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Minhf ko vieets ddc daaus banj owi!

help me!

1 The Temple of Literature is about 10 minutes away from Hoan Kiem Lake

2 There are five courtyards

3 It''s used for street signs of Ha Noi

4 We can find the stone tablets above tortoise backs with thenames of doctors in the third courtyard

EX4 

1 : My town was visited by hundreds of tourists last year

Đề bài là gì bạn?

\(\text{≌₰⇴⩸⨙⩸※◡⨦}\)

a) \(\dfrac{-8}{15}-\left(\dfrac{1}{9}+\dfrac{7}{15}-\dfrac{5}{18}\right)\)

\(=\dfrac{-8}{15}-\dfrac{1}{9}-\dfrac{7}{15}+\dfrac{5}{18}\\ =\left(-\dfrac{8}{15}-\dfrac{7}{15}\right)+\left(\dfrac{5}{18}-\dfrac{1}{9}\right)\\ =-1+\dfrac{1}{6}=-\dfrac{5}{6}\)

b) \(\dfrac{1}{2}.\dfrac{-1}{14}+\dfrac{1}{2}.\dfrac{9}{14}+\dfrac{-5}{14}.\dfrac{1}{2}\)

\(=\dfrac{1}{2}.\left(\dfrac{-1}{14}+\dfrac{9}{14}+\dfrac{-5}{14}\right)\\ =\dfrac{1}{2}.\dfrac{3}{14}=\dfrac{3}{28}\)

Câu 1: 

1:A

2:C

Câu 1:

3:D

4:C

Câu 1: 

5:C

6:D

a, \(P=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\right)\)

\(=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+2}{\sqrt{x}+1}\right):\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{x-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

b, \(P< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+2}< \dfrac{1}{2}\)

\(\Leftrightarrow2\sqrt{x}-2< \sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}< 4\)

\(\Leftrightarrow0\le x< 16\)

Vậy \(0\le x< 16;x\ne1;x\ne4\).

a: ta có: \(P=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\)

\(=\dfrac{x+\sqrt{x}-x-2}{\left(\sqrt{x}+1\right)}:\dfrac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}-2}{1}\cdot\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

By whom this letter was written?