Original formula =3X+2Y+4X+3Y=7X+5Y.
When x = 5 and y = 3.
Original formula =5*7+(-3)*5+20.
2.(5a 2-3b 2)+(a 2+b 2)-(5a 2+3b 2), where a =- 1 and b = 1.
=5a^2-3b^2+a^2+b^2-5a^2-3b^2
=a^2-5b^2
=(- 1)^2-5* 1^2
= 1-5
=-4
3.2 (A2B+AB 2)-2 (A2B-1)-2AB 2-2 where a =-2 and b = 2.
=2a^2b+2ab^2-2a^2b+2-2ab^2-2
=0
1. It is known that A and B are two real roots of the equation x 2+2x-5 = 0.
Find the value of (a 2+2ab+2a) (b 2+2ab+2b).
Derived from the fact that A and B are two real numbers of the equation x 2+2x-5 = 0:
AB=-5，A+B=-2
A^2+2AB+2A)(B^2+2AB+2B)
=AB(A+2B+2)(B+2A+2)
=-5(-2+B+2)(-2+A+2)
=-5AB
=25
2, 1/2(x+y+z) squared+1/2 (x-y-z) (x-y+z)-z (x+y), where x-y=6 and xy=2 1. Be specific.
Simplify:
1/2(x+y+z) squared+1/2 (x-y-z) (x-y+z)-z (x+y) =1
1/2[(x+y) square +2z(x+y)+z square ]+ 1/2[(x-y) square -z square ]-z(x+y)=]
1/2(x+y) square+1/2(x-y) square =x square +y square.
From x-y=6, xy=2 1, x+y = (x-y)+2xy = 78.
3. Find the value of 2ab-2a 2-4b 2-7 from a 2-ab+2b 2 = 3.
2ab-2a^2-4b^2-7
=2(ab-a^2-2b^2)-7
=-2(a^2-ab+2b^2)-7
=(-2)*3-7
=-6-7=- 13
Mathematics simplified evaluation topic in grade three, ~ [2/(x+ 1)+4/(x? - 1)]\x/(x- 1)
=[(2/(x+ 1)+4/(x- 1)(x+ 1)]÷x/(x- 1)
=[2(x- 1)/(x+ 1)(x- 1)+4/(x- 1)(x+ 1)]x/(x- 1)
=[2(x- 1)+4/(x- 1)(x+ 1)]x/(x- 1)
=[2(x+ 1)/(x- 1)(x+ 1)]x/(x- 1)
=[2/(x- 1)]\x/(x- 1)
=[2/(x- 1)]×(x- 1)/x
=2/x
When x=2
The above formula = 1
Simplify the math evaluation questions in grade three, first simplify and then evaluate. 【(2/x+ 1)+(4/x- 1)】÷x/x- 1x = 2，
2/(x+ 1)+4/(x？ - 1)]\x/(x- 1)
=[(2/(x+ 1)+4/(x- 1)(x+ 1)]÷x/(x- 1)
=[2(x- 1)/(x+ 1)(x- 1)+4/(x- 1)(x+ 1)]x/(x- 1)
=[2(x- 1)+4/(x- 1)(x+ 1)]x/(x- 1)
=[2(x+ 1)/(x- 1)(x+ 1)]x/(x- 1)
=[2/(x- 1)]\x/(x- 1)
=[2/(x- 1)]×(x- 1)/x
=2/x
When x=2
The above formula = 1
Pictures of simplified math evaluation questions in grade three (3X+2Y)+(4X+3Y), where X=5, Y+3.
Original formula =3X+2Y+4X+3Y=7X+5Y.
When x = 5 and y = 3.
Original formula =5*7+(-3)*5+20.
2.(5a 2-3b 2)+(a 2+b 2)-(5a 2+3b 2), where a =- 1 and b = 1.
=5a^2-3b^2+a^2+b^2-5a^2-3b^2
=a^2-5b^2
=(- 1)^2-5* 1^2
= 1-5
=-4
3.2 (A2B+AB 2)-2 (A2B-1)-2AB 2-2 where a =-2 and b = 2.
=2a^2b+2ab^2-2a^2b+2-2ab^2-2
=0
1. It is known that A and B are two real roots of the equation x 2+2x-5 = 0.
Find the value of (a 2+2ab+2a) (b 2+2ab+2b).
