求数学手写答案
\u6570\u5b66\uff0c\u6c42\u7b54\u6848(\u9ebb\u70e6\u6700\u597d\u624b\u5199)绛旓細锛1锛 CD//AB 鍥犱负 鈻矨OB涓庘柍COD鏄綅浼煎浘褰 鎵浠 鈭燗=鈭燙 鎵浠 CD//AB锛堝唴閿欒鐩哥瓑锛屼袱鐩寸嚎骞宠锛夛紙2锛 鍥犱负OA:OC=1:3锛孉C=12 鎵浠C=12梅锛1+3锛*3=9 OA=12梅锛1+3锛=3
绛旓細鍘熷紡=2-1+0-1=0 鍘熷紡=鏍瑰彿3/2+鏍瑰彿2/4 锛堟噦鐨勫惂锛侊級婊℃剰璇烽噰绾筹紝璋㈣阿锛
绛旓細1+sin2x+cos2x/1+sin2x-cos2x =锛2sinxcosx+2cos^2x锛/(2sinxcosx+2sin^2x)=(tanx+1)/(tanx+tan^2x)=(3+1)/(3+9)=1/3 娉:1+cos2x=2cos^2x 1-cos2x=2sin^2x
绛旓細鐪嬪浘锛岋紙1锛夋湭鎺ユ眰灏辨槸浜嗭紝闈炲父绠鍗
绛旓細f '(x)=3ax²-b-1/x锛宖 '(1)=0锛3a-b-1=0锛宐-1=3a-2锛屼护g(a)=3a-2-lna锛屽垯g '(a)=3-1/a锛0 1/3鏃秅 '(a)>0锛屽锛屽洜姝(a)鏈灏忓间负g(1/3)=ln3-1>0锛屾墍浠ユ亽鏈 3a-2>lna銆傞 B
绛旓細f(x)=sinxcosx+sinxsinx=1/2sin2x+(1-cos2x)/2=1/2+1/2sin2x-1/2cos2x =1/2+鏍瑰彿2/2(cospi/4sinx-sinpi/4cos2x)=1/2+鏍瑰彿2/2sin(x+pi/4)鏈灏忔鍛ㄦ湡=2pi 鐢诲浘鐭ワ紝鏇茬嚎鐢眣=sinx宸︾Щpi/4锛屽啀涓婄Щ1/2寰楀埌 0<=x<=pi/2 鎵浠 x=pi/4鏃, f(x)鏈澶=1/2+1=3/2 ...
绛旓細鏍瑰彿y锛塣2=0 (鏍瑰彿x-3鍊嶆牴鍙穣锛夛紙鏍瑰彿x+鏍瑰彿y锛=0 鍥犱负鏍瑰彿x锛0锛屾牴鍙穣锛0 鎵浠ユ牴鍙穢+鏍瑰彿y锛0 鍒欐牴鍙穢-3鍊嶆牴鍙穣=0 鏍瑰彿x=3鍊嶆牴鍙穣 鍗硏=9y 鎺ヤ笅鏉ュ皢x=9y浠ｅ叆鍒版墍姹傜殑寮忓瓙涓究鍙 濡傛灉浣犺鍙垜鐨勫洖绛旓紝鏁鍙婃椂閲囩撼 鍦ㄦ垜鍥炵瓟鐨勫彸涓婅鐐瑰嚮銆愰噰绾绛旀銆戣嫢鏈夌枒闂紝鍙户缁拷闂紝璋㈣阿 ...
绛旓細瑙ｅ涓嬪浘鎵绀
绛旓細DE=AB=DC 瑙扐DE+瑙扚DC=瑙扚DC+瑙扗CF=90搴 鎵浠ヨADE=瑙扗CF 瑙扐DE+瑙扐ED=瑙扚DC+瑙扗CF=90搴 涓婇潰鏈夎ADE=瑙扗CF 鎵浠ユ湁 瑙扐ED=瑙扚DC 鏍规嵁鍏ㄧ瓑涓夎褰㈡潯浠讹紝瑙掕竟瑙掕瘉鏄庝簡锛屼笁瑙掑舰ADE鍏ㄧ瓑涓夎褰CD 鎵浠D=FC 寰楄瘉
绛旓細鍥炵瓟锛氬洜涓簂gx/lg5=0 鎵浠 lgx=0 x1=1 璁緔=3鐨剎娆 鍗虫湁 y鐨勫钩鏂=2y+3 (y-3)(y+1)=0 鏈:y=3 y=-1(涓嶅瓨鍦) 鍗:x2=1 鎵浠:鍜屼负2
