Bài toán được mô tả như hình sau:

Sửa đề: chuyển cm thành m

Độ dài đáy bé được mở rộng thêm là:

\(600:\dfrac{1}{2}:30=40\left(m\right)\)

Độ dài đáy bé của thửa ruộng là:

\(160-40=120\left(m\right)\)

Diện tích thửa ruộng là:

\(\dfrac{\left(160+120\right)\times30}{2}=4200\left(m^2\right)=0,42\left(ha\right)\)

Năm 2023 số tiền hoa anh Nam thu được trên thửa ruộng đó là:

\(0,42\times650000000=273000000\) (đồng)

Số Lan dùng để nhân với 9 là:

   \(1242:6=207\)

Tích đúng là:

   \(207\times9=1863\)

Hình minh họa:))

\(tan\alpha=\dfrac{CD}{BC}=\dfrac{3}{\sqrt{3}}\Rightarrow\alpha=60^o\)

Vậy góc giữa đg chéo và cạnh ngắn hơn của hcn là 60^o

13. \(sin^2\alpha+cos^2\alpha=1\Rightarrow sin^2\alpha=1-cos^2\alpha=1-0,8^2=\dfrac{9}{25}\Rightarrow sin\alpha=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)

\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{3}{5}:0,8=\dfrac{3}{4}\)

\(cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{4}{3}\)

14. Ta có \(sin^2\alpha+cos^2\alpha=1\Rightarrow\left(\dfrac{5}{13}\right)^2+cos^2\alpha=1\Rightarrow cos^2\alpha=1-\left(\dfrac{5}{13}\right)^2=\dfrac{144}{169}\Rightarrow cos\alpha=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}\)

\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{5}{13}:\dfrac{12}{13}=\dfrac{5}{12}\)

\(cot\alpha=\dfrac{1}{tan\alpha}=1:\dfrac{5}{12}=\dfrac{12}{5}\)

15. \(A=sin^210^o+sin^220^o+...sin^260^o+sin^280^o=sin^210^o+sin^220^o+sin^230^o+sin^240^o+cos^240^o+cos^230^o+cos^220^o+cos^210^o=\left(sin^210^o+cos^210^o\right)+\left(sin^220^o+cos^220^o\right)+\left(sin^230^o+cos^230^o\right)+\left(sin^240^o+cos^240^o\right)=1+1+1+1=4\)

\(B=cos^215^o-cos^225^o+cos^235^o-cos^245^o+cos^255^o-cos^265^o+cos^275^o=cos^215^o-sin^265^o+cos^235^o-cos^245^o+sin^235^o-cos^265^o+sin^215^o=\left(sin^215^o+cos^215^o\right)+\left(sin^235^o+cos^235^o\right)-\left(sin^265^o+cos^265^o\right)-cos^245^o=1+1-1-cos^245^o=1-\left(\dfrac{\sqrt{2}}{2}\right)^2=1-\dfrac{1}{2}=\dfrac{1}{2}\)\(C=tan20^o.tan40^o.tan50^o.tan60^o.tan70^o=tan20^o.tan40^o.tan50^o.cot40^o.cot20^o=\left(tan20^o.cot20^o\right)\left(tan40^o.cot40^o\right)tan50^o=1.1.tan50^o=tan50^o\simeq1,19\)

Ta có: \(22\equiv1\left(\text{mod }3\right)\Rightarrow22^{22}\equiv1^{22}\equiv1\left(\text{mod }3\right)\)

\(\Rightarrow22^{22}-1\equiv0\left(\text{mod }3\right)\)

hay \(22^{22}-\text{1 } \vdots \text{ 3}\)

10. \(A=sin25^o+cos25^o-sin65^o-cos65^o=sin25^o+sin\left(90^o-25^o\right)-sin65^o-sin\left(90^o-65^o\right)=sin25^o+sin65^o-sin65^o-sin25^o=0\)

11. Áp dụng tỉ số lượng giác vào \(\Delta ABH\) vuông tại H: \(tan\alpha=\dfrac{AH}{BH}=\dfrac{2}{5}\Rightarrow\alpha\simeq21^o\)

15: \(\left\{{}\begin{matrix}x-y=4\\-2x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-2y=8\\-2x+2y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-2y-2x+2y=8+1\\2x-2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0x=9\\x-y=4\end{matrix}\right.\)

