課時(shí)規(guī)范練33基本不等式及其應(yīng)用基礎(chǔ)鞏固組1.下列不等式一定成立的是()A.lgx2+>lg x(x>0)B.sin x+1sinx2(xk,kZ)C.x2+12|x|(xR)D.1x2+1<1(xR)2.若a,b都是正數(shù),則1+1+4ab的最小值為()A.7B.8C.9D.103.已知a>0,b>0,a+b=2,則y=1a+4b的最小值是()A.B.4C.D.54.(2018江西南昌測(cè)試三,10)若正數(shù)x,y滿足x+4y-xy=0,則3x+y的最大值為()A.B.C.D.15.(2018江西新余四中適應(yīng)性考試,9)設(shè)正數(shù)x,y滿足x>y,x+2y=3,則1x-y+9x+5y的最小值為()A.B.3C.D.2336.(2018遼寧遼南協(xié)作校一模擬,6)若lg a+lg b=0且ab,則2a+1b的取值范圍為()A.22,+)B.(22,+)C.22,3)(3,+)D.(22,3)(3,+)7.(2018天津十二中學(xué)聯(lián)考一,12)已知a>b>0,則2a+3a+b+2a-b的最小值為()A.22+23B.2+3C.22+3D.2+328.(2018河北唐山遷安三中期中,9)設(shè)x,y均為正實(shí)數(shù),且32+x+32+y=1,則xy的最小值為()A.4B.43C.9D.169.若對(duì)于任意x>0,xx2+3x+1a恒成立,則a的取值范圍是.10.已知x,yR且滿足x2+2xy+4y2=6,則z=x2+4y2的取值范圍為.11.(2018河北唐山二模,23)已知a>0,b>0,c>0,d>0,a2+b2=ab+1,cd>1.(1)求證:a+b2;(2)判斷等式ac+bd=c+d能否成立,并說(shuō)明理由.12.已知a>0,b>0,a+b=1,求證:(1)1a+1b+1ab8;(2)1+1a1+9.綜合提升組13.(2018湖北宜昌一中適應(yīng)性考試,11)若P是面積為1的ABC內(nèi)的一點(diǎn)(不含邊界),PAB,PAC和PBC的面積分別為x,y,z,則y+zx+1y+z的最小值是()A.3B.3+23C.D.23+1314.(2018廣東廣州仲元中學(xué)期末,11)已知x,yR+,且滿足x+2y=2xy,則x+4y的最小值為()A.3-2B.3+22C.3+2D.4215.(2018湖南澧縣一中一檢,14)已知二次函數(shù)f(x)=ax2+2x+c(xR)的值域?yàn)?,+),則a+1c+c+1a的最小值為.創(chuàng)新應(yīng)用組16.(2018河南信陽(yáng)二模,11)點(diǎn)M(x,y)在曲線C:x2-4x+y2-21=0上運(yùn)動(dòng),t=x2+y2+12x-12y-150-a,且t的最大值為b,若a>0,b>0,則1a+1+1b的最小值為()A.1B.2C.3D.4參考答案課時(shí)規(guī)范練33基本不等式及其應(yīng)用1.C當(dāng)x>0時(shí),x2+2x=x,所以lgx2+lg x(x>0),故選項(xiàng)A不正確;運(yùn)用基本不等式時(shí)需保證“一正”“二定”“三相等”,而當(dāng)xk,kZ時(shí),sin x的正負(fù)不定,故選項(xiàng)B不正確;由基本不等式可知,選項(xiàng)C正確;當(dāng)x=0時(shí),有1x2+1=1,故選項(xiàng)D不正確.2.Ca,b都是正數(shù),1+1+4ab=5+4ab5+2ba4ab=9,當(dāng)且僅當(dāng)b=2a>0時(shí)取等號(hào).故選C.3.C依題意,得+=+(a+b)= 5+4ab5+2ba4ab=,當(dāng)且僅當(dāng)a+b=2,ba=4ab,a>0,b>0,即a=23,b=43時(shí)取等號(hào),即1a+4b的最小值是92.4.A因?yàn)閤+4y-xy=0,化簡(jiǎn)可得x+4y=xy,左右兩邊同時(shí)除以xy,得+=1,求3x+y的最大值,即求x+y3=+的最小值,所以+1=+=x3y+4y3x+2x3y4y3x+3,當(dāng)且僅當(dāng)x3y=4y3x時(shí)取等號(hào),所以3x+y的最大值為,所以選A.5.A因?yàn)閤+2y=3,所以2x+4y=6,所以(x-y)+(x+5y)=6,所以1x-y+9x+5y=1x-y+9x+5y6=1x-y+9x+5y(x-y) +(x+5y)= 10+x+5yx-y+9(x-y)x+5y (10+29)=,當(dāng)且僅當(dāng)x=2,y=時(shí)取最小值.