Đặt \(k=\dfrac{x}{2}=\dfrac{y}{2}=\dfrac{z}{-3}\)

\(\Rightarrow x=2k;y=2k;z=-3k\)

Ta thay \(x=2k;y=2k;z=-3k\) vào \(xyz=240\)

\(\Rightarrow\left(2k\right)\left(2k\right)\left(-3k\right)=240\)

\(\Rightarrow\left[2.2.\left(-3\right)\right]\left(k.k.k\right)=240\)

\(\Rightarrow\left(-12k^3\right)=240\)

\(\Rightarrow k=-\sqrt[3]{20}\)

Vậy \(\left\{{}\begin{matrix}x=y=-2\sqrt[3]{20}\\z=3\sqrt[3]{20}\end{matrix}\right.\)

\(x-y-z=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y+z\\y=x-z\\z=x-y\end{matrix}\right.\) 

Lại có :\(C=\left(\dfrac{1-z}{x}\right)\left(\dfrac{1-x}{y}\right)\left(\dfrac{1+y}{z}\right)\)

\(C=\left(\dfrac{x-z}{x}\right)\left(\dfrac{y-x}{y}\right)\left(\dfrac{z+y}{z}\right)\)

\(C=\dfrac{y}{x}\cdot\) \(\left(-\dfrac{z}{y}\right)\) \(\dfrac{x}{z}\)

\(C=\dfrac{-xyz}{xyz}=-1\)

\(\dfrac{x}{3}=\dfrac{y}{5}\) \(\Leftrightarrow x=\dfrac{3}{5}y\)

Thay vào B có

\(B=\dfrac{5x^2+3y^2}{10x^2-3y^2}=\dfrac{5\cdot\left(\dfrac{3}{5}y\right)^2+3y^2}{10\cdot\left(\dfrac{3}{5}y\right)^2-3y^2}=\dfrac{5\cdot\dfrac{9}{25}y^2+3y^2}{10\cdot\dfrac{9}{25}y^2-3y^2}\)

\(=\dfrac{\dfrac{9}{5}y^2+3y^2}{\dfrac{18}{5}y^2-3y^2}=\dfrac{y^2\left(\dfrac{9}{5}+3\right)}{y^2\left(\dfrac{18}{5}-3\right)}=\dfrac{\dfrac{9}{5}+3}{\dfrac{18}{5}-3}=\dfrac{\dfrac{24}{5}}{\dfrac{3}{5}}\)

\(=\dfrac{24}{5}\cdot\dfrac{5}{3}=8\)

\(\dfrac{2}{1.3}+\dfrac{3}{3.6}+\dfrac{6}{5.11}+\dfrac{9}{11.20}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{20}\)

\(=1-\dfrac{1}{20}\)

\(=\dfrac{19}{20}\)

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: \(\widehat{t'Oy}+\widehat{tOy}=\widehat{t'Ot}\)

\(\Rightarrow\widehat{t'Oy}=180^o-35^o=145^o\)

  \(\dfrac{-3}{7}\) + \(\dfrac{15}{4}\) - \(\dfrac{2}{13}\) - \(\dfrac{3}{7}\)

= \(\dfrac{-3}{7}\)- \(\dfrac{3}{7}\) + \(\dfrac{15}{4}\) - \(\dfrac{2}{13}\) 

=  \(\dfrac{-6}{7}\) + \(\dfrac{15}{4}\) - \(\dfrac{2}{13}\)

= \(\dfrac{-312}{364}\) + \(\dfrac{1365}{364}\) - \(\dfrac{56}{364}\)

= \(\dfrac{997}{364}\)

= \(\dfrac{19}{7}\)