a) Ta có :\(\hept{\begin{cases}a\perp c\\b\perp c\end{cases}}\Rightarrow a//b\)

b) Ta có : \(\widehat{C_1}=\widehat{C_2}=50^{\text{o}}\)(2 góc đối đỉnh) 

mà \(\widehat{C_1}+\widehat{D_1}=180^{\text{o}}\)(a//b)

=> \(\widehat{D_1}=130^{\text{o}}\)

=> \(\widehat{D_2}=50^{\text{o}}\)

c) Ta có \(\widehat{ODM}+\widehat{D_1}=180^{\text{o}}\)

=> \(\widehat{ODM}=50^{\text{o}}\)

mà \(\hept{\begin{cases}\widehat{ODM}+\widehat{OMD}=100^{\text{o}}\\\widehat{OMD}+\widehat{M_1}=180^{\text{o}}\end{cases}}\Rightarrow\widehat{M_1}-\widehat{ODM}=80^{\text{o}}\Rightarrow\widehat{M_1}=130^{\text{o}}\)

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

\(\frac{x}{5}=\frac{y}{6}=\frac{z}{7}=\frac{x-y+z}{5-6+7}=\frac{36}{6}=6\)

\(\Rightarrow\hept{\begin{cases}x=6.5=30\\y=6.6=36\\z=6.7=42\end{cases}}\)

Đặt 4x = 6y = 9z = k

\(\Rightarrow\hept{\begin{cases}x=\frac{k}{4}\\y=\frac{k}{6}\\z=\frac{k}{9}\end{cases}}\)

Theo bài ta có :

\(\frac{k}{4}+\frac{k}{6}-\frac{k}{9}=22\)

\(\Leftrightarrow\frac{9k+6k-4k}{36}=\frac{792}{36}\)

\(\Leftrightarrow11k=792\Leftrightarrow k=72\)

\(\Rightarrow\hept{\begin{cases}x=\frac{72}{4}=18\\y=\frac{72}{6}=12\\z=\frac{72}{9}=8\end{cases}}\)

TL

d = \(\frac{\left(a-1\right)c+a}{a}\)

HT

TL:

d = \(\frac{\left(a-1\right)c+a}{a}\)

~HT~

\(a)\)\(\frac{x}{0,9}=\frac{5}{6}\)

\(\Leftrightarrow\)\(x\times6=5\times0,9\)

\(\Leftrightarrow\)\(x\times6=4,5\)

\(\Leftrightarrow\)\(x=4,5\div6\)

\(\Leftrightarrow\)\(x=0,75\)

\(b)\)\(\frac{-6}{x}=\frac{9}{-15}\)

\(\Leftrightarrow\)\(x\times9=\left(-15\right)\times\left(-6\right)\)

\(\Leftrightarrow\)\(x\times9=90\)

\(\Leftrightarrow\)\(x=90\div9\)

\(\Leftrightarrow\)\(x=10\)

SORY TÔI MỚI LỚP BA

2/3-(1\6+3/8)

\(\frac{x}{2}=\frac{y}{3}\)

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

\(\frac{x}{2}=\frac{y}{3}=\frac{7x-2y}{7.2-2.3}=\frac{40}{8}=5\)

\(\Leftrightarrow\hept{\begin{cases}x=5.2=10\\y=5.3=15\end{cases}}\)