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(1)计算:|22?3|+(%2)2+8%2sin30°;(2)解方...

(1)解:原式=-4-(π-3)+1+1=-4-π+3+1+1=-π+1.(2)原式=a(a?1)?a?1(a?1)2=a.

(1)(-2) 2 - 4 +( 1 3 ) -1 - 1 2 sin30°,=4-2+3- 1 2 × 1 2 ,=5- 1 4 ,= 19 4 ;(2)方程两边都乘以(x+1)(1-2x)得,(x-1)(1-2x)+2x(x+1)=0,整理得,5x=1,解得x= 1 5 ,检验:当x= 1 5 时,(x+1)(1-2x)=( 1 5 +1)...

(1)原式=2× 1 2 +2 3 × 3 - 1 2 =1+6- 1 2 =6 1 2 ;(2)分解因式得: 1 2 x(x+1)=0, 1 2 x=0,x+1=0,x 1 =0,x 2 =-1.

(1)原式=2×12+4×12+1=212;(2)原式=22+32-322=722;(3)∵2x2-x-2=0,∴x2-12x-1=0,∴(x-14)2=1716,∴x-14=±174,∴x1=1+174,x2=1?174.

(1)原式=3-1+12-1+12,=3-1(2)2xx+2?3x?2=2,去分母得:2x(x-2)-3(x+2)=2(x-2)(x+2),去括号得:2x2-4x-3x-6=2x2-8,移项得:2x2-2x2-3x-4x=-8+6,合并同类项得:-7x=-2,把x的系数化为1:x=27,检验:把x=27代入最简公分母(x-2)...

(1)原式=12+1?22×22?3×33=12+1?12?1=0;(2)原方程变形得:x2-9x+20=0方程左边分解因式得:(x-4)(x-5)=0∴x-4=0或x-5=0,∴x1=4,x2=5;(3)①根据题意列表如下: 1 2 3 4 1 (1,2) (1,3) (1,4) 2 (2,1) (2,3) (2,4) 3 (...

(1)原式=4-1-32+23+1-32=4;(2)方程变形得:x2-2x-2=0,这里a=1,b=-2,c=-2,∵△=4+8=12>0,∴x=2±232=1±3,则x1=1+3,x2=1-3.

(1)原式=112+1+(22)-1=2+1+2=3+2;(2)方程两边同时乘以x(x-1)得:2x-1=-x(x-1)+x2,即2x-1=-x2+x+x2移项、合并同类项得:x=1,当x=1时,x(x-1)=0,故x=1不是方程的解,故方程无解;(3)原式=a?1a+2?(a+2)(a?2)(a?1)2?(a+1)(a-1...

(1)原式=1+2-2×32-2+3=1+2-3-2+3=1;(2)方程两边同时乘以(x+2)(x-2)得,2+x(x+2)=x2-4,去括号得,2+x2+2x=x2-4,移项、合并同类项得,2x=-6,系数化为1得,x=-3.经检验:x=-3是原方程的解.

(1)原式=22×22-1-(2-1)+4=2-1-2+1+4=6-2;(2)去分母得:2+2x-4=x+1,解得:x=3,经检验x=3是分式方程的解.

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