Câu 39: D

Câu 40: B

Câu 41:A

(x-3)  =    -4

(x) =   -4  +  3

(x)  =    -1

x  = 1

\(a,ĐK:x>0;x\ne1\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b,A>0\Leftrightarrow\sqrt{x}-1>0\left(\sqrt{x}>1\right)\\ \Leftrightarrow x>1\)

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

lâm ơi chưa biết nó kể về cái gì sao làm bài

\(A=\dfrac{7}{19}\times\dfrac{8}{11}+\dfrac{7}{19}\times\dfrac{3}{11}+\dfrac{12}{19}\)

\(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\)

\(=\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{19}{19}=1\)

a: \(A=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}=\dfrac{7}{19}+\dfrac{12}{19}=1\)

b: \(B=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{5}{117}\right)\cdot\dfrac{4-3-1}{12}=0\)

a)\(A=\dfrac{7}{19}.\dfrac{8}{11}+\dfrac{7}{19}.\dfrac{3}{11}+\dfrac{12}{19}\) 

       \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\)

       \(=\dfrac{7}{19}.\dfrac{11}{11}+\dfrac{12}{19}\)

       \(=\dfrac{7}{19}+\dfrac{12}{19}\)

       \(=\dfrac{19}{19}=1\)

b)\(B=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)

       \(=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{5}{39}\right)\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)

       \(=\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{5}{39}\right).0=0\)

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

a: Xét ΔABM và ΔACM có 

AB=AC
AM chung

BM=CM

Do đó: ΔABM=ΔACM

b: Ta có: ΔABC cân tại A

mà AM là đường trung tuyến

nên AM là đường cao

c: Xét ΔHBM vuông tại H và ΔKCM vuông tại K có 

MB=MC

\(\widehat{B}=\widehat{C}\)

Do đó: ΔHBM=ΔKCM

Suy ra: MH=MK