Ta có:\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\) ( tổng 4 góc của tứ giác ) 

Mà \(\widehat{C}=3\widehat{D}\)

\(\Rightarrow30^o+130^o+3\widehat{D}+\widehat{D}=360^o\)

\(\Rightarrow4\widehat{D}=200^o\)

\(\Rightarrow\widehat{D}=50^o\)

\(\Rightarrow\widehat{C}=3\widehat{D}=3.50^o=150^o\)

\(D=\dfrac{1-x-\sqrt{x^3}}{\sqrt{x}+1}\) \(\left(x\ge0\right)\)

\(D=\dfrac{1-x-x\sqrt{x}}{\sqrt{x}+1}\)

\(D=\dfrac{1-x\left(1+\sqrt{x}\right)}{\sqrt{x}+1}\)

\(D=-\dfrac{x\left(1+\sqrt{x}\right)}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}+1}\)

\(D=-x+\dfrac{1}{\sqrt{x}+1}\)

ABCD là hình thang `=>AB////CD`  ( cả 3 hình )

`a.`\(\widehat{A}+\widehat{B}=180^o\) ( trong cùng phía )

\(\Rightarrow x=180^o-80^o=100^o\)

\(\widehat{C}+\widehat{B}=180^o\) ( trong cùng phía )

\(\Rightarrow y=180^o-40^o=140^o\)

`b.`\(\widehat{A}=\widehat{D}=70^o\) ( đồng vị )

\(\Rightarrow x=70^o\)

\(\widehat{B}=\widehat{C}=50^o\) ( so le trong )

\(\Rightarrow y=50^o\)

`c.`\(\widehat{B}=x=180^o-\widehat{C}=180^o-90^o=90^o\)

\(\widehat{A}+\widehat{D}=180^o\)

\(\Rightarrow y=180^o-65^o=115^o\)

`a.` `yxx3+yxx2=6665`

       `=>yxx5=6665`

         `=>`y=6665/5`

         `=>y=1333`

`b.` `yxx9-yxx5=9484`

       `=>yxxx4=9484`

       `=>y=2371`

`a.`\(A=\dfrac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\dfrac{x}{x+2\sqrt{x}}+\dfrac{x}{\sqrt{x}+2}\right)\) \(\left(x>0\right)\)

\(A=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{x}{\sqrt{x}+2}\right)\)

\(A=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\dfrac{x+x\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\)

\(A=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}.\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x\left(1+\sqrt{x}\right)}\)

\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

`b.`\(A\ge\dfrac{1}{3\sqrt{x}}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}\ge\dfrac{1}{3\sqrt{x}}\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+2\right)\le3\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}+2\le3\)

\(\Leftrightarrow\sqrt{x}\le1\)

\(\Leftrightarrow x\le1\)

Vậy \(S=\left\{x|0< x\le1\right\}\)

`a.` \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) 

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1                                   0,1     ( mol )

\(Fe_xO_y+2yHCl\rightarrow xFeCl_{\dfrac{2y}{x}}+yH_2O\)

\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

\(\Rightarrow m_{Fe_xO_y}=6,4-5,6=0,8\left(g\right)\)

Ta có: 3,2g hh + H₂ `->` 0,1g H₂O

            \(\Rightarrow\) 6,4g hh + H₂  `->` 0,2g H₂O

\(n_{H_2O}=\dfrac{0,2}{18}=\dfrac{1}{90}\left(mol\right)\)

\(Fe_xO_y+yH_2\rightarrow\left(t^o\right)xFe+yH_2O\)

\(\Rightarrow n_{O\left(Fe_xO_y\right)}=n_{H_2O}=\dfrac{1}{90}\left(mol\right)\)

Ta có:\(m_{Fe_xO_y}=56x+16.\dfrac{1}{90}=0,8\)

\(\Leftrightarrow x=\dfrac{1}{90}\)

\(\Rightarrow x:y=\dfrac{1}{90}:\dfrac{1}{90}=1:1\)

\(\Rightarrow CTHH:FeO\)

`a.`Xét tam giác ABC vuông A, đcao AH:

\(AC^2=\sqrt{BC^2-AB^2}=\sqrt{20^2-12^2}=16\left(cm\right)\)

\(AH=\sqrt{\dfrac{AB^2.AC^2}{AB^2+AC^2}}=\sqrt{\dfrac{12^2.16^2}{12^2+16^2}}=9,6\left(cm\right)\)

\(\sin B=\dfrac{AC}{BC}=\dfrac{16}{20}=\dfrac{4}{5}\)  \(\Rightarrow\widehat{B}\simeq53^o\)

`b.`Xét tam giác AHC vuông H, đcao HF:

\(AH^2=AF.AC\) (1)

Xét tam giác AHB vuông H, đcao HE:

\(AH^2=AE.AB\) (2)

\(\left(1\right);\left(2\right)\Rightarrow AF.AC=AE.AB\)

\(\Leftrightarrow AF=AE.\dfrac{AB}{AC}\)

\(\Leftrightarrow AF=AE.\tan C\) ( vì \(\tan C=\dfrac{AB}{AC}\) )

Tóm tắt: 

\(R_1=20\Omega\)   \(R_2=30\Omega\)    \(U=90V\)   

\(R_{tđ}=?\)           \(I=?\)

Giải:

`a.` \(R_{tđ}=R_1+R_2=20+30=50\Omega\)

`b.` \(I=\dfrac{U}{R_{tđ}}=\dfrac{90}{50}=1,8A\)

Vì \(R_1ntR_2\) \(\Rightarrow I=I_1=I_2=1,8A\)

`c.` Ptr đường thẳng (d) có dạng: \(y=ax+b\left(a\ne0\right)\)

Tọa đọa giao điểm (d₃) và (d₄):

\(x-2=3x+4\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\) (1)

\(\Rightarrow y=-3-2=-5\) (2)

Ta có: \(\left(d\right)\perp\left(d_2\right):y=-\dfrac{x}{2}-1\) 

\(\Rightarrow a.-\dfrac{1}{2}=-1\)

\(\Rightarrow a=2\) (3)

\(\left(1\right);\left(2\right);\left(3\right)\Rightarrow\left(d\right):-5=2.\left(-3\right)+b\)

                                    \(\Leftrightarrow b=1\)

\(\Rightarrow\left(d\right):y=2x+1\)