Derived from the fact that A and B are two real numbers of the equation x 2+2x-5 = 0:
AB=-5，A+B=-2
A^2+2AB+2A)(B^2+2AB+2B)
=AB(A+2B+2)(B+2A+2)
=-5(-2+B+2)(-2+A+2)
=-5AB
=25
2, 1/2(x+y+z) squared+1/2 (x-y-z) (x-y+z)-z (x+y), where x-y=6 and xy=2 1. Be specific.
Simplify:
1/2(x+y+z) squared+1/2 (x-y-z) (x-y+z)-z (x+y) =1
1/2[(x+y) square +2z(x+y)+z square ]+ 1/2[(x-y) square -z square ]-z(x+y)=]
1/2(x+y) square+1/2(x-y) square =x square +y square.
From x-y=6, xy=2 1, x+y = (x-y)+2xy = 78.
3. Find the value of 2ab-2a 2-4b 2-7 from a 2-ab+2b 2 = 3.
2ab-2a^2-4b^2-7
=2(ab-a^2-2b^2)-7
=-2(a^2-ab+2b^2)-7
=(-2)*3-7
=-6-7=- 13
4. If A = 2x 2+3xy-2x-3, B =-x 2+xy+2, and the value of 3A+6B has nothing to do with X, find the value of y..
3a+6b=6x^2+9xy-6x-9-6x^2+6xy+ 12
= 15xy-6x+3
=x( 15y-6)+3
5.9x+6x 2-3 (x-2/3x 2), where x=-2.
9x+6x -3(x-2/3x)
=9x+6x-3x+2x
=8x+6x
=8×(-2)+6×(-2)
=32- 12
=20
6.1/4 (-4x2+2x-8)-(1/2x-1), where x= 1/2.
1/4(-4x+2x-8)-( 1/2x- 1)
=-x+ 1/2x-2- 1/2x+ 1
=-x- 1
=-( 1/2)- 1
=- 1/4- 1
=-5/4
7.3x'y-[2x'y-(2-x'z)-4x'z]-, where x =-2, y =-3, z = 1,
:3x'y-[2x'y-(2-x'z)-4x'z]-
=3x'y-2x'y+2-x'z+4x'z-
=x'y-+3x'z
=4*(-3)-2*3* 1+3*4* 1
=- 12-6+ 12
=-6
8.(5a 2-3b 2)+(a 2+b 2)-(5a 2+3b 2), where a =- 1 and b = 1.
=5a^2-3b^2+a^2+b^2-5a^2-3b^2
=a^2-5b^2
=(- 1)^2-5* 1^2
= 1-5
=-4
9,2 (A2B+AB 2)-2 (A2B-1)-2AB 2-2, where a =-2 and b = 2.
=2a^2b+2ab^2-2a^2b+2-2ab^2-2
=0
10 squared, (x-21y-1) (x-21y+1)-(x-21y-1).
Where x = 1.7 and y = 3.9 (simplify first and then evaluate).
[(X-2 is 1Y)- 1][(X+2 is 1Y)+ 1]-(X-2 is1y-0/) square.
= (x+ 1Y) square-1-(x- 1Y) square +2 (x- 1Y)- 1.
= (x+ 1Y) square -(x- 1Y) square +2 (x- 1y) -(X-2
=2XY+2X-Y-2
=3.9*2.4+ 1.4
= 10.76
Junior high school math exam simplified evaluation special training questions 1. (3X+2Y)+(4X+3Y), where X=5 and Y+3.
Original formula =3X+2Y+4X+3Y=7X+5Y.
When x = 5 and y = 3.
Original formula =5*7+(-3)*5+20.
2.(5a 2-3b 2)+(a 2+b 2)-(5a 2+3b 2), where a =- 1 and b = 1.
=5a^2-3b^2+a^2+b^2-5a^2-3b^2
=a^2-5b^2
=(- 1)^2-5* 1^2
= 1-5
=-4
3.2 (A2B+AB 2)-2 (A2B-1)-2AB 2-2 where a =-2 and b = 2.
=2a^2b+2ab^2-2a^2b+2-2ab^2-2
=0
1. It is known that A and B are two real roots of the equation x 2+2x-5 = 0.
Find the value of (a 2+2ab+2a) (b 2+2ab+2b).
Derived from the fact that A and B are two real numbers of the equation x 2+2x-5 = 0:
AB=-5，A+B=-2
A^2+2AB+2A)(B^2+2AB+2B)
=AB(A+2B+2)(B+2A+2)
=-5(-2+B+2)(-2+A+2)
=-5AB
=25
2, 1/2(x+y+z) squared+1/2 (x-y-z) (x-y+z)-z (x+y), where x-y=6 and xy=2 1. Be specific.
