a.

\(SA\perp\left(ABCD\right)\Rightarrow d\left(S;\left(ABCD\right)\right)=SA=7\left(cm\right)\)

b.

\(CD||AB\left(gt\right)\Rightarrow DC||\left(SAB\right)\)

\(\Rightarrow d\left(DC;\left(SAB\right)\right)=d\left(D;\left(SAB\right)\right)\)

Ta có: \(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp AD\\AB\perp AD\left(gt\right)\end{matrix}\right.\) \(\Rightarrow AD\perp\left(SAB\right)\)

\(\Rightarrow d\left(D;\left(SAB\right)\right)=AD=5\left(cm\right)\)

c.

\(SA\perp\left(ABCD\right)\Rightarrow AB\) là hình chiếu vuông góc của SB lên (ABCD)

\(\Rightarrow\widehat{SBA}\) là góc giữa SB và (ABCD)

\(tan\widehat{SBA}=\dfrac{SA}{AB}=\dfrac{7}{5}\Rightarrow\widehat{SBA}\approx54^028'\)

1.

\(y'=\left(2x^4\right)'+\left(4x^2\right)'+\left(1\right)'=8x^3+8x\)

2.

\(y'=\left(\dfrac{1}{3}x^3\right)'-\left(\dfrac{1}{2}x^2\right)'-\left(x\right)'+\left(3\right)'=x^2-x-1\)

3.

\(y'=\left(2x^3+1\right)'\left(4x-2\right)+\left(2x^3+1\right)\left(4x-2\right)'\)

\(=6x^2\left(4x-2\right)+4\left(2x^3+1\right)\)

\(=24x^3-12x^2+8x^3+4\)

\(=32x^3-12x^2+4\)

4.

\(y'=\dfrac{\left(x^4-x^3\right)'.\left(2x^2+1\right)-\left(2x^2+1\right)'\left(x^4-x^3\right)}{\left(2x^2+1\right)^2}\)

\(=\dfrac{\left(4x^3-3x^2\right)\left(2x^2+1\right)-4x\left(x^4-x^3\right)}{\left(2x^2+1\right)}\)

\(=\dfrac{8x^5+4x^3-6x^4-3x^2-4x^5+4x^2}{\left(2x^2+1\right)^2}\)

\(=\dfrac{4x^5-6x^4+4x^3+4x^2}{\left(2x^2+1\right)^2}\)

5.

\(y'=2cos\left(4x-\dfrac{\pi}{5}\right).\left(cos\left(4x-\dfrac{\pi}{5}\right)\right)'\)

\(=2cos\left(4x-\dfrac{\pi}{5}\right).\left(-sin\left(4x-\dfrac{\pi}{5}\right)\right).\left(4x-\dfrac{\pi}{5}\right)'\)

\(=-8.cos\left(4x-\dfrac{\pi}{5}\right).sin\left(4x-\dfrac{\pi}{5}\right)\)

\(=-4sin\left(8x-\dfrac{2\pi}{5}\right)\)

\(\lim\limits_{x\rightarrow6}f\left(x\right)=\lim\limits_{x\rightarrow6}\dfrac{x^2-36}{6-x}=\lim\limits_{x\rightarrow6}\dfrac{\left(x-6\right)\left(x+6\right)}{-\left(x-6\right)}=\lim\limits_{x\rightarrow6}\dfrac{x+6}{-1}=-12\)

\(f\left(6\right)=12\)

\(\Rightarrow\lim\limits_{x\rightarrow6}f\left(x\right)\ne f\left(6\right)\)

\(\Rightarrow\) Hàm gián đoạn tại \(x=6\)

refer

a.

P là trung điểm B'C', Q là trung diểm C'D'

\(\Rightarrow PQ\) là đường trung bình tam giác B'C'D'

\(\Rightarrow PQ||B'D'\Rightarrow PQ||BD\)

\(\Rightarrow\widehat{\left(MN;PQ\right)}=\widehat{\left(MN;BD\right)}\)

Lại có MN là đường trung bình tam giác ABC \(\Rightarrow MN||AC\)

Mà \(AC\perp BD\Rightarrow MN\perp BD\)

\(\Rightarrow\widehat{\left(MN;PQ\right)}=90^0\)

b.

