1樓:徐少
2(a+b)
解析:(√a+√b)2+(√a-√b)2
=[a+b+2√(ab)]+[a+b-2√(ab)]=2(a+b)
求20道二次根式計算題,含過程和答案
2樓:樊老溼
1.x(2-√2)=(√2-2)
x(2-√2)=(√2-2)
x(2-√2)=-1(2-√2)
x=12.(√5+√
2)2-(√5-√2)2
=(√5+√2+√5-√2)(√5+√2-√5+√2)=2√5*2√2
=4√10
3.√8+3√(1/3)-1/(√2)+(√3)/2=2√2+3*(√3)/3-(√2)/2+(√3)/2=2√2+√3-(√2)/2+(√3)/2=(3/2)√3+(3/2)√2
4.(√3+√2+√5)(√3-√2-√5)=(√3+√2+√5)[√3-(√2+√5)]=(√3)2-(√2+√5)2
=3-(2+5+2√10)
=3-7-2√10
=-4-2√10
5.√8-2√32+√50
=5*3√2-2*4√2+5√2
=√2(15-8+5)
=12√2
6.√6-√3/2-√2/3
=√6-√6/2-√6/3
=√6/6
7. (√45+√27)-(√4/3+√125)=(3√5+3√3)-(2√3/3+5√5)=-2√5+7√5/3
8.(√4a-√50b)-2(√b/2+√9a)=(2√a-5√2b)-2(√2b/2+3√a)=-4√a-6√2b
9.(√32-3√3)(4√2+√27)
=(4√2-3√3)(4√2+3√3)
=(4√2)^2-(3√3)^2
=32-27
=5 10.1+√2-√3)(1-√2+√3)=[1+(√2-√3)][1-(√2-√3)]=1-(√2-√3)^2
=1-(2+3+2√6)
=-4-2√6
美女,10道夠不夠啊...球採納啊!!
求20道二次根式的計算題附答案盒過程,不要知道里有過的!!!
3樓:莪與幸福有個約
1.x(2-√
2)=(√2-2)
x(2-√2)=(√2-2)
x(2-√2)=-1(2-√2)
x=12.(√5+√2)2-(√5-√2)2=(√5+√2+√5-√2)(√5+√2-√5+√2)=2√5*2√2
=4√10
3.√8+3√(1/3)-1/(√2)+(√3)/2=2√2+3*(√3)/3-(√2)/2+(√3)/2=2√2+√3-(√2)/2+(√3)/2=(3/2)√3+(3/2)√2
4.(√3+√2+√5)(√3-√2-√5)=(√3+√2+√5)[√3-(√2+√5)]=(√3)2-(√2+√5)2
=3-(2+5+2√10)
=3-7-2√10
=-4-2√10
5.√8-2√32+√50
=5*3√2-2*4√2+5√2
=√2(15-8+5)
=12√2
6.√6-√3/2-√2/3
=√6-√6/2-√6/3
=√6/6
7. (√45+√27)-(√4/3+√125)=(3√5+3√3)-(2√3/3+5√5)=-2√5+7√5/3
8.(√4a-√50b)-2(√b/2+√9a)=(2√a-5√2b)-2(√2b/2+3√a)=-4√a-6√2b
9.(√32-3√3)(4√2+√27)
=(4√2-3√3)(4√2+3√3)
=(4√2)^2-(3√3)^2
=32-27
=5 10.1+√2-√3)(1-√2+√3)=[1+(√2-√3)][1-(√2-√3)]=1-(√2-√3)^2
=1-(2+3+2√6)
=-4-2√6
美女,10道夠不夠啊...球採納啊!!
求20道二次根式習題及20道一元二次方程,都要帶答案,儘量簡單的。
4樓:匿名使用者
15√62616964757a686964616fe58685e5aeb9313333326363368-2√32 √50 =5*3√2-2*4√2 5√2 =√2(15-8 5) =12√2 2√6-√3/2-√2/3 =√6-√6/2-√6/3 =√6/6 3(√45 √27)-(√4/3 √125) =(3√5 3√3)-(2√3/3 5√5) =-2√5 7√5/3 4(√4a-√50b)-2(√b/2 √9a) =(2√a-5√2b)-2(√2b/2 3√a) =-4√a-6√2b 5√4x*(√3x/2-√x/6) =2√x(√6x/2-√6x/6) =2√x*(√6x/3) =2/3*|x|*√6 6(x√y-y√x)÷√xy =x√y÷√xy-y√x÷√xy =√x-√y 7(3√7 2√3)(2√3-3√7) =(2√3)^2-(3√7)^2 =12-63 =-51 8(√32-3√3)(4√2 √27) =(4√2-3√3)(4√2 3√3) =(4√2)^2-(3√3)^2 =32-27 =5 9(3√6-√4) =(3√6)^2-2*3√6*√4 (√4)^2 =54-12√6 4 =58-12√6 10(1 √2-√3)(1-√2 √3) =[1 (√2-√3)][1-(√2-√3)] =1-(√2-√3)^2 =1-(2 3 2√6) =-4-2√6
二次根式計算題30道帶答案
5樓:櫻空雪
1/6√
1又3/5×(-5√3又√3/5)
=1/6√(8/5)×(-5/3√(3/5)=-5/18√(24/25)
=-5/18×2/5√6