Simplify:
1/2(x+y+z) squared+1/2 (x-y-z) (x-y+z)-z (x+y) =1
1/2[(x+y) square +2z(x+y)+z square ]+ 1/2[(x-y) square -z square ]-z(x+y)=]
1/2(x+y) square+1/2(x-y) square =x square +y square.
From x-y=6, xy=2 1, x+y = (x-y)+2xy = 78.
3. Find the value of 2ab-2a 2-4b 2-7 from a 2-ab+2b 2 = 3.
2ab-2a^2-4b^2-7
=2(ab-a^2-2b^2)-7
=-2(a^2-ab+2b^2)-7
=(-2)*3-7
=-6-7=- 13
4. If A = 2x 2+3xy-2x-3, B =-x 2+xy+2, and the value of 3A+6B has nothing to do with X, find the value of y..
3a+6b=6x^2+9xy-6x-9-6x^2+6xy+ 12
= 15xy-6x+3
=x( 15y-6)+3
5.9x+6x 2-3 (x-2/3x 2), where x=-2.
9x+6x -3(x-2/3x)
=9x+6x-3x+2x
=8x+6x
=8×(-2)+6×(-2)
=32- 12
=20
6.1/4 (-4x2+2x-8)-(1/2x-1), where x= 1/2.
1/4(-4x+2x-8)-( 1/2x- 1)
=-x+ 1/2x-2- 1/2x+ 1
=-x- 1
=-( 1/2)- 1
=- 1/4- 1
=-5/4
7.3x'y-[2x'y-(2-x'z)-4x'z]-, where x =-2, y =-3, z = 1,
:3x'y-[2x'y-(2-x'z)-4x'z]-
=3x'y-2x'y+2-x'z+4x'z-
=x'y-+3x'z
=4*(-3)-2*3* 1+3*4* 1
=- 12-6+ 12
=-6
8.(5a 2-3b 2)+(a 2+b 2)-(5a 2+3b 2), where a =- 1 and b = 1.
=5a^2-3b^2+a^2+b^2-5a^2-3b^2
=a^2-5b^2
=(- 1)^2-5* 1^2
= 1-5
=-4
9,2 (A2B+AB 2)-2 (A2B-1)-2AB 2-2, where a =-2 and b = 2.
=2a^2b+2ab^2-2a^2b+2-2ab^2-2
=0
10 squared, (x-21y-1) (x-21y+1)-(x-21y-1).
Where x = 1.7 and y = 3.9 (simplify first and then evaluate).
[(X-2 is 1Y)- 1][(X+2 is 1Y)+ 1]-(X-2 is1y-0/) square.
= (x+ 1Y) square-1-(x- 1Y) square +2 (x- 1Y)- 1.
= (x+ 1Y) square -(x- 1Y) square +2 (x- 1y) -(X-2
=2XY+2X-Y-2
=3.9*2.4+ 1.4
= 10.76
Mathematics in Grade Three, simplify first, then evaluate [(5/(a-2))-a-2] = (3+a) (3-a)/(a-2).
Eliminating similar terms indicates that the original formula is equal to
(- 1)/2(3+a)=(- 1)/2√3 =-√3/6
Hope to adopt
〔(a？ -B? /a？ -2ab+b？ ) +(a/b-a)] divided by (b? /a？ -ab)
=a/b
Grade three math simplification or simplification evaluation (A 2+B 2) 2 minus (A 2+B 2) minus 6 = 0 (A 2+B 2) 2-(A 2+B 2)-6 = 0.
(a^2+b^2-3)(a^2+b^2+2)=0
A 2+b 2 = 3, or A 2+b 2 =-2.
The result is a 2+b 2 = 3.
Math simplification in grade three (n+1) (n+2)/1+(n+2) (n+3)/1+(n+3) (n+4)/1.
=(n+ 1)(n+2)+(n+2)(n+3)+(n+3)(n+4)
=(n+2)(n+ 1+n+3)+n^2+7n+ 12
=(n+2)(2n+4)+n^2+7n+ 12
=2(n+2)^2+n^2+7n+ 12
=2(n^2+4n+4)+n^2+7n+ 12
=3n^2+ 15n+20
Math Problems in Grade Three (Fraction Simplification) (2009a/A2-2A+1)/[(1/A-1)+1]
=2009a/(a- 1)^2/[( 1+a- 1)/(a- 1)]
=2009a/(a- 1)^2*(a- 1)/a
=2009/(a- 1)