x = 0 nhé

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TL: x = 0

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Đào Nguyễn Thùy Dương à

đặt x/8=y/5=z/12=k

suy ra x=8k,y=5k,z=12k

-x+y-z=60

-8k+5k-12k=60

k(-8+5-12)=60

k=-4

thay k=-4 vào x=8k,y=5k,z=12k

x=8.-4=-32

y=5.-4=-20

z=12.-4=-48

ta có:x/7=y/4=z/2=3z/6=x-3z/7-6=9

suy ra x/7=9 suy ra x=63

y/4=9 suy ra y=36

z/2=9 suy ra z=18

\(\frac{x}{9}=\frac{y}{5}=\frac{z}{10}\)\(=\frac{x-y+z}{9-5+10}\)\(=5\)

---> x = 9.5 = 45

---> y = 5.5 = 25

---> z = 10.5 = 50

học tốt nhoa bạn

              Áp dụng tính chất dãy tỉ số bằng nhau ta có:

           \(\frac{x}{9}=\frac{y}{5}=\frac{z}{10}=\frac{x-y+z}{9-5+10}=\frac{70}{14}=5\)

           \(\frac{x}{9}=5\Rightarrow x=45\) 

          \(\frac{y}{5}=5\Rightarrow y=25\) 

           \(\frac{z}{10}=5\Rightarrow z=50\)

          Vậy x = 45; y = 25; z = 50

ta có : \(\hept{\begin{cases}\widehat{mOx}=\widehat{mOy}\\\widehat{m'Ot}=180^0-90^0-\widehat{mOy}\\\widehat{m'Oz}=180^0-90^0-\widehat{mOx}\end{cases}}\Rightarrow\widehat{m'Ot}=\widehat{m'Oz}\) Vậy m'O là tia phân giac của tOz

b. ta có : \(\widehat{mOz}=\widehat{mOx}+\widehat{xOz}=50^0+90^0=140^0\)

\(\widehat{m'Oy}=\widehat{m'Ot}+\widehat{tOy}=40^0+90^0=130^0\)

vậy \(\widehat{mOz}>\widehat{m'Oy}\)

\(B=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2018^2}\right)\)

\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}.....\frac{2018^2-1}{2018^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{2017.2019}{2018.2018}\)

\(=\frac{\left(1.2.3.....2017\right).\left(3.4.5.....2019\right)}{\left(2.3.4.....2018\right).\left(2.3.4.....2018\right)}\)

\(=\frac{1.2019}{2018.2}=\frac{2019}{4036}\)