3. celebrated

4. entertainment

5. excitement

6. festive

4:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;9\right\}\end{matrix}\right.\)

\(A=\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

\(=\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

\(=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

b: A=căn 3

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=\sqrt{3}\)

=>\(\sqrt{x}+2=\sqrt{3}\cdot\sqrt{x}-3\sqrt{3}\)

=>\(\sqrt{x}\left(1-\sqrt{3}\right)=-3\sqrt{3}-2\)

=>\(\sqrt{x}=\dfrac{3\sqrt{3}+2}{\sqrt{3}-1}=\dfrac{11+5\sqrt{3}}{2}\)

=>\(x=\dfrac{98+55\sqrt{3}}{2}\)

c: Để A nguyên thì \(\sqrt{x}-3+5⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3\in\left\{1;-1;5;-5\right\}\)

=>\(\sqrt{x}\in\left\{4;2;8;-2\right\}\)

=>\(\sqrt{x}\in\left\{4;2;8\right\}\)

=>\(x\in\left\{16;4;64\right\}\)

h)\(\dfrac{x+2}{x^2+2x+1}\ge0\)

⇔\(\dfrac{x+2}{\left(x+1\right)^2}\ge0\)

⇔\(\dfrac{x+2}{\left(x+1\right)\left(x+1\right)}\ge\dfrac{0.\left(x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x+1\right)}\)

⇒\(x+2\ge0\)

⇔\(x+2-2\ge0-2\)

⇔\(x\ge-2\)

i)\(\dfrac{x-1}{x-3}>1\)

⇔\(\dfrac{x-1}{x-3}>\dfrac{1.\left(x-3\right)}{1.x-3}\)

⇒\(x-1>x-3\)

⇔\(x-x>-3+1\)

⇔\(0x>-2\)

Bạn cần hỗ trợ bài nào vậy?

sao nó lại mờ thế?

nhiều thế này ai mà làm nổi

nhỏ xí à

a: Xét ΔABD và ΔAED có 

AB=AE

\(\widehat{BAD}=\widehat{EAD}\)

AD chung

Do đó: ΔABD=ΔAED

b: Xét ΔBDF và ΔEDC có 

\(\widehat{BDF}=\widehat{EDC}\)

DB=DE

\(\widehat{DBF}=\widehat{DEC}\)

Do đó: ΔBDF=ΔEDC

đã viết

đã viết

wrote là QUÁ KHỨ ĐƠN của write.

4:

a: =>x+2,8=1,5

=>x=-1,3

b: =>5x=-5,05

=>x=-1,01

c: =>-0,6x=-8,844

=>x=14,74

d: =>x^2=1,44

=>x=1,2 hoặc x=-1,2

3:

a: =-8,43+8,43+9,1-9,1=0

b: =18,7-18,7+5,6=5,6

c: =-4,23(19,5+80,5)

=-4,23*100=-423

d:=A*(-2,3+2,3)=0