A person who make friend with you through letters is call a penpal

chỗ cần điền là penpal

A person who make friend with you through letters is call a _penpal

Rewrite the sentence ,use the word in parentheses:

There are two fridges,a table,four chairs in Thuong's  kitchen. - ............Thuong's kitchen has 2 fridges, a table,  four chairs..............

=> Thuong's kitchen has two fridges , a table , four chairs .

Thuong*s kitchen has two fridges, a table, four chairs.

đúng đó bn ạ

Đề bài là gì bạn?

Câu 1: 

1:A

2:C

Câu 1:

3:D

4:C

Câu 1: 

5:C

6:D

Bất đẳng thức là cái j vậy các anh chụy?

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

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

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

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

Câu 2 bị che mất r

Mình cũng không chắc về kết quả

Chúc bạn học tốt !

Câu 3:

Đặt \(f_{\left(x\right)}=2x^3+3x-a+10\)

Để \(f_{\left(x\right)}⋮x-3\)

thì \(\Rightarrow f_{\left(x\right)}:x-3\text{ }dư\text{ }0\)

\(\Rightarrow\) Theo định lí \(Bê-du:f_{\left(3\right)}=0\)

\(\Rightarrow2\cdot3^3+3\cdot3-a+10=0\\ \Rightarrow73-a=0\\ \Rightarrow a=73\)

Vậy để \(2x^3+3x-x+10⋮x-3\) thì \(a=73\)

Không chắc lắm.