1, Nam isn't so short as Hưng

2,Lan drives to work

3,How many fans are there in the classroom ?

4,It took 15 minutes 

5,Unless she work hard ,giữ nguyên

vì be là đường cao của tam giác abc :

=)góc bec = góc bea =90

vì ch là đường cao của tam giác abc (gt)

=) góc bch = góc cha =90

xét tam giác ahc và tam giác aeb ta có 

góc ahc =góc aeb( =90)

góc a chung

=) tam giác ahc ~TAM GIÁC EAB(g.g)

b, gọi i là giao điểm của be và ch 

xét tam giác hib và tam giác eic ta có 

h2 =e2 =90

i1=i2 ( 2gocs đối đỉnh)

=)hib ~eic (g.g)

do mk đang bận nên chỉ giải đc đến đây thôi ạ

1:

a: BC=8-3=5cm

b: MN=MC+CN=1/2(CA+CB)

=1/2*AB=4cm

2: 

a: Có 2 tia là OA và OB

b: AB=OB+OA=11cm

c: AC=BC=11/2=5,5cm

3.14:

Ta thấy $\widehat{xNM}=\widehat{xQP}=45^0$. Mà 2 góc này ở vị trí đồng vị nên $MN\parallel PQ$

3.15

$EF\parallel NP$ do cùng vuông góc với $MH$

3.16: Bạn tự vẽ hình nhé.

3.17:

Ta thấy $\widehat{yKH}+\widehat{KHx}=130^0+50^0=180^0$. Mà 2 góc này ở vị trí trong cùng phía nên $Ky\parallel Hx$

\(1,=3ab\left(1-2a+b\right)\\ 2,=\left(x-y\right)\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(x-y-7\right)\\ 3,=\left(a-5\right)\left(5a-2\right)\\ 4,=5x\left(x-3\right)-\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(4x-3\right)\\ 5,=9a^2-\left(b-2\right)^2=\left(3a-b+2\right)\left(3a+b-2\right)\\ 6,=2x^2-4x+3x-6=\left(x-2\right)\left(2x+3\right)\\ 7,=3x^2\left(2x-5\right)\\ 8,=\left(3x-5\right)\left(3x+5\right)\\ 9,=4x^2\left(x-y\right)-x\left(x-y\right)=x\left(4x-1\right)\left(x-y\right)\)

Giúp em bài số 4 ạ😓

1 Having slept

2 not being invited

3 Having had

4 having

5 talking

6 succeeded - launching

7 Having travelled

8 Have - considered - trying

9 Having seen - had - to go

10 Being invited

11 Being found

12 having

13 taken - being photographed 

14 to fix

15 living

16 Having waited - to deliver - decided to cancel

17 Having photocopied 

18 to have happen

19 to give

20 spoiling

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\)

\(\sqrt{x^2-x-2}-\sqrt{x-2}=0\\ \Leftrightarrow\sqrt{x^2-x-2}=\sqrt{x-2}\\ \Leftrightarrow x^2-x-2=x-2\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

\(a,ĐK:x\ge2\\ PT\Leftrightarrow x^2-x-2=x-2\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=2\\ b,ĐK:\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\\ PT\Leftrightarrow\sqrt{x^2-1}=x^2-1\\ \Leftrightarrow x^2-1=\left(x^2-1\right)^2\\ \Leftrightarrow\left(x^2-1\right)\left(x^2-1-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\\x=\sqrt{2}\left(tm\right)\\x=-\sqrt{2}\left(tm\right)\end{matrix}\right.\)

\(c,ĐK:\left[{}\begin{matrix}x\le-2\\x\ge1\end{matrix}\right.\\ PT\Leftrightarrow\sqrt{x^2-x}=-\sqrt{x^2+x-2}\\ \Leftrightarrow x^2-x=x^2+x-2\\ \Leftrightarrow2x=2\\ \Leftrightarrow x=1\left(tm\right)\)

\(d,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{\left(x^2-1\right)^2}=x-1\\ \Leftrightarrow\left|x^2-1\right|=x-1\\ \Leftrightarrow\left[{}\begin{matrix}x^2-1=x-1\left(x\le-1;x\ge1\right)\\x^2-1=1-x\left(-1< x< 1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(e,PT\Leftrightarrow\left|x+2\right|+\left|x-4\right|=0\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

\(g,\Leftrightarrow x\in\varnothing\left(\sqrt{x-2}+\sqrt{x-3}\ge0>-5\right)\\ f,\Leftrightarrow\left|x-1\right|+\left|x-3\right|=1\\ \Leftrightarrow\left[{}\begin{matrix}1-x+3-x=1\left(x< 1\right)\\x-1+3-x=1\left(1\le x< 3\right)\\x-1+x-3=1\left(x\ge3\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(ktm\right)\\0x=-1\left(ktm\right)\\x=\dfrac{5}{2}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x\in\varnothing\)

Bài 6 : 

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

       1          2             1           1

     0,1                       0,1        0,1

a) \(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

b) \(n_{FeCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

⇒ \(m_{FeCl2}=0,1.127=12,7\left(g\right)\)

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