Solving fractional equation 1 / X & sup3; + 2x & sup2; + X + 1 / X & sup2; + 2x + 1 = 5 / 2x & sup2; + 2 Solving the fractional equation: 1 / (X & sup3; + 2x & sup2; + x) + 1 / (X & sup2; + 2x + 1) = 5 / (2x & sup2; + 2) It was like this

Solving fractional equation 1 / X & sup3; + 2x & sup2; + X + 1 / X & sup2; + 2x + 1 = 5 / 2x & sup2; + 2 Solving the fractional equation: 1 / (X & sup3; + 2x & sup2; + x) + 1 / (X & sup2; + 2x + 1) = 5 / (2x & sup2; + 2) It was like this

1/(x³+2x²+x)+1/(x²+2x+1)=1/X/(X+1)^2+1/(X+1)^2=1/X/(X+1)^2+X/X/(X+1)^2=1/X/(X+1)=5/(2x²+2)
(3x-1)(x-2)=0
x1=1/3
x2=2
dear... Can you add two brackets? It's a bit confusing where to cut this
1/(x³+2x²+x)= 1/x(x²+2x+1)
1 / X (X & sup2; + 2x + 1) and 1 / (X & sup2; + 2x + 1) are divided into 1 / X (x + 1) on the left and 5 / 2x & sup2; + 2 on the right
The rest should be OK
Solution equation: 3 (2x + 5) 2 + 4 (2x + 1) 2 = 7 (2x + 5) (2x + 1)
We get 3 (2x + 1) 2 + 4 (2x + 5) 2 = 7 (2x + 5) (2x + 1) by removing the denominator, 12x2 + 12x + 3 + 16x2 + 80x + 100 = 28x2 + 84x + 35 by removing the bracket, 8x = 32 by shifting the term, and x = 4 by solving. It is proved that x = 4 is the solution of the fractional equation
When m is equal to what, the fractional equation x / X-2 = 2 - (M / X-2) will have an increasing root when the denominator is removed?
x/x-2=2-(m/x-2)
(m+x)/(x-2)=2
m+x=2x-4
x=m+4
If the increasing root is x = 2, then M + 4 = 2, M = - 2
When m = - 2, the equation has an increasing root
X / X-2 = 2 - (M / X-2) (x + m) / (X-2) = 2 (x + m) = 2 (X-2) if there is an increasing root, the increasing root is 2. If there is an increasing root, the increasing root is 2. If x = 2 generations, we get: (2 + m) = 2 * 0, M = - 2
When solving the fractional equation [2 / (X-2)] + [(MX + 1) / (x ^ 2-4)] = 0, there will be an increasing root, then the coefficient M=
2/(x-2)+(mx+1)/(x^2-4)=0
2/(x-2)+(mx+1)/[(x+2)(x-2)]=0
2(x+2)+(mx+1)=0
2x+mx=-5
(m+2)x=-5
When x = 2 or - 2, the equation has increasing roots
So m = - 9 / 2 or 1 / 2
It is known that the square of the equation 2x + kx-1 = 0 about X
It's better not to use discriminant. We haven't learned it yet
Divide two sides by two
x²+kx/2=1/2
x²+kx/2+k²/16=1/2+k²/16
(x+k/4)²=(8+k²)/16
Because (8 + K & # 178;) / 16 > 0
So he has two square roots
So x + K / 4 = ± √ (8 + K & # 178;) / 4
Then x = - K / 4 ± √ (8 + K & # 178;) / 4
So there are two unequal real roots
2x²＋kx－1＝0
x²＋k/2 x＝1/2
﹙x＋k/4﹚²＝1/2＋k²/16
∵ k²≥0
∴ 1/2＋k²/16＞0
The equation has two unequal real roots.
2X plus one third x = one third 56, find X
2X is regarded as 6x out of 3, and the sum is 7x out of 3. 7x out of 3 equals 56 out of 3. That is 7x equals 56, x = 8
For non-zero natural numbers a and B, the meaning of the symbol ⁃ is defined as a ⁃ B = ma + b2ab (M is a certain integer). If 1 ⁃ 4 = 2 ⁃ 3, then 3 ⁃ 4=______ .
1 * 4 = 2 * 3 & nbsp; & nbsp; m + 42 (1 × 4) = 2m + 32 × (2 × 3) & nbsp; & nbsp; & nbsp; m + 48 = 2m + 31212 (M + 4) = 8 (2m + 3) & nbsp; & nbsp; 12m + 48 = 16m + 24 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 4m = 24 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; m = 6; 3 * 4 = 6 × 3 + 42 × (3 × 4) = 2224 = 1112, so the answer is: 1112
It is known that the equation 2x2 + kx-1 = 0 for X. (1) prove that the equation has two unequal real roots; (2) if one root of the equation is - 1, find the other root and K value
It is proved that: (1) ∵ a = 2, B = k, C = - 1 ∵ a = k2-4 × 2 × (- 1) = K2 + 8, ∵ no matter what the value of K is, K2 ≥ 0, ∵ K2 + 8 > 0, that is ∵ 0, ∵ equation 2x2 + kx-1 = 0 has two unequal real roots. (2) substituting x = - 1 into the original equation, 2-k-1 = 0 ∵ k = 1 ∵ the original equation is transformed into 2x2 + X-1 = 0, and the solution is X1 = - 1, X2 = 12, that is, the other root is 12
2X + 3 x = 3 56
2X + 3 x = 3 56
2X 3 / 56
7x / 3 = 56 / 3
X = 56 / 3 divided by 7 / 3
x =8
For non-zero natural numbers a and B, the symbol * is defined as a * b = ma + B / 2A (M is a certain integer). If 1 * 4 = 2 * 3, what is 3 * 4 equal to?
The meaning of the specified symbol @ is: a@b=2ab Ma + B (M is a definite integer). This is a calculation formula, in which m must be solved before it can be calculated
If 1@4=2@3 Using this condition, we can list the equation: SO 2 × 1 × 4 parts m + 4 = 2 × 2 × 3 parts 2m + 3, and solve the equation M = 6;
After calculating the value of M, the formula becomes: a@b=2ab If the values of a and B are known, they can be substituted into the calculation 3@4= ?
It's equivalent to a = 3, B = 4. After substituting, it's: so 3@4=2 X 3 × 4 / 3 × 6 + 4 = 12 / 11
1*4=2*3
Substituting 1 * 4 into a * B equation 1 * 4 = m + 2
Substituting 2 * 3 into a * B equation 2 * 3 = 2m + 3 / 4
Solve the value of M and substitute it into the equation of 3 * 4
3 * 4 = 3M + 2 / 3, you can get the correct result. I probably calculated 53 / 12, you can calculate it again.