Mn ơi giúp mình với ạ
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HV
7 tháng 12 2021
1 his illness, he cannot come
2 her busyness, she couldn't help us
3 his illness, he tries to go to school on time
4 the bad weather, we tried to finish the work on the road
5 the bad weather, we got to the station late
6 the old house, she liked it
7 not wearing any shoes, Carol ran outside to see what was happening
8 being afraid of flying, Fiona had to get on the plane
DT
Đỗ Thanh Hải
CTVVIP
7 tháng 6 2021
9 Although the girl works hard from morning till night, her step mother isn't happy
10 Rice is grown in tropical countries
11 The countryside is more peaceful than the city, but I prefer living in the city
12 A lot of Chung Cakes are made on Tet holiday
13 Although Little Pea works hard from morning till night, her mother isn't happy
14 An ordinary car goes more slowly than a sport car
15 An's sister performs better than him
16 Many fesstivals are held by the Thai every year
17 I enjoy hanging out with my best friends
18 If you don't hurry up, you will be late for the race
25 tháng 10 2021
\(a,A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\\ b,A=\dfrac{2\left(\sqrt{x}+1\right)-3}{\sqrt{x}+1}=2-\dfrac{3}{\sqrt{x}+1}\in Z\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(3\right)=\left\{1;3\right\}\left(\sqrt{x}+1\ge1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\\ \Leftrightarrow x\in\left\{0;4\right\}\left(tm\right)\)
25 tháng 10 2021
a) \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(\Rightarrow A=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{\left(2x-2\sqrt{x}\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
1:
a: \(P=B:A\)
\(=\dfrac{3\sqrt{x}-3-2\sqrt{x}+4+\sqrt{x}+2}{x-4}:\dfrac{4\sqrt{x}-1}{x-4}\)
\(=\dfrac{2\sqrt{x}+3}{x-4}\cdot\dfrac{x-4}{4\sqrt{x}-1}=\dfrac{2\sqrt{x}+3}{4\sqrt{x}-1}\)
b: Để P nguyên thì \(2\sqrt{x}+3⋮4\sqrt{x}-1\)
=>\(4\sqrt{x}+6⋮4\sqrt{x}-1\)
=>\(4\sqrt{x}-1+7⋮4\sqrt{x}-1\)
=>4căn x-1 thuộc {1;-1;7;-7}
=>căn x thuộc {1/2;0;2}
mà x nguyên và x>=0 và x<>4
nên x thuộc {0}