1) Ta có: \(\dfrac{1}{\sqrt{3}-1}+\dfrac{1}{4+2\sqrt{3}}-\dfrac{2}{\sqrt{3}}-\dfrac{3}{2}\)

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

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

\(=-\dfrac{2\sqrt{3}}{3}\)

3) Ta có: \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}-\dfrac{3\sqrt{5}}{4}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\sqrt{5}}{4}\)

\(=\dfrac{2\sqrt{5}-3\sqrt{5}}{4}\)

\(=\dfrac{-\sqrt{5}}{4}\) 

a: =1/2*2/3*...*9/10=1/10

b: =17/32*32/11*11/34=1/2

\(2x+\frac{\pi}{6}=\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow2x=\frac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=\frac{\pi}{6}+\frac{k\pi}{2}\)

Bài 5 hình 1: (tự vẽ hình nhé bạn)
a) Xét ΔABD và ΔACB ta có:
\(\widehat{BAD}\)= \(\widehat{BAC}\) (góc chung)
\(\widehat{ABD}\)= \(\widehat{ACB}\) (gt)
=> ΔABD ~ ΔACB (g-g)
=> \(\dfrac{AB}{AC}\) = \(\dfrac{BD}{CB}\) = \(\dfrac{AD}{AB}\) (tsđd)
b) Ta có: \(\dfrac{AB}{AC}\) = \(\dfrac{AD}{AB}\) (cm a)
=> \(AB^2\) = AD.AC
=> \(2^2\) = AD.4
=> AD = 1 (cm)
Ta có: AC = AD + DC (D thuộc AC)
      => 4   =   1   + DC
      => DC = 3 (cm)
c) Xét ΔABH và ΔADE ta có: 
   \(\widehat{AHB}\) = \(\widehat{AED}\) (=\(90^0\))
   \(\widehat{ADB}\) = \(\widehat{ABH}\) (ΔABD ~ ΔACB)
=> ΔABH ~ ΔADE
=> \(\dfrac{AB}{AD}\) = \(\dfrac{AH}{AE}\) = \(\dfrac{BH}{DE}\) (tsdd)
Ta có: \(\dfrac{S_{ABH}}{S_{ADE}}\) = \(\left(\dfrac{AB}{AD}\right)^2\)= \(\left(\dfrac{2}{1}\right)^2\)= 4
=> đpcm

Tiếp bài 5 hình 2 (tự vẽ hình)
a) Xét ΔABC vuông tại A ta có:
\(BC^2\) = \(AB^2\) + \(AC^2\)
\(BC^2\) = \(21^2\) + \(28^2\)
BC = 35 (cm)
b) Xét ΔABC và ΔHBA ta có:
\(\widehat{BAC}\) = \(\widehat{AHB}\) ( =\(90^0\))
\(\widehat{ABC}\) = \(\widehat{ABH}\) (góc chung)
=> ΔABC ~ ΔHBA (g-g)
=> \(\dfrac{AB}{BH}\) = \(\dfrac{BC}{AB}\) (tsdd)
=> \(AB^2\) = BH.BC
=> \(21^2\) = 35.BH
=> BH = 12,6 (cm)
c) Xét ΔABC ta có:
BD là đường p/g (gt)
=> \(\dfrac{AD}{DC}\) = \(\dfrac{AB}{BC}\) (t/c đường p/g)
Xét ΔABH ta có: 
BE là đường p/g (gt)
=> \(\dfrac{HE}{AE}\) = \(\dfrac{BH}{AB}\) (t/c đường p/g)
Mà: \(\dfrac{AB}{BC}\) = \(\dfrac{BH}{AB}\) (cm b)
=> đpcm
d) Ta có: \(\left\{{}\begin{matrix}\widehat{HBE}+\widehat{BEH}=90^0\\\widehat{ABD}+\widehat{ADB=90^0}\\\widehat{HBE}=\widehat{ABD}\end{matrix}\right.\)
=> \(\widehat{BEH}=\widehat{ADB}\)
Mà \(\widehat{BEH}=\widehat{AED}\) (2 góc dd)
Nên \(\widehat{ADB}=\widehat{AED}\)
=> đpcm

B

B

B

suy ra 1 trong 2 so la so am 

ma (x^2-8) lon hon (x^2-39) nen (x^2 -39) la am

xong thi tu lap luan

\(\left(x^2-8\right)\left(x^2-39\right)< 0\)

TH1 : Cả 2 cùng âm

\(\hept{\begin{cases}x^2-8< 0\\x^2-39< 0\end{cases}=>\hept{\begin{cases}x< 2,828.....\\x< 6,244........\end{cases}}}\)

TH2 : Cả 2 cùng dương

\(\hept{\begin{cases}x^2-8>0\\x^2-39>0\end{cases}=>\hept{\begin{cases}x>2,82.......\\x>6,244......\end{cases}}}\)

8-3x0+4:2
=8-0+2
=10

8-3x0+4:2

=8-3x0+2

=5x0+2

=0+2

=2

tick cho minh nha

8-3x0+4:2=8-0+2

=10

Mấy bài toán nào vậy

b chép đề ra đi để coi giải đc ko