(\dfrac{1}{4}x+\dfrac{4}{3}\right)=0\)
\(\Rightarrow\left(\dfrac{3}{4}x-\dfrac{1}{4}x\right)+\dfrac{1}{2}-\dfrac{4}{3}=0\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{3}{6}-\dfrac{8}{6}=0\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{5}{6}=0\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{3}\)
\(x-y-z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y+z\\y=x-z\\z=x-y\end{matrix}\right.\)
Lại có :\(C=\left(\dfrac{1-z}{x}\right)\left(\dfrac{1-x}{y}\right)\left(\dfrac{1+y}{z}\right)\)
\(C=\left(\dfrac{x-z}{x}\right)\left(\dfrac{y-x}{y}\right)\left(\dfrac{z+y}{z}\right)\)
\(C=\dfrac{y}{x}\cdot\) \(\left(-\dfrac{z}{y}\right)\) \(\dfrac{x}{z}\)
\(C=\dfrac{-xyz}{xyz}=-1\)