If 7 / a = 2 / b = 5 / 1, find the value of 3 / a-2b

If 7 / a = 2 / b = 5 / 1, find the value of 3 / a-2b

A = 1 / 5 × 7 = 7 / 5
B = 1 / 5 × 2 = 2 / 5
One third (a-2b)
=3 / 3 (7 / 5-4 / 5)
=3 / 3 (3 / 5)
=1 / 5
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Find the median of the following groups: (1) 7 + 3 √ 5 and 7-3 √ 5 (2) a ^ 4 + A ^ 2B ^ 2 and B ^ 4 + A ^ 2B ^ 2 (a ≠ 0, B ≠ 0)
Find the median of the following groups: (1) 7 + 3 √ 5 and 7-3 √ 5 (2) a ^ 4 + A ^ 2B ^ 2 and B ^ 4 + A ^ 2B ^ 2 (a ≠ 0, B ≠ 0)

A ^ 3 + A ^ 2B = - 2 A ^ 2B + B ^ 3 = 5 find the value of a ^ 3 + 2A ^ 2B + B ^ 3

It is proved that sin ^ (x + y) ≤ (x + y) ^ 2 D is any bounded closed domain
sin^2(x+y)≤(x+y)^2

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The known proposition p: "any x ∈ [1,2], X & # 178; - a ≥ 0", proposition q "exists x0 & # 178; + 2ax0 + 2-A = 0"
If the proposition "P and Q" is true, find the value range of real number a
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p: X & # 178; - a ≥ 0, then a ≤ X & # 178;, and X ∈ [1,2], so x & # 178; ∈ [1,4], then a ≤ 1;
q: Δ = (- 2A) & # 178; - 4 (2-A) ≥ 0, a & # 178; + A-2 ≥ 0, (A-1) (a + 2) ≥ 0, then a ≥ 1, or a ≤ - 2
If P and Q are true, then both P and Q are true
So the intersection of the two ranges is a ≤ - 2, or a = 1
That is, the value range of real number a is (- ∞, - 2] ∪ {1}

Let proposition p: for any x ∈ R, x ^ 2 + x > A; proposition q: there exists x0 ∈ R (0 is the corner mark of the lower right corner), so that x0 ^ 2 + 2ax0 + 2-A = 0. If proposition p is true and proposition q is false, the value range of a is obtained

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It is known that the proposition p: x2-2x + a ≥ 0 holds on R. the proposition q: ∃ x0 ∈ R, X02 + 2ax0 + 2 − a = 0. If P or q is true, P and Q is false, the value range of real number a is obtained

If P is a true proposition, then △ = 4-4a ≤ 0  a ≥ 1; & nbsp (3 points) if q is a true proposition, then the equation x2 + 2aX + 2-A = 0 has a real root, that is, a ≥ 1 or a ≤ - 2 According to the meaning of the question, when p is true and Q is false, a ∈ϕ; & nbsp When p is false and Q is true, a ≤ - 2 is obtained

Let P: given the function f (x) = x ^ 2-mx + 1, all x0 belong to R, and there exists Y0 > 0,
Let the proposition p: known function f (x) = x ^ 2-mx + 1, all x0 belong to R, exist Y0 > 0, such that f (x0) = Y0, proposition q: inequality x ^ 2 less than 9-m ^ 2 has real number solution. If it is not p and Q is true proposition, then the value range of real number m?

P true: for any x0, f (x0) > 0, equivalent delta = m ^ 2-4

The negation of the proposition "there exists x0 ∈ r such that 2x0 ≤ 0" is______ ．

Because the negation of a special proposition is a universal proposition, the negation of the proposition "exist x0 ∈ R, make 2x0 ≤ 0" is: any x ∈ R, make 2x > 0. So the answer is: any x ∈ R, make 2x > 0

It is known that the proposition "existence x0 belongs to R, ax ^ 2-2ax0-3 > 0" is a false proposition, and the range of a is obtained

The proposition "existence x0 belongs to R, ax0 & # 178; - 2ax0-3 > 0" is a false proposition,
Then the proposition "for any x belongs to R, ax & # 178; - 2ax-3 ≤ 0" is true
When a = 0, the inequality is changed to - 3 ≤ 0;
When a ≠ 0, there is a