=>\(\left(x,y\right)\in\varnothing\)

16: Vì \(\dfrac{\dfrac{1}{2}}{-1}=\dfrac{1}{-2}\ne\dfrac{-2}{1}\)

nên hệ phương trình vô nghiệm

17: Vì \(\dfrac{10}{2}=\dfrac{-5}{-1}\ne\dfrac{5}{2}\)

nên hệ phương trình vô nghiệm

18: Vì \(\dfrac{1}{-2}=\dfrac{-2}{4}=\dfrac{4}{-8}\left(=-\dfrac{1}{2}\right)\)

nên hệ có vô nghiệm có dạng như sau:

\(\left\{{}\begin{matrix}x-2y=4\\-2x+4y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y=4\\x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0y=0\\x=2y+4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y\in R\\x=2y+4\end{matrix}\right.\)

19: \(\left\{{}\begin{matrix}4x-2y=\dfrac{1}{2}\\2x-y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=0,25\\2x-y=0,25\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}0x=0\\y=2x-0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in R\\y=2x-0,25\end{matrix}\right.\)

20: Vì \(\dfrac{1}{2}=\dfrac{-3}{-6}=\dfrac{\sqrt{2}}{2\sqrt{2}}\left(=\dfrac{1}{2}\right)\)

nên hệ phương trình có vô số nghiệm 

\(\left\{{}\begin{matrix}x-3y=\sqrt{2}\\2x-6y=2\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3y=\sqrt{2}\\x-3y=\sqrt{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}0y=0\\x-3y=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\in R\\x=3y+\sqrt{2}\end{matrix}\right.\)

21: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\-5x+8y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2y=6\\-5x+8y=-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x-8y=24\\-5x+8y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=21\\3x-2y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3\\2y=3x-6=3\cdot3-6=9-6=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=\dfrac{3}{2}\end{matrix}\right.\)

22: \(\left\{{}\begin{matrix}\dfrac{x-1}{6}+\dfrac{2x-y}{4}=1\\\dfrac{x+y}{2}-\dfrac{y-x-1}{3}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)+3\left(2x-y\right)=12\\3\left(x+y\right)-2\left(y-x-1\right)=12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-2+6x-3y=12\\3x+3y-2y+2x+2=12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}8x-3y=14\\5x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-3y=14\\15x+3y=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23x=44\\5x+y=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{44}{23}\\y=10-5x=10-5\cdot\dfrac{44}{23}=\dfrac{10}{23}\end{matrix}\right.\)

23: 

\(\left\{{}\begin{matrix}3\left(x+1\right)+2\left(x+2y\right)=4\\4\left(x+1\right)-\left(x+2y\right)=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3\left(x+1\right)+2\left(x+2y\right)=4\\8\left(x+1\right)-2\left(x+2y\right)=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11\left(x+1\right)=22\\4\left(x+1\right)-\left(x+2y\right)=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+1=2\\x+2y=4\cdot2-9=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\2y=-1-x=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

24: \(\left\{{}\begin{matrix}2\left(x+y\right)-\left(x-y\right)=1\\\left(x+y\right)-3\left(x-y\right)=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+2y-x+y=1\\x+y-3x+3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=1\\-2x+4y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+6y=2\\-2x+4y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+6y-2x+4y=2-5\\-x+2y=-\dfrac{5}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}10y=-3\\x-2y=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{10}\\x=2y+\dfrac{5}{2}=-\dfrac{3}{5}+\dfrac{5}{2}=\dfrac{19}{10}\end{matrix}\right.\)

25: \(\left\{{}\begin{matrix}2x+1=x+2y\\x-y=2x+y+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-x-2y=-1\\x-y-2x-y=1\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}-4y=0\\x=2y-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-1\end{matrix}\right.\)

26: \(\left\{{}\begin{matrix}2\left(x-2y\right)+3\left(x+2y\right)=4\\\left(x-y\right)+2\left(x+y\right)=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y+3x+6y=4\\x-y+2x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=4\\3x+y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x+2y=4\\6x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x=2\\3x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1-3x=1-3\cdot\left(-2\right)=7\end{matrix}\right.\)

