kb nha

em kết bạn với nha.

Ta có;

\(\widehat{xOt}-\widehat{xOz}=40độ\)

Mà \(\widehat{xOt}+\widehat{xOz}=180độ\)

=>\(\widehat{xOt}-\widehat{xOz}+\widehat{xOt}+\widehat{xOz}=40độ+180độ\)

                                          \(2\widehat{xOt}=220độ\)

                                            \(\widehat{xOt}=110độ\)

=>\(\widehat{zOy}=110độ\)

Vậy \(\widehat{xOt}=\widehat{zOy}=110độ\)

Thanks bạn

\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)

\(x=\left(\frac{3}{7}\right)^7:\left(\frac{3}{5}\right)^2\)

\(x=\frac{3^7}{7^7}:\frac{3^2}{5^2}\)

\(x=\frac{3^7.5^2}{7^7.3^2}\)

\(x=\frac{3^5.5^2}{7^7}=\frac{6075}{823543}\)

c) Theo bài ra: \(\frac{2x}{3}=\frac{y}{5}\Rightarrow\frac{x}{\frac{3}{2}}=\frac{5y}{25}\)

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

\(\frac{x}{\frac{3}{2}}=\frac{5y}{25}=\frac{x+5y}{\frac{3}{2}+25}=\frac{3}{\frac{53}{2}}=\frac{6}{53}\)

\(\Rightarrow x=\frac{9}{53};y=\frac{30}{53}\)

a) Ta có: \(\frac{x}{-5}=\frac{2y}{3}\Rightarrow\frac{x}{-5}=\frac{y}{\frac{3}{2}}\Rightarrow\frac{x}{-5}=\)\(\frac{5y}{\frac{15}{2}}\)

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

\(\frac{x}{-5}=\frac{5y}{\frac{15}{2}}=\frac{x+5y}{\left(-5\right)+\frac{15}{2}}=\)\(\frac{-1}{\frac{5}{2}}=\frac{-2}{5}\)

\(\Rightarrow x=2;y=-0,6\)

a)    (a+b)-(-c+a+b)

=      a+b+c-a-b

= c

b)     -(x+y)+(-z+x+y)

=       -x-y-z+x+y

=       x

c)      ( m-n+p)+(-m+n+p)

=       m-n+p-m+n+p

=       p+p

=      2p

a,a+b+c-a-b

= c

b,= -x-y-z+x+y

= -z

c,= m-n+p-m+n+p

= 2p