故選A.6.Alg a+lg b=0且ab,lg ab=0,即ab=1.+ab=2b+a22ab=22,當(dāng)且僅當(dāng)a=2b=2時(shí)取等號(hào).+的取值范圍為22,+),故選A.7.Aa>b>0,2a+3a+b+2a-b=a+b+a-b+3a+b+2a-b,a+b+3a+b23,當(dāng)且僅當(dāng)a+b=3時(shí)取等號(hào);a-b+2a-b22,當(dāng)且僅當(dāng)a-b=2時(shí)取等號(hào).聯(lián)立a+b=3,a-b=2,解得a=3+22,b=3-22.當(dāng)a=3+22,b=3-22時(shí),a+b+a-b+3a+b+2a-b22+23,即2a+3a+b+2a-b取得最小值22+23.8.D將等式化簡(jiǎn)可得xy-8=x+y2xy,解得xy4,所以xy16,所以最小值為16.故選D.9.,+xx2+3x+1=13+x+1x,因?yàn)閤>0,所以x+1x2(當(dāng)且僅當(dāng)x=1時(shí)取等號(hào)),則13+x+1x13+2=15,即xx2+3x+1的最大值為15,故a15.10.4,122xy=6-(x2+4y2),而2xyx2+4y22,6-(x2+4y2)x2+4y22,x2+4y24(當(dāng)且僅當(dāng)x=2y時(shí)取等號(hào)).(x+2y)2=6+2xy0,即2xy-6,z=x2+4y2=6-2xy12(當(dāng)且僅當(dāng)x=-2y時(shí)取等號(hào)).綜上可知4x2+4y212.11.(1)證明 由題意得(a+b)2=3ab+13a+b22+1,當(dāng)且僅當(dāng)a=b時(shí),取等號(hào).解得(a+b)24,又a,b>0,所以a+b2.(2)解 不能成立.ac+bda+c2+b+d2,因?yàn)閍+b2,所以ac+bd1+c+d2,因?yàn)閏>0,d>0,cd>1,所以c+d=c+d2+c+d2c+d2+cd>c+d2+1,故ac+bd=c+d不能成立.12.證明 (1)a+b=1,a>0,b>0,1a+1b+1ab=1a+1b+a+bab=21a+1b=2a+ba+a+bb=2ba+ab+44baab+4=8當(dāng)且僅當(dāng)a=b=12時(shí),等號(hào)成立,1a+1b+1ab8.(2)1+1a1+1b=1a+1b+1ab+1,由(1)知1a+1b+1ab8.1+1a1+1b9.13.Ax+y+z=1,y+zx+1y+z=1-xx+11-x=1-xx+x+1-x1-x=1-xx+x1-x+121-xxx1-x+1=3,當(dāng)且僅當(dāng)x=時(shí)取等號(hào),y+zx+1y+z的最小值為3,故選A.14.B由題意可得(2y-1)(x-1)=1,變形為(x-1)(4y-2)=2,所以2=(x-1)(4y-2)x+4y-32,所以x+4y22+3,當(dāng)且僅當(dāng)x-1=4y-2時(shí),等號(hào)成立,即x=2+1,y=2+24,選B.15.4由題意知,a>0,=4-4ac=0,ac=1,c>0,則a+1c+c+1a=+=+2+21ac=2+2=4,當(dāng)且僅當(dāng)a=c=1時(shí)取等號(hào).a+1c+c+1a的最小值為4.16.A曲線C:x2-4x+y2-21=0可化為(x-2)2+y2=25,表示圓心為A(2,0),半徑為5的圓.t=x2+y2+12x-12y-150-a=(x+6)2+(y-6)2-222-a,(x+6)2+(y-6)2可以看作點(diǎn)M到點(diǎn)N(-6,6)的距離的平方,圓C上一點(diǎn)M到N的距離的最大值為|AN|+5,即點(diǎn)M是直線AN與圓C的離點(diǎn)N最遠(yuǎn)的交點(diǎn),所以直線AN的方程為y=-34(x-2),由y=-34(x-2),(x-2)2+y2=25,解得x1=6,y1=-3或x2=-2,y2=3(舍去),當(dāng)x=6,y=-3時(shí),t取得最大值,且tmax=(6+6)2+(-3-6)2-222-a=b,a+b=3,(a+1)+b=4,1a+1+1b=141a+1+1b(a+1)+b=14ba+1+a+1b+21,當(dāng)且僅當(dāng)ba+1=a+1b,且a+b=3,即a=1,b=2時(shí)等號(hào)成立.故選A.