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The square of (x + 1) + the square of (2x + 1) is equal to the square of (3x + 2) (2x + 1)
Because the square of (x + 1) + the square of (2x + 1) is equal to the square of (3x + 2)
therefore
x^2+2x+1+4x^2+4x+1=9x^2+12x+4
The result of simplification
4x^2+6x+2=0
2x^2+3x+1=0
Factorization
（x+1)(2x+1）=0
The square of 2x plus 3x equals 3
Because x ^ 2 + 3x + 3 = 1006, so x ^ 2 + 3x = 1003, then 2x ^ 2 + 6x-4 = 2 (x ^ 2 + 3x) - 4 = 2 × 1003-4 = 2006-4 = 2002
A △ B = 1 / A + 1 / b = (a + b) / AB, substituting a and B into the above formula, i.e. (3-4) / 3 × (- 4) = - 1 / (- 12) = 1 / 12
What is (3x squared + 2x + 1) (2x squared + 3x-1) equal to
=The fourth power of 6x + 9x? - 3x? + 4x? + 6x? - 2x + 2x? + 3x-1
=6X to the fourth power + 13X? + 5x? + X-1
1. 2x square + 4x + 3 is greater than 0 2. - 3x square - 2x + 8 is greater than or equal to 0
2 (x ^ 2 + 2x + 1) + 1 > 0,2 (x ^ 2 + 2x + 1) + 1 > 0,2 (x + 1) ^ 2 > - 1, so the value range of X is r.2, and both sides are multiplied by - 1, we get that 3x ^ 2 + 2x - 8 is less than or equal to 0, (3x-4) (x + 2) is less than or equal to 0, and the solution of - 2 is less than or equal to X and less than or equal to 4 / 3
If the square of 2x + 1 plus the square of x-3 equals the square of 3x-2, what is x equal to?
4x²+4x+1+x²-6x+9=9x²-12x+4
4x²-10x-6=0
2x²-5x-3=0
(x-3)(2x+1)=0
x=3,x=-1/2
Given that the square of (3x-2) plus 2x minus y minus 3 is equal to 0, find the value of 5 (2x minus y) minus 2 (3x minus y plus 1). Fast, wait!
The absolute value of complete sum of squares is nonnegative
therefore
3x-2=0
x=2/3
And
2x-y-3=0
2x-y=3
y=2×2/3-3=-5/3
5(2x-y)-2(3x-y+1)=5×3-2(3×2/3+5/3+1）=15-6-10/3=17/3
If x squared minus 3x plus 2 equals 0, what is the value of 2x squared minus 6x plus 1
x²-3x+2=0
x²-3x=-2
2x²-6x+1=2(x²-3x)+1=2*(-2)+1=-4+1=-3
Hope to help you
2X square minus 3x plus 1 / 1 equals 2x square plus x minus 1 / 3,
1/(2x^2-3x+1)=1/(2x^2+x-3)
∵ the molecules are the same
Then the denominator is equal
That is, 2x ^ 2-3x + 1 = 2x ^ 2 + x-3
x+3x=1+3
4x=4
X=1
Test: put x = 1 into (2x ^ 2-3x + 1) (2x ^ 2 + x-3) = 0
The original fractional equation has no solution
If the square of 4x minus 3x minus 2 = 4, then the square of 2x minus 3 / 2 times x + 5 equals?
It's very simple, as follows
Because 4x ^ 2-3x-2 = 4
So 4x ^ 2-3x = 6
So 2x ^ 2 - (3x) / 2 = 3
So 2x ^ 2 - (3x) / 2 + 5 = 3 + 5 = 8
Given that the square of x minus 3x plus one equals zero, find the value of the square of x plus one part of the square of X
From x ^ 2-3x + 1 = 0, X ≠ 0
Each item is divided by X to obtain:
x-3+1/x=0.
x+1/x=3.
(x+1/x)^2=x^2+2+1/x^2=9.
x^2+1/x^2=9-2=7.
∴x^2+1/x^2=7.
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