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

a)\(x-5\sqrt{x}+6=x-3\sqrt{x}-2\sqrt{x}+6\)

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

b) ĐKXĐ:\(x\ge0,x\ne9,x\ne4\)

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

M=\(\dfrac{x-4-x+3\sqrt{x}-\sqrt{x}+3-3\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

d) để M<-1

\(\Rightarrow\dfrac{-1}{\sqrt{x}-3}< -1\Rightarrow-1>-1\left(\sqrt{x}-3\right)\)

\(\Rightarrow-1>3-\sqrt{x}\)\(\Rightarrow-\sqrt{x}< -4\Rightarrow x>16\)

e) để \(M\in Z\)\(\Rightarrow\)\(\sqrt{x}-3\inƯ\left(-1\right)=\left\{1;-1\right\}\)

vậy x=4 thì \(M\in Z\)

bn tự vẽ hình giúp mik nha

a) áp dụng pytago đảo ta có

\(AB^2+AC^2=6^2+8^2=100\)

\(BC^2=10^2=100\)

vậy \(\Delta ABC\) vuông tại A

b) trong \(\Delta ABC\) vuông tại A có

sinB=\(\dfrac{AC}{BC}=\dfrac{8}{10}\)=0,8\(\Rightarrow\widehat{B}\)=53

\(\widehat{C}=90-\widehat{B}=90-53\)=37

hình như thiếu câu c á bn nên mik k lm câu d đc

Câu 1

a)=\(8\sqrt{3}-10\sqrt{3}+15\sqrt{3}=13\sqrt{3}\)

b)=\(4\sqrt{x}+6\sqrt{x}-6\sqrt{x}=4\sqrt{x}\)

c)=\(21\sqrt{2}+8\sqrt{2}-28\sqrt{2}=\sqrt{2}\)

d)\(\Rightarrow\)\(8\sqrt{2\sqrt{3}}-\sqrt{5\sqrt{3}}-4\sqrt{5\sqrt{3}}\)

\(\Rightarrow\)\(8\sqrt{2\sqrt{3}}-5\sqrt{5\sqrt{3}}\)

câu 2

a)\(\Rightarrow4x=64\)\(\Rightarrow x=16\)

b)\(\Rightarrow9x\le36\)\(\Rightarrow x\le4\)

BCABBAAD

d)=\(4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)=\(4\sqrt{b}-5\sqrt{10b}\)

e)=\(6+\sqrt{15}-2\sqrt{15}\)=\(6-\sqrt{15}\)

f)=\(5\sqrt{10}+10-5\sqrt{10}\)=10

g)=\(\sqrt{196}-\sqrt{84}-\sqrt{49}+2\sqrt{21}\)=14-\(2\sqrt{21}\)-7+\(2\sqrt{21}\)=7

h)=\(\sqrt{1089}-\sqrt{198}-\sqrt{121}+3\sqrt{22}\)

=\(33-3\sqrt{22}-11+3\sqrt{22}\)=22

trong \(\Delta ABH\) vuông tại H có

AH=AB.cosA=5.cos60=2,5

BH=\(\sqrt{AB^2-AH^2}\)(pytago)=\(\sqrt{5^2-2,5^2}\)=4,3

trong \(\Delta BHC\) vuông tại H có

\(HB^2=BC.BF\)(dl1)\(\Rightarrow BF=\dfrac{HB^2}{BC}\)=\(\dfrac{4,3^2}{5\sqrt{3}}\)=2,1

HF=\(\sqrt{HB^2-BF^2}\)=\(\sqrt{4,3^2-2,1^2}\)=3,8

bạn tự vẽ hình giúp mik nha

a) Trong \(\Delta EDI\) vuông tại I có

EI=ED.sinD=7.sin45=4,9

ID=ED.cosD=7.cos45=4,9

b) ta có IF=DF-DI=12-4,9=7,1

trong \(\Delta EIF\) vuông tại I có

EF=\(\sqrt{EI^2+IF^2}\)=\(\sqrt{4,9^2+7,1^2}\)=8,6

sinF=\(\dfrac{EI}{EF}\)=\(\dfrac{4,9}{8,6}\)\(\Rightarrow\widehat{EFI}\)=34