\(n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\)

PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

_____0,1_____0,2_______0,1 (mol)

a, \(\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,1.22,4}{6,72}.100\%\approx33,33\%\\\%V_{CH_4}\approx66,67\%\end{matrix}\right.\)

b, \(n_{CH_4}=\dfrac{6,72}{22,4}-0,1=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,2.16}{0,2.16+0,1.26}.100\%\approx55,17\%\\\%m_{C_2H_2}\approx44,83\%\end{matrix}\right.\)

c, \(m_{C_2H_2Br_4}=0,1.346=34,6\left(g\right)\)

Để hệ có nghiệm duy nhất thì \(\dfrac{2}{a+2}\ne\dfrac{a-2}{-2}\)

=>\(\left(a+2\right)\left(a-2\right)\ne-4\)

=>\(a^2\ne0\)

=>\(a\ne0\)

\(\left\{{}\begin{matrix}2x+\left(a-2\right)y=a+1\\\left(a+2\right)x-2y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(2a+4\right)x+\left(a^2-4\right)y=\left(a+1\right)\left(a+2\right)\\\left(2a+4\right)x-4y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a^2y=a^2+3a+2-6=a^2+3a-4\\2x+\left(a-2\right)y=a+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{a^2+3a-4}{a^2}\\2x=a+1-\dfrac{\left(a-2\right)\left(a^2+3a-4\right)}{a^2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{a^2+3a-4}{a^2}\\2x=\dfrac{a^3+a^2-a^3-3a^2+4a+2a^2+6a-8}{a^2}=\dfrac{10a-8}{a^2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5a-4}{a^2}\\y=\dfrac{a^2+3a-4}{a^2}\end{matrix}\right.\)

\(x+y=\dfrac{a^2+3a-4+5a-4}{a^2}=\dfrac{a^2+8a-8}{a^2}\)

\(=1+\dfrac{8}{a}-\dfrac{8}{a^2}\)

\(=-8\left(\dfrac{1}{a^2}-\dfrac{1}{a}-\dfrac{1}{8}\right)\)

\(=-8\left(\dfrac{1}{a^2}-2\cdot\dfrac{1}{a}\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{3}{8}\right)\)

\(=-8\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+3< =3\forall a\ne0\)
Dấu '=' xảy ra khi a=2

\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)

\(=\left(\dfrac{-\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{2}\right)\)

\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\)

\(=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(5-2\right)=-3\)

a: Xét tứ giác BCEF có \(\widehat{BFC}=\widehat{BEC}=90^0\)

nên BCEF là tứ giác nội tiếp

a: Xét tứ giác ADHE có \(\widehat{ADH}+\widehat{AEH}=90^0+90^0=180^0\)

nên ADHE là tứ giác nội tiếp

Xét tứ giác BEDC có \(\widehat{BEC}=\widehat{BDC}=90^0\)

nên BEDC là tứ giác nội tiếp

b: Xét ΔADB vuông tại D và ΔAEC vuông tại E có

\(\widehat{DAB}\) chung

Do đó: ΔADB~ΔAEC

=>\(\dfrac{AD}{AE}=\dfrac{AB}{AC}\)

=>\(AD\cdot AC=AB\cdot AE\)

help meeeeee

Pt hoành độ giao điểm:

\(\dfrac{1}{2}x^2=2x-m+1\Leftrightarrow x^2-4x+2m-2=0\) (1)

(d) cắt (P) tại 2 điểm pb nằm về 2 phía trục tung khi và chỉ khi (1) có 2 nghiệm pb trái dấu

\(\Leftrightarrow x_1x_2=2m-2< 0\)

\(\Leftrightarrow m< 1\)

43

43 là sao ạ