???

Bài nào :>

Bài 4 nào bài 8 đâu em 

là sao

what is her mother going to prepare for her bỉthdat party

there are three sticks of butter in the cupboard

chỉ có thể thả tim chia buồn cho bạn tui bó tay -_-

Bài 12: 

a: Xét ΔABM và ΔACN có

AB=AC

\(\widehat{ABM}=\widehat{ACN}\)

BM=CN

Do đó: ΔABM=ΔACN

Có thể lm bài 11 đc ko ạ🥺😅

Bài 13:

a: Ta có: F là trung điểm của DA

=>\(DF=\frac12\cdot DA\)

=>\(S_{BDF}=\frac12\cdot S_{BDA}\)

Ta có: E là trung điểm của DC

=>\(DE=\frac12\cdot DC\)

=>\(S_{BDE}=\frac12\cdot S_{BDC}\)

Ta có: \(S_{BDF}+S_{BDE}=S_{BFDE}\)

=>\(S_{BFDE}=\frac12\left(S_{BDA}+S_{BDC}\right)=\frac12\cdot S_{ABCD}\)

b: Ta có: \(DF=\frac12\cdot DA\)

=>\(S_{DFK}=\frac12\cdot S_{DKA}\)

Ta có: \(DE=\frac12\cdot DC\)

=>\(S_{DEK}=\frac12\cdot S_{DKC}\)

Ta có: \(S_{DFK}+S_{DEK}=S_{DFKE}\)

=>\(S_{DFKE}=\frac12\left(S_{DKA}+S_{DKC}\right)=\frac12\cdot S_{ADC}\)

bài 7

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

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

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

A=\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

bài 8

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

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

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

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

bài 9

P=\(\left[\dfrac{2\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}\right].\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

P=\(\dfrac{4\sqrt{xy}-\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

P=\(\dfrac{2\sqrt{xy}-x-y}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

P=\(\dfrac{-\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}.\dfrac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

P=\(\dfrac{-\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)

bài 10

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

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

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

cảm ơn bạn nha

\(\sqrt{9x+9}-2\sqrt{\dfrac{x+1}{4}}=4\left(đk:x\ge-1\right)\)

\(\Leftrightarrow3\sqrt{x+1}-\sqrt{x+1}=4\)

\(\Leftrightarrow2\sqrt{x+1}=4\)

\(\Leftrightarrow\sqrt{x+1}=2\Leftrightarrow x+1=4\Leftrightarrow x=3\left(tm\right)\)