plsssss hlep meeeeeeeeeee:<<<<<
cho tam giác ABC có đường cao AD cắt đường cao BE tại H. Gọi M là trung điểm của đoạn thẳng BC. Trên tia HM lấy Q sao cho HM=MQ.
a) Chứng minh tứ giác HCQB là hình bình hành.
b) Chứng minh CQ vuông góc với AC và BQ vuông góc với AB.
c) Trên tia HD lấy P sao cho HD=DP. Chứng minh DM là đường trung bình của tam giác PHQ từ đó chứng minh tứ giác BPQC là hình thang cân
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Bài 13
a) Ta có: OAt+AOy=1800
Mà hai góc này ở vị trí trong cùng phía
⇒ At//Oy
Bài 2 :
\(a,\left(x+2\right)\left(x^2+3x-2\right)=2\left(x+2\right)x^2\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+3x-2\right)-2x^2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+3x-2-2x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-x^2+3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\-x^2+x+2x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\-x\left(x-1\right)+2\left(x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left(x-1\right)\left(-x+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left[{}\begin{matrix}x-1=0\\-x+2=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{-2;2;1\right\}\)
\(b,9x^2-\left(6x+2\right)\left(x-5\right)=1\)
\(\Leftrightarrow9x^2-\left(6x^2-30x+2x-10\right)-1=0\)
\(\Leftrightarrow9x^2-6x^2+30x-2x+10-1=0\)
\(\Leftrightarrow3x^2+28x+9=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-9\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{3};-9\right\}\)
\(c,\dfrac{x}{3x-2}-\dfrac{x}{2+3x}=\dfrac{6x^2}{9x^2-4}\left(dkxd:x\ne\pm\dfrac{2}{3}\right)\)
\(\Leftrightarrow\dfrac{x}{3x-2}-\dfrac{x}{3x+2}-\dfrac{6x^2}{\left(3x-2\right)\left(3x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(3x+2\right)-x\left(3x-2\right)-6x^2}{\left(3x-2\right)\left(3x+2\right)}=0\)
\(\Leftrightarrow3x^2+2x-3x^2+2x-6x^2=0\)
\(\Leftrightarrow4x-6x^2=0\)
\(\Leftrightarrow-2x\left(-2+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\-2+3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmdk\right)\\x=\dfrac{2}{3}\left(ktmdk\right)\end{matrix}\right.\)
Vậy \(S=\left\{0\right\}\)
Bài 1 :
\(a,P=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\left(dkxd:x\ne0,x\ne\pm1\right)\)
\(=\dfrac{x^2+x}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x^2+x}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\left(dpcm\right)\)
\(b,P=-\dfrac{1}{2}\Rightarrow\dfrac{x^2}{x-1}=-\dfrac{1}{2}\)
\(\Rightarrow2x^2=-\left(x-1\right)\)
\(\Rightarrow2x^2=-x+1\)
\(\Rightarrow2x^2+x-1=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy \(P=-\dfrac{1}{2}\) thì \(x=\dfrac{1}{2};x=-1\)
\(c,\) Để P nhận giá trị nguyên dương thì \(P\ge0\)
\(\Leftrightarrow\dfrac{x^2}{x-1}\ge0\Leftrightarrow x\ge0\)
Sorry, mình không biết làm phần Task 1 nhaaa :<<<
Task 2: Complete the sentences using PAST SIMPLE TENSE.
1. didn't like
2. phoned
3. went
4. Did they leave
5. Did you like
6. didn't finish
7. Did she go
8. stayed / got
9. went
10. Did you travel
Chúc bạn học tốt!! ^^
a) \(=3\left(xy-4\right)\)
b) \(=x^2\left(x-y\right)+4\left(x-y\right)=\left(x-y\right)\left(x^2+4\right)\)
c) \(=x^2-\left(y^2-12y+36\right)=x^2-\left(y-6\right)^2=\left(x-y+6\right)\left(x+y-6\right)\)
d) \(=\left(4p^2-36p+81\right)-25=\left(2p-9\right)^2-25=\left(2p-9-5\right)\left(2p-9+5\right)=4\left(p-7\right)\left(p-2\right)\)
1 saw - didn't see - was looking
2 met - went - went - had - were waiting
3 was cycling - stepped - was going - managed - didn't hit
4 was waiting - arrived
5 were you doing
6 Did you go
7 wore
8 were you driving - happened
9 took - wasn't looking
10 were - didn't know