高考数学100个热点题型秒解技巧之高考中常见的线性规划题型完整汇总

化难为易化繁为简2019年4月版——选择、填空篇——◆例（2016山东理7）函数f(x)(3sinxcosx)(3cosxx)的最小正周期是()22【秒解】根据口诀：和差不变,积商减半,易知2四大特色助快速解题◎100个秒解技巧◎80个精妙二级结论◎10年高考真题为例◎700个例题深入剖析12f速解“奇函数＋Cf1aS与nnn210022、线性规划高考中常见的线性规划题型完整汇总,x、y、如x2x3y20而2x3y102x3y10.;.axbyb0axbyzz;z中l：0zaxbyb中0laxbyzz.：0z2xyz3xy①②;zxayxy2x③z22x1④zy2xyzz|xy2|等⑤⑥xy22.2x3y30yz2xy◆例1理x,2x3y30则y30）．15．9．1．9,32x3y302x3y30,2x3y302x3y30,y30y30,.2xy,y则l2xy0B(在z0选yAx2x3y302x3y30y30BCxy10◆例2x,y则zxy.xy30x3y303xy,zy3xzy3xC(y3xz,得zz3xy.xy10Bx3y30CxDxy30x0yy30则zx2y()◆例3xx2y0C.)D.[4,).所以直线过点C(选D．yx2y01yx2C4xxy30xya,xay◆例4I文,xy且zyx则a()55或33yxy1a1a1(,)22xzxaya1xy1a1a1a11由,x,y.(①当az,)xya2222xayAa1a(a7a2a02,22aa5.②当a1选xy◆例5y2x3y则xy2()2x(xa)(yb)P(x,y)Q(a,b)zz22.,yxA(3,xy((x0)(y0))(x,y),222225(xy)(0)(1)选22222yx◆例6,y则s(x1)(y1).xy422y2ABC(1,1)B(8,2)．x10y则◆例71理yxy0.xxy40yazP(x,y)zQb,a)xby,xAy故点xy4ACBOx014x0◆例8、yy,xxy12x2y3x1则z.x2y3y1令ty112z,t与x1x1x1113故z．,t42y6x2y0xy◆例9y则）2y20x2xy2x05552,52102[[[,][]10255aby(x,y)xa,b2(x,y)与的cos=|a||b|y2xy11x222222cos如上图可知450∴2cos即2即2cos选yx2y2011)24y3xx25y25求z|3x4y3|◆例xx2y0x1|34|34z|3x4y3|54y30而(x,y)3x55则zz5.y2xz,y2x,y2x,2xzy2xzzyzz2xz2即z25.y得2217yz2AxO20-2y0xa则xy◆例x2xyyxz,2z21xzyyx,zzy在24.yxy0x(xy0yx2z210xy◆例x,y214则xy.xyxy6252492C.12D.14:A2,6,B4,2若xyAB或.yBA1Cy=-2x+7Oxx2x10当x52252y5即若P上则xyy=-2x+10.若PAB,P152xx712x.则在当xx,0,2228252y5AC选x4y3OPOA5y3x◆例(x,y)P及则x10.OA.OPOA·cosAOPOP在OPOAx4y3，，M2)由OA3x5