\(D'P||DN\Rightarrow\widehat{\left(CM;D'P\right)}=\widehat{\left(CM;DN\right)}\)

Ta có:

\(\overrightarrow{CM}=\overrightarrow{CB}+\overrightarrow{BM}=\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{BA}\)

\(\overrightarrow{DN}=\overrightarrow{DC}+\overrightarrow{CN}=-\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{CB}\)

\(\Rightarrow\overrightarrow{CM}.\overrightarrow{DN}=\left(\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{BA}\right)\left(-\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{CB}\right)\)

\(=-\overrightarrow{CB}.\overrightarrow{BA}+\dfrac{1}{2}CB^2-\dfrac{1}{2}BA^2+\dfrac{1}{4}\overrightarrow{BA}.\overrightarrow{CB}\)

\(=\dfrac{1}{2}CB^2-\dfrac{1}{2}BA^2=0\)

\(\Rightarrow CM\perp DN\)

\(\Rightarrow\widehat{\left(CM;D'P\right)}=90^0\)

a. Do ABCD.A'B'C'D' là hình lập phương

\(\Rightarrow C'D||AB'\)

\(\Rightarrow\) Góc giữa A'B và C'D bằng góc giữa A'B và AB'

Mà \(A'B\perp AB'\) (hai đường chéo hình vuông)

\(\Rightarrow\widehat{\left(A'B;C'D\right)}=90^0\)

b.

Do \(AD'||BC'\) (t/c lập phương)

\(\Rightarrow\widehat{\left(A'B;AD'\right)}=\widehat{\left(A'B;BC'\right)}=\widehat{A'BC'}\)

Mà \(A'B=BC'=A'C'=a\sqrt{2}\) (với a là cạnh lập phương)

\(\Rightarrow\Delta A'BC'\) đều

\(\Rightarrow\widehat{A'BC'}=60^0\)

Kẻ \(AE\perp BD\) , \(AF\perp SE\Rightarrow AF\perp\left(SBD\right)\)

Dễ dàng chứng minh \(AD\perp\left(SAB\right)\) ; \(AB\perp\left(SAD\right)\) 

Từ đó ta có: \(\alpha=\widehat{FAD}\) ; \(\beta=\widehat{FAB}\) ; \(\gamma=\widehat{FAS}\)

\(\dfrac{1}{AF^2}=\dfrac{1}{SA^2}+\dfrac{1}{AE^2}=\dfrac{1}{SA^2}+\dfrac{1}{AB^2}+\dfrac{1}{AD^2}=\dfrac{2}{a^2}+\dfrac{1}{b^2}=\dfrac{a^2+2b^2}{a^2b^2}\)

\(\Rightarrow AF=\dfrac{ab}{\sqrt{a^2+2b^2}}\)

\(\Rightarrow T=cos\alpha+cos\beta+cos\gamma=\dfrac{AF}{AD}+\dfrac{AF}{AB}+\dfrac{AF}{AS}=\dfrac{ab}{\sqrt{a^2+2b^2}}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\)

\(\Rightarrow T=\dfrac{\sqrt{3}ab}{\sqrt{\left(1+2\right)\left(a^2+2b^2\right)}}\left(\dfrac{a+2b}{ab}\right)\le\dfrac{\sqrt{3}ab}{a+2b}\left(\dfrac{a+2b}{ab}\right)=\sqrt{3}\)

Dấu "=" xảy ra khi và chỉ khi \(a=b\)

\(y'=\dfrac{4.3}{8}.x^3+\dfrac{5.3}{6}x^2-\dfrac{2}{2\sqrt{x}}+3=\dfrac{3x^3}{2}+\dfrac{5x^2}{2}-\dfrac{1}{\sqrt{x}}+3\)

3-D

Đề bài yêu cầu làm gì bạn nhỉ?