a: góc C=180-110-40=30 độ

Xét ΔABC có AB/sinC=BC/sinA=AC/sinB

=>AB/sinC=BC/sinA

=>AB/sin30=12/sin110

=>\(AB\simeq6,39\left(cm\right)\)

b: BC/sinA=AC/sinB

=>AC/sin40=12/sin110

=>\(AC\simeq8,21\left(cm\right)\)

\(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne1\right)\\ =\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}+3+x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}+3+\sqrt{x}\left(\sqrt{x}-1\right)-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}-9+x-\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)

\(3,\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\\ =\sqrt{\dfrac{9}{4}.6}+\sqrt{4.\dfrac{2}{3}}-\sqrt{16.\dfrac{3}{2}}\\ =\sqrt{\dfrac{27}{2}}+\sqrt{\dfrac{8}{3}}-\sqrt{24}\\ =\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-2\sqrt{6}\\ =\dfrac{9\sqrt{6}+4\sqrt{6}-12\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)

\(4,5\sqrt{x}+6\sqrt{\dfrac{x}{4}}-x\sqrt{\dfrac{4}{x}}\\ =5\sqrt{x}+\sqrt{\dfrac{36x}{4}}-\sqrt{\dfrac{4x^2}{x}}\\ =5\sqrt{x}+\sqrt{9x}-\sqrt{4x}\\ =5\sqrt{x}+3\sqrt{x}-2\sqrt{x}\\ =6\sqrt{x}\)

a: AD=căn 6^2+8^2=10cm

b: Xét ΔBAD vuông tại B có sin BAD=BD/DA=4/5

=>góc BAD=53 độ

Xét ΔBAC vuông tại B có tan BAC=BC/BA=1/2

=>góc BAC=27 độ

a: \(A=\dfrac{2\sqrt{x}-9-x+9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

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

b: A<1

=>A-1<0

=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

=>4/(căn x-3)<0

=>căn x-3<0

=>0<=x<9 và x<>1

Hình bạn tự vẽ nha .

Xét : \(\Delta ABC\) đều có đường cao là AH.

\(\Rightarrow AH\) cũng là đường trung tuyến của \(\Delta ABC\)

\(\Rightarrow HC=\dfrac{1}{2}BC=\dfrac{1}{2}a\)

Xét \(\Delta AHC\) vuông tại H :

\(\Rightarrow AH^2=AC^2-HC^2\)

\(\Rightarrow AH^2=a^2-\dfrac{1}{4}a^2=\dfrac{3}{4}a^2\)

\(\Rightarrow AH=a\sqrt{\dfrac{3}{4}}\)

Tam giác đều ABC \(\Rightarrow A=B=C=60^o\)

⇒ Δ ABH là Δ nửa đều

\(\Rightarrow HB=\dfrac{a}{2}\Rightarrow AH=\dfrac{a\sqrt[]{3}}{2}\)

3: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)

\(=\sqrt{6}\left(\dfrac{3}{2}+\dfrac{2}{3}-1\right)=\dfrac{7}{6}\sqrt{6}\)

4: \(=5\sqrt{x}+3\sqrt{x}-2\sqrt{x}=6\sqrt{x}\)

Gọi (d): y=ax+b là phương trình đường thẳng cần tìm

(d) đi qua A(2;-2) và B(-1;3) nên ta có hệ:

2a+b=-2 và -a+b=3

=>a=-5/3 và b=4/3

camon

a: \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

=>\(13\sqrt{2x}=28\)

=>căn 2x=28/13

=>2x=784/169

=>x=392/169

b: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

=>2*căn x-5=4

=>căn x-5=2

=>x-5=4

=>x=9

c: =>\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

=>x-2=0 hoặc x+2=1

=>x=-1 hoặc x=2

\(1,5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}+\sqrt{15}\\ =\sqrt{\dfrac{25}{5}}+\sqrt{\dfrac{20}{4}}+\sqrt{5}.\sqrt{3}\\ =\sqrt{5}+\sqrt{5}+\sqrt{5}.\sqrt{3}\\ =2\sqrt{5}+\sqrt{3}.\sqrt{5}\\ =\left(2+\sqrt{3}\right).\sqrt{5}\\\)

\(2,\sqrt{\dfrac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}\\ =\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{9}{2}}+\sqrt{\dfrac{25}{2}}\\ =\dfrac{\sqrt{2}}{2}+\dfrac{3\sqrt{2}}{2}+\dfrac{5\sqrt{2}}{2}\\ =\dfrac{\sqrt{2}+3\sqrt{2}+5\sqrt{2}}{2}=\dfrac{9\sqrt{2}}{2}\)