a) Ta có: BC//AD(gt)

Mà \(AD\perp AB\)(gt)

=> BC⊥AB

\(\Rightarrow\widehat{ABC}=90^0\)

b) Ta có: BC//AD(gt)

\(\Rightarrow\widehat{C_2}=\widehat{D_1}=110^0\)(đồng vị)

\(\Rightarrow\widehat{C_1}=180^0-\widehat{C_2}=180^0-110^0=70^0\)(l=kề bù)

c) Ta có: \(\widehat{BCD}=\widehat{C_2}=110^0\)(đối đỉnh)

\(\Rightarrow\widehat{BCI}=\dfrac{1}{2}\widehat{BCD}=55^0\)(do CI là phân giác góc BCD)

Ta có:BC//AD(gt)

\(\Rightarrow\widehat{BCI}+\widehat{AIC}=180^0\)(trong cùng phía)

\(\Rightarrow\widehat{AIC}=180^0-\widehat{BCI}=180^0-55^0=125^0\)

Bài 1:

\(a,\Rightarrow-30x=12\cdot25=300\Rightarrow x=-10\\ b,\Rightarrow24x=\dfrac{7}{5}\cdot12=\dfrac{84}{5}\Rightarrow x=\dfrac{7}{10}\\ c,\Rightarrow\dfrac{4}{3}\cdot\dfrac{5}{4}=\dfrac{2}{3}:\dfrac{1}{10}x\\ \Rightarrow\dfrac{5}{3}=\dfrac{20}{3}x\Rightarrow x=\dfrac{1}{4}\\ d,\Rightarrow-2,5x=-40\Rightarrow x=16\\ e,\Rightarrow x^2=100\Rightarrow\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

Bài 1 : 

a, \(=20x^5y^4\)

b, \(=\left(-2-\dfrac{1}{2}+8\right)x^5y^2=\dfrac{11}{2}x^5y^2\)

c, \(=-\dfrac{9}{2}x^2z^4\)

1) 75kg

Lỗi rồi ạ

lỗi r cj

Giups mik vs ai có câu trl mik sẽ tick cho ạ!

Bài 3: 

a: a=140

b: a=270

giúp với ,mai mik nộp r

      \(\frac{2^{10}\cdot55+2^{10}\cdot26}{28\cdot27}\)

\(=\frac{2^{10}\cdot\left(55+26\right)}{28\cdot27}\)

\(=\frac{2^{10}\cdot81}{28\cdot27}\)

\(=\frac{1024\cdot81}{756}\)

\(=\frac{20736}{189}\)

\(\text{Nếu còn rút gọn được thì bạn tự rút gọn nha !}\)