Ảnh đẹp thì

jdhjdhshfsjsxhxhxx                  udjdghxhjxhg

sao dạo này toàn người cho toán lớp 9 nhỉ khó qué

d: Cặp đường song song là y=x-3 và y=x+4 vì a=1=a'=1

Cặp đường cắt nhau là y=x-3 và y=-2x+5 vì a=1<>a'=-2

a) \(\left\{{}\begin{matrix}\left(d\right):y=-2x-5\\\left(d'\right):y=-x\end{matrix}\right.\)

b) \(\left(d\right)\cap\left(d'\right)=M\left(x;y\right)\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=5\end{matrix}\right.\)

\(\Rightarrow M\left(-5;5\right)\)

c) Gọi \(\widehat{M}=sđ\left(d;d'\right)\)

\(\left(d\right):y=-2x-5\Rightarrow k_1-2\)

\(\left(d'\right):y=-x\Rightarrow k_1-1\)

\(tan\widehat{M}=\left|\dfrac{k_1-k_2}{1+k_1.k_2}\right|=\left|\dfrac{-2+1}{1+\left(-2\right).\left(-1\right)}\right|=\dfrac{1}{3}\)

\(\Rightarrow\widehat{M}\sim18^o\)

d) \(\left(d\right)\cap Oy=A\left(0;y\right)\)

\(\Leftrightarrow y=-2.0-5=-5\)

\(\Rightarrow A\left(0;-5\right)\)

\(OA=\sqrt[]{0^2+\left(-5\right)^2}=5\left(cm\right)\)

\(OM=\sqrt[]{5^2+5^2}=5\sqrt[]{2}\left(cm\right)\)

\(MA=\sqrt[]{5^2+10^2}=5\sqrt[]{5}\left(cm\right)\)

Chu vi \(\Delta MOA:\)

\(C=OA+OB+MA=5+5\sqrt[]{2}+5\sqrt[]{5}=5\left(1+\sqrt[]{2}+\sqrt[]{5}\right)\left(cm\right)\)

\(\Rightarrow p=\dfrac{C}{2}=\dfrac{5\left(1+\sqrt[]{2}+\sqrt[]{5}\right)}{2}\left(cm\right)\)

\(\Rightarrow\left\{{}\begin{matrix}p-OA=\dfrac{5\left(1+\sqrt[]{2}+\sqrt[]{5}\right)}{2}-5=\dfrac{5\left(\sqrt[]{2}+\sqrt[]{5}-1\right)}{2}\\p-OB=\dfrac{5\left(1+\sqrt[]{2}+\sqrt[]{5}\right)}{2}-5\sqrt[]{2}=\dfrac{5\left(-\sqrt[]{2}+\sqrt[]{5}+1\right)}{2}\\p-MA=\dfrac{5\left(1+\sqrt[]{2}+\sqrt[]{5}\right)}{2}-5\sqrt[]{5}=\dfrac{5\left(\sqrt[]{2}-\sqrt[]{5}+1\right)}{2}\end{matrix}\right.\)

\(p\left(p-MA\right)=\dfrac{5\left(1+\sqrt[]{2}+\sqrt[]{5}\right)}{2}.\dfrac{5\left(1+\sqrt[]{2}-\sqrt[]{5}\right)}{2}\)

\(\Leftrightarrow p\left(p-MA\right)=\dfrac{25\left[\left(1+\sqrt[]{2}\right)^2-5\right]}{4}=\dfrac{25.2\left(\sqrt[]{2}-1\right)}{4}=\dfrac{25\left(\sqrt[]{2}-1\right)}{2}\)

\(\left(p-OA\right)\left(p-OB\right)=\dfrac{25\left[5-\left(\sqrt[]{2}-1\right)^2\right]}{4}\)

\(\Leftrightarrow\left(p-OA\right)\left(p-OB\right)=\dfrac{25.2\left(\sqrt[]{2}+1\right)}{4}=\dfrac{25\left(\sqrt[]{2}+1\right)}{4}\)

Diện tích \(\Delta MOA:\)

\(S=\sqrt[]{p\left(p-OA\right)\left(p-OB\right)\left(p-MA\right)}\)

\(\Leftrightarrow S=\sqrt[]{\dfrac{25\left(\sqrt[]{2}-1\right)}{2}.\dfrac{25\left(\sqrt[]{2}+1\right)}{2}}\)

\(\Leftrightarrow S=\sqrt[]{\dfrac{25^2}{2^2}}=\dfrac{25}{2}=12,5\left(cm^2\right)\)

*) Đồ thị:

b) Phương trình hoành độ giao điểm của (d) và (d'):

\(-2x-5=-x\)

\(\Leftrightarrow-2x+x=5\)

\(\Leftrightarrow x=-5\) \(\Rightarrow y=-\left(-5\right)=5\)

Vậy tọa độ giao điểm của (d) và (d') là \(M\left(-5;5\right)\)

c) Ta có:

\(tanB=\dfrac{OA}{OB}=\dfrac{-5}{-\dfrac{5}{2}}=2\)

\(\Rightarrow\widehat{B}\simeq63^0\)

Mà góc tạo bởi d với trục hoành là \(\widehat{OBM}\)

\(\Rightarrow\widehat{OBM}\simeq180^0-63^0=117^0\)

d) Ta có:

\(OM^2=5^2+5^2=50\) 

\(\Rightarrow OM=5\sqrt{2}\left(cm\right)\)

\(AM^2=5^2+10^2=125\)

\(\Rightarrow AM=5\sqrt{5}\left(cm\right)\)

Chu vi \(\Delta MOA\):

\(5\sqrt{2}+5\sqrt{5}+5=5\left(\sqrt{2}+\sqrt{5}+1\right)\left(cm\right)\)

Diện tích \(\Delta MOA\)

\(S_{MOA}=\dfrac{MH.OA}{2}=\dfrac{5.5}{2}=25\left(cm^2\right)\)