23.17-23.14

=23.(17-14)

=23.3 = 69

23.(17-14)=23.3=69

\(d\left(G;\left(ABCD\right)\right)=\dfrac{1}{3}d\left(S;\left(ABCD\right)\right)=\dfrac{1}{3}.\dfrac{a\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{6}\)

\(S_{\Delta ACD}=\dfrac{1}{2}S_{ABCD}=\dfrac{a^2}{2}\)

\(\Rightarrow V=\dfrac{1}{3}.\dfrac{a^2}{2}.\dfrac{a\sqrt{3}}{6}=\dfrac{a^3\sqrt{3}}{36}\)

a,mấy đoạn dấu : dấu+ trong đề hơi khó nhìn

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

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

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

b, \(P>0=>\dfrac{x-1}{\sqrt{x}}>0=>x-1>0< =>x>1\)(tm)

Vậy \(x>1\) .....

\(\)

lỗi ảnh bạn ơi

trả lời nhanh cho mình và đầy đủ nhá

=\(\left(3\sqrt{3}-3\sqrt{3}+2\sqrt{6}\right):3\sqrt{3}\)
\(=1-\dfrac{\sqrt{6}}{2}+\dfrac{2\sqrt{2}}{3}\)
=\(\dfrac{6}{6}-\dfrac{3\sqrt{6}}{6}+\dfrac{4\sqrt{2}}{6}\)
=\(\dfrac{6+\sqrt{6}}{6}\)

a)\(42-15+28-5+20\)
\(=20-\left(15+5\right)+\left(42+28\right)\)
\(=20-20+70\)
\(=0+70\)
\(=70\)
b)\(\left(8\times5-40\right):\left(2+4+6+8+...+32+34\right)\)
\(=\left(40-40\right):\left(2+4+6+8+...+32+34\right)\)
\(=0:\left(2+4+6+8+...32+34\right)\)
\(=0\)

a: =42+28-20+20=70

b: =(40-40):A=0

a) 

b)

Đây nha bạn!!!!

ĐÚNG R Ạ 

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