$b)$trong $(SAD)$kẻ $AH\bot SD=H$

$AC\bot BD,AC \bot SB,SB \cap BD=B,SB,BD\subset (SBD)$

$=>AC\bot (SBD)$

$=>AC\bot SD$

mà $AH\bot SD,AC\cap AH =A, AC,AH\subset (AHC)$

$=>SD\bot (AHC)$

$=>SD\bot HC$

$=>((SAD),(SCD))=(AH,HC)$

$\Delta SAB=\Delta SBC(cgc)$

$=>SA=SC$

$\Delta SAD= \Delta SBC(ccc)$

$=>AH=HC$

gợi ý làm nốt nhé. dựa vào $\Delta SAD$ vuông tại $A$ có$\frac{1}{AH^2}=\frac{1}{SA^2}+\frac{1}{AD^2}$

$AC=a\sqrt2$

$cos\widehat{AHC}=\frac{AH^2+CH^2-AC^2}{2AH.HC}$

$XONG$

$XONG$

$3(sin^2x-cos^2x)(sin^6x+sin^4xcos^2x+sin^2xcos^4x+cos^6x)+4(cos^2x-sin^2x)(cos^4x+cos^2xsin^2x+sin^4x)+4sin^4x$

$=-3cos2x(sin^6x+3sin^2xcos^2x+cos^6x-2sin^2xcos^2x)+4coss2x(cos^4x+sin^4x+2sin^2xcos^2x-sin^2xcos^2x)+6sin^4x$

$=-3cos2x[(sin^2x+cos^2x)^3-2sin^2xcos^2x]+4cos2x[(cos^2x+sin^2x)-sin^2xcos^2x]+6sin^4x$

$=-3cos2x(1-2sin^2xcos^2x)+4cos2x(1-sin^2xcos^2x)+6sin^4x$

$=cos2x-2sin^2xcos^2x+6sin^4x$

$=(1-2sin^2x)-2sin^2x(1-sin^2x)+6sin^4x$

$=1-2sin^2x-2sin^2x+2sin^4x+6sin^4x$

$=1-4sin^2x+8sin^4x$

$\mathop {\lim }\limits_{x \to 1}\frac{x^3-1+x^2-1+x-1}{(x-1)(x-2)}$

$=\mathop {\lim }\limits_{x \to 1}\frac{x^2+x+1+x+1+1}{x-2}$

$=\mathop {\lim }\limits_{x \to 1}\frac{x^2+2x+3}{x-2}$

$=\frac{1+2+3}{1-2}=-6$

$\mathop {\lim }\limits_{x \to 1^+}\frac{x-2}{x+1}=\frac{1-2}{1+1}=\frac{-1}{2}$

$XONG$

$b)\mathop {\lim }\limits_{x \to 1}\frac{x^3-6x^2+11x-6}{x^2-3x+2}$

$=\mathop {\lim }\limits_{x \to 1}\frac{(x-1)(x^2-5x+6)}{(x-1)(x-2)}$

$=\mathop {\lim }\limits_{x \to 1}\frac{x^2-5x+6}{x-2}$

$=\mathop {\lim }\limits_{x \to 1}(x-3)$

$=1-3=-2$

$a)\mathop {\lim }\limits_{x \to \sqrt3}\left| {x^2-8} \right|=\left| {3-8} \right|=5$

có $5+6+7=18$ viên

có $C^8_{18}$cách chọn 8 viên bi

có $C^8_{11}+C^8_{12}+C^8_{13}$ cách chọn 8 viên bi có $2$ màu

$=>C^8_{18}-C^8_{11}-C^{8}_{12}-C^{8}_{13}$ cách chọn 8 viên bi có đủ $3$ màu

$\mathop {\lim }\limits_{x \to +\infty}u_n=\mathop {\lim }\limits_{x \to +\infty}\frac{\sqrt{1+\frac{2}{x^6}}}{3-\frac{1}{x^2}}=\frac{1}{3}$

$XONG$