6: Qua C, kẻ tia CM nằm giữa hai tia CA và CD sao cho CM//DE//AB

CM//DE

=>\(\hat{MCD}=\hat{CDE}\) (hai góc so le trong)

=>\(\hat{MCD}=60^0\)

Ta có: tia CM nằm giữa hai tia CA và CD

=>\(\hat{ACM}+\hat{DCM}=\hat{ACD}\)

=>\(\hat{ACM}=110^0-60^0=50^0\)

Ta có: CM//AB

=>\(\hat{BAC}=\hat{ACM}\) (hai góc so le trong)

=>\(\hat{BAC}=50^0\)

BÀi 5:

Qua B, kẻ tia BM nằm giữa hai tia BA và BC sao cho BM//Ax

BM//Ax

=>\(\hat{xAB}+\hat{ABM}=180^0\) (hai góc trong cùng phía)

=>\(\hat{ABM}=180^0-120^0=60^0\)

Ta có: tia BM nằm giữa hai tia BA và BC

=>\(\hat{ABM}+\hat{CBM}=\hat{ABC}\)

=>\(\hat{CBM}=140^0-60^0=80^0\)

Ta có: \(\hat{CBM}+\hat{BCy}=80^0+100^0=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên BM//Cy

mà BM//Ax

nên Ax//Cy

Câu 1:

Cacbon: C

Sắt: Fe

Oxi: O2

Magie: Mg

Photpho : P

Canxi: Ca

Hidro: H2

Natri: Na

Lưu huỳnh: S

Clo: Cl2

cảm ơn nhiều

⇒ Đáp án: C. Nghề trồng rau

Mìnk nghĩ lè C.nghề trồng rau

Câu 3 : 

\(n_{SO_3}=\dfrac{8}{80}=0.1\left(mol\right)\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(0.1....................0.1\)

b) Cho quỳ tím vào => quỳ tím hóa đỏ

\(m_{H_2SO_4}=0.1\cdot98=9.8\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{0.1}{0.25}=0.4\left(M\right)\)

Câu 2 : 

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe_2O_3+3H_2\underrightarrow{^{t^0}}2Fe+3H_2O\)

\(0.1.........0.3.........0.2\)

\(m_{Fe_2O_3}=0.1\cdot160=16\left(g\right)\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

a) Ta có: \(\widehat{BEC}\) là góc ngoài tại đỉnh E của ΔABE

nên \(\widehat{BEC}=\widehat{A}+\widehat{ABE}=90^0+\widehat{ABE}>90^0\)

hay \(\widehat{BEC}\) là góc tù

b) \(\widehat{BEA}=180^0-110^0=70^0\)

\(\Leftrightarrow\widehat{ABE}=20^0\)

\(\Leftrightarrow\widehat{ABC}=40^0\)

\(\Leftrightarrow\widehat{ACB}=50^0\)

\(\text{ #ID 072019}\)

\(205m^2=2dam^25m^2\\ 408dam^2=4hm^28dam^2\\ 2006dam^2=20hm^2600m^2\)

g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)

h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)

\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)

\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)