ai giúp mk với ạ!!!!!!!
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a, Phép so sánh: có từ ''như''
=> SS ko ngang bằng
b, Phép so sánh: có từ ''như''
=> SS ko ngang bằng
c, Phép so sánh: có từ ''như''
=> SS ko ngang bằng
d, Phép so sánh: có từ ''như''
=> SS ko ngang bằng
e, Phép so sánh: ''bao nhiêu-bấy nhiêu''
=> SS ngang bằng
a, Phép so sánh: có từ ''như''
=> SS ko ngang bằng
b, Phép so sánh: có từ ''như''
=> SS ko ngang bằng
c, Phép so sánh: có từ ''như''
=> SS ko ngang bằng
d, Phép so sánh: có từ ''như''
=> SS ko ngang bằng
360. Scientists ( để -s vì có have ) have researched science subjects.
361. His father is an architect.
362. Japan is an idustrial.
363. The solution of the book..
364. Every success...
365. Can you imagine .... ( Câu này chưa chắc )
366. The United Nations organization...
I.
1 D
2 B
3 D
4 A
5 C
6 A
7 A
8 C
II.
1 am listening
2 Are you playing
3 am reading/is watching
4 is cooking
5 is riding
II.
1 am listening (keep silent! . DHNB thì HTTD)
2 Are you playing ( now : DHNB thì HTTD)
3 am reading/is watching ( at the moment )
4 is cooking (at the moment DNNB thì HTTD)
5 is riding ( look)
1 love
2 did you do - did
3 Did she get
4 stopped - bought
5 Did they participate
6 drove
7 did you live
8 had you taught
9 tried
10 was skating
11 was walking
12 were dancing
13 was having - clocked
14 was sleeping
15 was walking
17 was raining
18 was she writing
\(f\left(x\right)=\dfrac{x+4}{\left(x-3\right)\left(x+3\right)}-\dfrac{2}{x+3}+\dfrac{4x}{x\left(x-3\right)}\)
\(f\left(x\right)=\dfrac{x\left(x+4\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{2x\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}+\dfrac{4x\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}\)
\(f\left(x\right)=\dfrac{3x^2+22x}{x\left(x-3\right)\left(x+3\right)}\)
\(f\left(x\right)< 0\Leftrightarrow\left\{{}\begin{matrix}x< -\dfrac{22}{3}\\-3< x< 0\\0< x< 3\end{matrix}\right.\) \(\Rightarrow x_{max}=2\)
Bài 2:
\(\sqrt{x^2+2x+4}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+2x+4=x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x=-2\end{matrix}\right.\)(vô nghiệm)
Vậy pt vô nghiệm
\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}=\dfrac{\sqrt{32}-\sqrt{12}}{-\left(\sqrt{48}-\sqrt{18}\right)}=-\dfrac{\sqrt{2}\left(\sqrt{16}-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{16}-\sqrt{6}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}=\dfrac{-\sqrt{6}}{3}\)
\(A=45,98:0,01\times4,2-6,27:0,5+3,9\)
\(A=4598\times4,2-12,64+3,9\)
\(A=19311,6-16,54\)
\(A=19295,06\)
A=4598 x 4,2 - 12,54+3,9
A=19311,6-12,54+3,9
A=19299,06+3,9
A=19302,96