1.x-y-z=1 2x+y-3z=4 3x-2y-z=-1 2.a+0.5b=c 0.5a-b-c=1 3a+b=-2 one x-y-z=1 2x+y-3z=4 3x-2y-z=-1 two a+0.5b=c 0.5a-b-c=1 3a+b=-2

1.x-y-z=1 2x+y-3z=4 3x-2y-z=-1 2.a+0.5b=c 0.5a-b-c=1 3a+b=-2 one x-y-z=1 2x+y-3z=4 3x-2y-z=-1 two a+0.5b=c 0.5a-b-c=1 3a+b=-2

1. X-Y-Z = 1. ① ⑥ ⑦ 2x + y-3z = 4.. ② 3x-2y-z = - 1. ③ ③ - ① 2x-y = - 2. ④ ③ * 3 - ② 7x-7y = - 7, X-Y = - 1. ⑤ ④ - ⑤ x = - 1, y = 0. Substitute ① z = - 2X = - 1y = 0z = - 22. A + 0.5B = C ① 0.5a-b-c = 1 ② 3A + B = - 2. ③ - ①

Given polynomial a = 3x ^ 2-2x + 7, B = x ^ 3-8x ^ 2-2, find 3a-2b

3A-2B
=9x^2-6x+21-2x^3+16x^2+4
=-2x^3+25x^2-6x+25

Given that a = 3x ^ 2 + 2x + 1, B = 2x ^ 3-x ^ 2 + 3x-7, find: a + B =? 3A + 2B +?
Today is the day

A+B
=3x^2+2x+1+2x^3-x^2+3x-7
=2x^3+2x^2+5x-6
3A+2B
=3(3x^2+2x+1)+2(2x^3-x^2+3x-7)
=9x^2+6x+3+4x^3-2x^2+6x-14
=4x^3+7x^2+12x-11

Remove the brackets and calculate the following questions: 1. (2x + 1) + (- x + 2) 2. (x + 2) - (3-6x) 3. (3a + 2b) + (a-2b) 4. (3x + 6) + (2x-7)
5. (4ab-b Square) - 2 (a square + 2ab-b Square) please

1. X + 3 2.7x-1 3.4A 4.5x-1 5. B-2a

There are two points a (- 1,0), B (1,0) on the plane, and point P is on the circumference (x-3) 2 + (y-4) 2 = 4. Find the coordinates of point P when ap2 + bp2 is the minimum

According to the meaning of the title, if we make the symmetric point Q of point P about the origin, then the quadrilateral paqb is a parallelogram. According to the properties of parallelogram, there are ap2 + bp2 = 12 (4op2 + AB2), that is, when OP is the minimum, ap2 + bp2 takes the minimum, opmin = 5-2 = 3, PX = 3 × 35 = 95, py = 3 × 45 = 125, P (95125)

If M + 2 of 2 and M + 3 of 2 are equal to 96, then M is equal to 96

Is it 2 ^ (M + 2) + 2 ^ (M + 3) = 96

It is known that the bottom of p-abcd is rhombic, and E is the midpoint of PA

To link AC BD is to draw the diagonal of the diamond
We know that the two diagonals of the diamond bisect each other, that is, the intersection is the midpoint. Let this point be f
that
In the triangular APC, e is the midpoint of AP and F is the midpoint of AC
The median line theorem, EF is parallel to PC
F is also the midpoint of BD, so EF is in BDE
PC parallel to EF is naturally parallel to BDE

3 ^ 2004-3 ^ 2003 and (- 2) ^ 101 + (- 2) ^ 100
Factorization
Thank you

3^2004-3^2003=3^2003*(3-1)=2*3^2003
(-2)^101+(-2)^100=(-2)^100*(-2+1)=-(-2)^100=-2^100

As shown in the figure, ⊙ o center angle ∠ AOB = 90 °, the distance from point O to chord AB is 4, then the diameter length of ⊙ o is______ ．

As shown in the figure, the crossing point O is OC ⊥ AB, the perpendicular foot is C, ∵ ∠ AOB = 90 °, a = ∠ AOC = 45 °, OC = AC, ∵ co = 4, ∵ AC = 4, ∵ OA = ac2 + CO2 = 42, and the diameter length of ∵ o is 82

If MX2 + 5x-15 = 5x2 + 5m is a linear equation of one variable about X, then its solution is () A.8B. - 8c.5d. - 5
Come on, me

Because it is a system of linear equations with one variable about X, it should not contain quadratic terms. So m = 5
The solution is x = 8
Choose a