=-1/9√6
(2)√8/a×√2a/b
=√(8/a×2a/b)
=√(16/b)
=4/b(√b)
(3)√2x乘以√2y乘以√x
=√(2x*2y*x)
=2x√y
(4)2√a÷4√b
=√a/2√b
=1/2b√ab
(5)5√xy÷√5x^3
=5√(xy/5x3)
=1/x√5y
(6)√x-y÷√x+y
=1/(x+y)√(x2-y2)
(7)√x(x+y)÷√xy^2/x+y(x>0,y>0)=√[x(x+y)÷xy2/(x+y)]
=(x+y)/y
(8)√xy乘以√6x÷√3y
=√6x2y÷√3y
=x√2
(9)(√mn-√m/n)÷√m/n(n>0)=√mn÷m/n-√m/n÷m/n
=n-1(10)√3/8-(-3/4√27/2+3√1/6)=1/4√6+3/8√6-1/2√6
=1/8√6
(11)2/3√9x+6√x/4-2x√1/x=2√3x+3/2√x-2√x
=5/2√x
(12)2/a√4a+√1/a-2a√1/a^3=1/a√a+1/a√a-2/a√a
=0(13)√0.2m+1/m√5m^3-m√125/m=1/5√5m+√5m-5√5m
=-19/5√5m
(14)√a+b/a-b-√a-b/a+b-√1/a^2-b^2(a>b>0)
=1/(a-b)√(a2-b2)-1/(a+b)√(a2-b2)-1/(a2-b2)√(a2-b2)
=(a+b-a+b-1)/(a2-b2)√(a2-b2)=(2b+1)/(a2-b2)√(a2-b2)解不等式
(15)2x+√32 2x-x<√2-4√2 x<-3√216)√3/8-(-3/4√27/2+3√1/6)=1/2√3/2 + 9/4√3/2 - 1/2√6=1/4√6 + 9/8√6 - 1/2√6=7/8√6 (17)√0.2m+1/m√5m^3-m√125/m=√1/5*m + 1/m√5m*m^2 - m√25*5m/m^2=1/5√5m+√5m-5√5m =-19/5√5m (18)(√45+√27)+(√1又1/3-√125)=3√5+3√3 + √4/3-5√5 =3√3 + 2/3√3 + 3√5 - 5√5=5√3 -2√5 (19)2/3√9x+6√x/4-2x√1/x=2√x+3√x-2√x =3√x 20 √40÷√5 =√8*√5÷√5 =√8=2√2 21 √32/√2 =√16*√2/√2 =√16 =422 √4/5÷√2/15 =√4/5*√15/2 =√(4/5*15/2) =√623 2√a^3b/√ab =2√a2√ab/√ab =2√a2 =2|a|(24)√18-√32+√2 =√2×9-√4×4×2+√2 =3√2-4√2+√2 =0(25)√75-√54+√96-√108=√5×5×3-√6×3×3+√6×4×4-√3×6×6=5√3-3√6+4√6-6√3 =√6-√3 =√3(√2-1) (26)(√45+√18)-(√8-√125)=√5×3×3+√2×3×3-√2×2×2+√5×5×5=3√5+3√2-3√2+5√5 =8√5 (27)1⁄2(√2+√3)-3⁄4(√2+√27)=1⁄4(2√2+2√3-√2-√27)此處通分,分子不變,分母都分別乘進去了,因為不好寫就省略了 =1⁄4(2√2+2√3-√2-√3×3×3)=1⁄4(√2-√3) (28)1⁄4根號下18ab×(-2/b根號下6a2/a)=1/4×(-2/b)×√(18ab×6a2/a)=-1/(2b)×3a√(2b) =-3a/(2b) √(2b) (29)根號下50a2b(a<0,b>0)=√(25a2×2b) =-5a√(2b) (30)根號18×3/2根號20×(-1/3根號15)=-1/3×3/2×√(18×20×15)=-1/2×√5400 =-1/2×30√6 =-15√6 6樓:匿名使用者 ni shi liu zhong de? 14 因為二次根式要有意義 x 6 5 x x 5 原式 x 6 x 5 x 解得x 113 原式 x x 2 5 x x 5 x 5 16 因為版面積為48 所以邊長為4 3再減去權面積為3的正方形,即邊長減去2 3 2 3且高為 3 容積為 2 3 2 3 3 12 3.6 x 2 4x 4化簡... 解方程 62616964757a686964616fe78988e69d83313333326330333 x 1 2 x 求x注 x全部不在根號內 1 2x 2 10 9x 2 1 4x 2 10 9x 2 49 36x 2 若x 0,7 6x 若x 0,7 6x a 4mb 2n 1 a 2mb... 解 m n 2，mn 3 解方程組得 m1 1 m2 3 x在正半軸，不合題意捨去 由m 1,得n 3 故a，b兩點座標分別為 1,0 3,0 又c是 0,3 知道3點，其中兩點是與x軸相交的。可以用交點式來求解析式 交點式是 y a x x1 x x2 代入 得 3 a 0 1 0 3 得a 1再...二次根式，應用題，填空題，難題，求20道 二次根式 題目
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