4.

\(\left(0,36\right)^8=\left(\left(0,6\right)^2\right)^8=\left(0,6\right)^{16}\)

\(\left(0,216\right)^4=\left(\left(0,6\right)^3\right)^4=\left(0,6\right)^{12}\)

5. 

a, \(\left(3\times5\right)^3=15^3=1125\)

b, \(\left(\frac{-4}{11}\right)^2=\frac{16}{121}\)

c, \(\left(0,5\right)^4\times6^4=\left(0,5\times6\right)^4=3^4=81\)

d, \(\left(\frac{-1}{3}\right)^5\div\left(\frac{1}{6}\right)^5=\left(\frac{-1}{3}\right)^5\times6^5=\left(\frac{-1}{3}\times6\right)^5=\left(-2\right)^5=-32\)

6.

a, \(\frac{6^2\times6^3}{3^5}=\frac{6^5}{3^5}=\frac{2^5\times3^5}{3^5}=2^5=32\)

b, \(\frac{25^2\times4^2}{5^5\times\left(-2\right)^5}=\frac{100^2}{\left(-10\right)^5}=\frac{10^4}{\left(-10\right)^5}=\frac{-1}{10}\)

c, Mình không nhìn rõ đề 

d, \(\left(-2\frac{3}{4}+\frac{1}{2}\right)^2=\left(\frac{-11}{4}+\frac{1}{2}\right)^2=\left(\frac{-9}{4}\right)^2=\frac{81}{16}\)

7.

a, \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\Rightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\Rightarrow m=4\)

b, \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\Rightarrow\left(\frac{3}{5}\right)^n=\left(\left(\frac{3}{5}\right)^2\right)^5\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\Rightarrow n=10\)

c, \(\left(-0,25\right)^p=\frac{1}{256}\Rightarrow\left(-0,25\right)^p=\left(\frac{1}{4}\right)^4\Rightarrow\left(-0,25\right)^p=\left(0,25\right)^4\Rightarrow p=4\)

8.

a, \(\left(\frac{2}{5}+\frac{3}{4}\right)^2=\left(\frac{23}{20}\right)^2=\frac{529}{400}\)

b, \(\left(\frac{5}{4}-\frac{1}{6}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

cam on ban nhieu

A x C B y z

Vẽ tia Cz // Ax

           Ax // By

\(\Rightarrow\)Cz // By (định lí 3 trong bài từ vuông góc đến song song)

Ta có: 

\(\widehat{ACz} = \widehat{A}\)(so le trong)

mà \(\widehat{A} = 40^O (gt)\)

\(\Rightarrow\)\(\widehat{ACz} = 40^O\)

\(\widehat{zCB} = \widehat{B}\)(so le trong)

mà \(\widehat{B} = 50^O (gt)\)

\(\Rightarrow\)\(\widehat{zCB} = 50^O\)

\(\widehat{ACB} = \widehat{ACz} +\widehat{zCB}\)(Cz nằm giữa hai tia CA và CB)

\(\widehat{ACB} = 40^O + 50^O (\widehat{ACz} = 40^O(cmt); \widehat{zCB} =50^O(cmt))\)

\(\widehat{ACB} = 90^O\)

Vậy \(\widehat{ACB} = 90^O\)

Qua C kẻ đường thẳng zt//Ax

Mà Ax//By nên zt//By

Vì zt//Ax nên góc zCA=góc xAC ( 2 góc so le trong)

Mà góc xAC=40 độ nên góc zCA=40 độ

Vì zt//By nên góc zCB=góc CBy ( 2 góc so le trong)

Mà góc CBy=50 độ nên góc zCB=50 độ

Mặt khác: góc ACB=góc zCA+góc zCB

=> góc ACB= 40 độ + 50 độ = 90 độ

Vậy góc ACB=90 độ

chẳng nhìn thấy j cả! Thông cảm mk bị cận!

Ta có \(\frac{x+1}{111}\) = \(\frac{y+2}{222}\) = \(\frac{z+3}{333}\) và 3\(x\) + 2\(y\) +\(z\) = 989

\(\Rightarrow\) \(\frac{3x+3}{333}\) = \(\frac{2y+4}{444}\) = \(\frac{z+3}{333}\) = \(\frac{3x+2+2y+4+z+3}{333+444+333}\)

= \(\frac{3x+2y+z+9}{1110}\) = \(\frac{989+9}{1110}\) = \(\frac{998}{1110}\) = \(\frac{499}{555}\)

Từ \(\frac{3x+3}{333}\) = \(\frac{499}{555}\) \(\Rightarrow\) 3\(x\) = \(\frac{499}{555}\) . 333 - 3= \(\frac{1482}{5}\) \(\Rightarrow\) \(x\) = 98,8

\(\frac{2y+4}{444}\) = \(\frac{499}{555}\) \(\Rightarrow\) 2\(y\) = \(\frac{499}{555}\) . 444 -4 = \(\frac{1976}{5}\) \(\Rightarrow\) \(y\) = 197,6

\(\frac{z+3}{333}\) = \(\frac{499}{555}\) \(\Rightarrow\) \(z+3\) = \(\frac{499}{555}\) . 333= \(\frac{1497}{5}\) \(\Rightarrow\) \(z\)= 296,4

uk

kẻ đường thẳng OK sao cho OK // a

Ta có góc A+KOA=180^o( hai góc trong cùng phía bù nhau)

=> góc KOA=180^o-110^o=70^o

=> góc KOB=140^o - 70^o = 70^o

Mà KOB+B=70^o+110^o=180^o

=> OK//b Mà OK//a; OK//b

=>a//b. tick giùm tui đi, please V_V

 khẩn cấp 

câu 1 là gì?

Xét \(\Delta ABC\) có \(\widehat{A} = 90^0\) (gt)

\(\Rightarrow\)\(\widehat{B} + \widehat{C} = 90^0\) (Định lí tam giác vuông)

mà \(\widehat{B} = \frac{1}{4}\widehat{C}\) (gt)

\(\Rightarrow\)\(\widehat{B} = 18^0\)

\(\widehat{C} = 72^0\)

Giải:

Ta có: \(\widehat{B}=\frac{1}{4}\widehat{C}\Rightarrow4\widehat{B}=\widehat{C}\)

Xét \(\Delta ABC\) có: \(\widehat{B}+\widehat{C}=90^o\) ( do \(\widehat{A}=90^o\) )

\(\Rightarrow\widehat{B}+4\widehat{B}=90^o\)

\(\Rightarrow5\widehat{B}=90^o\)

\(\Rightarrow\widehat{B}=18^o\)

\(\Rightarrow\widehat{C}=4.\widehat{B}=4.18^o=72^o\)

Vậy \(\widehat{B}=18^o,\widehat{C}=72^o\)

bạn lấy máy tính ra bấm cho nhanh