8.已知三角形一边长为 sqrt (2)cm, 这条边上的高为 sqrt (12)cm, 求该三角形的面积-|||-综合、运用、设断-|||-一、填空题-|||-后-|||-定义运算 https:/img.zuoyebang.cc/zyb_bf73b8ba0938757c1e3c7ab42a238df7.jpg+4=1 的运算法则为:x@ =sqrt (xy+4) 则2000-|||-10.已知矩形的长为 sqrt (5)cm, 宽为 sqrt (10)cm, 则面积为_ m-|||-11.比较大小 (1)3sqrt (2) __ sqrt (3) (2)5sqrt (2) __ sqrt (3)= (3)-2sqrt (2) __ sqrt (6)-|||-二、选择题-|||-12.若 sqrt ({a)^2b}=-asqrt (b) 立,则a,b满足的条件是( 在A-|||-A.lt 0且bgt 0 B leqslant 0且bgt 0 c lt 0|bgt 0| D.a b是号-|||-bigcirc (13)approx 4sqrt (2dfrac {3)(4)} 号外的因式移进根号内,结果等于(D)-|||-A. =sqrt (11) B. sqrt (18) c =sqrt (44) D sqrt (11)-|||-三、解答题-|||-14.计算: (1)5sqrt (3xy)cdot 3sqrt (6x)= __ (2) sqrt (27{a)^2+9(a)^2(b)^3}= __-|||-③ sqrt (12)-sqrt (2dfrac {2)(3)}cdot sqrt (1dfrac {1)(2)}= __ (4) sqrt (3)-(sqrt (3)+sqrt (12))= __-|||-15 ((x-y+2))^2=sqrt (x+y-2) 为相数。 (x+y)' 的值.-|||-解 ((x-y+2))^2+sqrt (xy-2)=0-|||-(- +(y))^n=((0.2))^n-|||--y+2=0 100-|||-+y-2=0-|||-拓广、探究。用考-|||-16.化简: (1)((sqrt {2)+1)}^10((sqrt {2)-1)}^10= __-|||-(2)sqrt ((sqrt {3)+1)cdot (sqrt (3)-1)}=-|||-1.把下列各式化成简二次极式-|||-(1)sqrt (12)= __, (2)sqrt (18x)= __, (4)sqrt (dfrac {y)(x)}=underline ( ) ___ -|||-cos 50sqrt (dfrac {2)(3)}= __00 sqrt (4dfrac {1)(2)}= ___ (7)sqrt ({x)^2+3(x)^2}= () __-|||-→(60),-|||-2.在横线上填出一个最简单的国式,使得它与所给二次短式相乘的结果为有理式,如 sqrt (2)-|||-sqrt (2)-|||-cos 2sqrt (3)= __-|||-(2)sqrt [3](32) __-|||-(3)sqrt (3a) __-|||-(4)sqrt (3{a)^2}= __-|||-_-|||-(5)sqrt (3{a)^2}= __-|||-二、选择题-|||-其sqrt (dfrac {1-x)(x)}+dfrac (sqrt {1-x)}(sqrt {x)} 立的条件是( )-|||-A. xlt xlt 1-|||-A.-|||-A. __ B. sqrt (dfrac {2y)(3x)}=dfrac (1)(3x)sqrt (4xy)-|||-C. sqrt ({(dfrac {1)(4))}^2-((dfrac {1)(5))}^2}=dfrac (1)(20) D.-|||-5.把 sqrt (dfrac {1)(32)} 元成-|||-A. sqrt (72) B. dfrac (1)(12)sqrt (32) C dfrac (1)(8)sqrt (2) D dfrac (1)(4)sqrt (2)-|||-三、计算题-|||-. (1)sqrt (dfrac {16)(25)}= 。sqrt (2dfrac {7)(9)}= o dfrac (sqrt {24)}(sqrt {3)}= (4) =5sqrt (75)+2sqrt (125)-|||-(5) )6sqrt (6)+3sqrt (3) (7) (n)sqrt (dfrac {1)(2)}+dfrac (1)(2)sqrt (0.125)-|||-综合,运用、诊断-|||-一、填空题-|||-7.化简二次根式: (1)sqrt (2)times sqrt (6)= (2)sqrt (dfrac {1)(8)}= (3)-sqrt (4dfrac {1)(3)}=-|||-8.计算下列各式,使得结果的分母中不含有二次应式:-|||-(1) dfrac (1)(sqrt {3)}= (2) sqrt (dfrac {2)(x)}- (3)dfrac (sqrt {2)}(2sqrt {3)}= )dfrac (x)(sqrt {5y)}=-|||-9.已知 sqrt (3)approx 1.732 __ sqrt (27)= __ (结果精确到0.001)-|||-二、选择题-|||-10.已知 =sqrt (3)+1. =dfrac (2)(sqrt {3)-1} 则a与b的关系为( )-|||-B.ab C.-|||-B.ab C.-|||-11.下列各式中,最简次根式是( ) a=-b D =-1-|||-A. sqrt (dfrac {1)(x-y)} B C sqrt ({x)^2+4} D. sqrt (5{a)^2b}-|||-三、解答题-|||-12.计算 (1)sqrt (dfrac {b)(a)}+sqrt (ab)times sqrt (dfrac {{a)^3}(b)} (2)sqrt (12xy)+dfrac (2)(3)sqrt (y) o dfrac (a+b)(sqrt {a-b)}