27: \(\left\{{}\begin{matrix}\left(x+4\right)\left(y+4\right)=xy+216\\\left(x+2\right)\left(y-5\right)=xy-50\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy+4x+4y+16=xy+216\\xy-5x+2y-10=xy-50\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+4y=200\\-5x+2y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y=100\\-5x+2y=-40\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7x=140\\x+y=50\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=50-x=50-20=30\end{matrix}\right.\)

`24)`

\(\left\{{}\begin{matrix}2\left(x+y\right)-\left(x-y\right)=1\\\left(x+y\right)-3\left(x-y\right)=-5\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}2x+2y-x+y=1\\x+y-3x+3y=-5\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x+3y=1\\-2x+4y=-5\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x=1-3y\\-2\left(1-3y\right)+4y=-5\end{matrix}\right.\) `<=>` \(\left\{{}\begin{matrix}x=1-3y\\-2+6y+4y=-5\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x=1-3y\\10y=-3\end{matrix}\right.\)  `<=>` \(\left\{{}\begin{matrix}x=1-3.\left(-\dfrac{3}{10}\right)\\y=-\dfrac{3}{10}\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x=\dfrac{19}{10}\\y=-\dfrac{3}{10}\end{matrix}\right.\)

Vậy hpt có nghiệm `(x;y)=((19)/(10) ; (-3)/(10))`

`25)`

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

`<=>` \(\left\{{}\begin{matrix}2x-x-2y=-1\\x-y-2x-y=1\end{matrix}\right.\) `<=>` \(\left\{{}\begin{matrix}x-2y=-1\\-x-2y=1\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}2x=-2\\-x-2y=1\end{matrix}\right.\) `<=>` \(\left\{{}\begin{matrix}x=-1\\-\left(-1\right)-2y=1\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)

Vậy hpt có nghiệm `(x;y)=(-1;0)`

`26)`

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

`<=>` \(\left\{{}\begin{matrix}2x-4y+3x+6y=4\\x-y+2x+2y=1\end{matrix}\right.\)   `<=>` \(\left\{{}\begin{matrix}5x+2y=4\\3x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+2y=4\\6x+2y=2\end{matrix}\right.\)  `<=>` \(\left\{{}\begin{matrix}-x=2\\6x+2y=2\end{matrix}\right.\)

`<=> {(x=-2),(6.(-2) + 2y=2):}`

`<=>{(x=-2),(y=7):}`

Vậy hpt có nghiệm `(x;y)=(-2;7)`

`27)`

\(\left\{{}\begin{matrix}\left(x+4\right)\left(y+4\right)=xy+216\\\left(x+2\right)\left(y-5\right)=xy-50\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}xy+4x+4y+16=xy+216\\xy-5x+2y-10=xy-50\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}xy+4x+4y-xy=216-16\\xy-5x+2y-xy=-50+10\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}4x+4y=200\\-5x+2y=-40\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x+y=50\\-5x+2y=-40\end{matrix}\right.\) `<=>` \(\left\{{}\begin{matrix}x=50-y\\-5\left(50-y\right)+2y=-40\end{matrix}\right.\)

`<=>` \(\left\{{}\begin{matrix}x=50-y\\-250+5y+2y=-40\end{matrix}\right.\) 

`<=>` \(\left\{{}\begin{matrix}x=50-y\\7y=210\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=50-30\\y=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=30\end{matrix}\right.\) 

Vậy hpt có nghiệm `(x;y)=(20;30)`

a: Xét ΔADB và ΔAEC có

AD=AE

\(\widehat{DAB}\) chung

AB=AC

Do đó: ΔADB=ΔAEC
=>BD=CE

b: Ta có: AE+EB=AB

AD+DC=AC

mà AE=AD và AB=AC

nên EB=DC

Xét ΔEBC và ΔDCB có

EB=DC

BC chung

EC=DB

Do đó: ΔEBC=ΔDCB

=>\(\widehat{OBC}=\widehat{OCB}\)

=>ΔOBC cân tại O

=>OB=OC

Ta có: OB+OD=BD

OC+OE=CE

mà BD=CE và OB=OC

nên OD=OE

=>ΔODE cân tại O

c: Xét ΔABC có \(\dfrac{AE}{AB}=\dfrac{AD}{AC}\)

nên ED//BC

Mình làm câu a r.

Trình bày câu `b)` thôi cũng được, nhưng làm câu b chi tiết với ạ.