The formula of a and B is a.3a = 4b, b.4a = 3 / B, C.3 / a = 4 / b A and B are proportional to each other A． 3A＝4B B. 4A = 3 / b C. A / 3 = 4 / b

The formula of a and B is a.3a = 4b, b.4a = 3 / B, C.3 / a = 4 / b A and B are proportional to each other A． 3A＝4B B. 4A = 3 / b C. A / 3 = 4 / b

Choose a

My classmates gave me an equation, which made me puzzled. The equation is as follows: 4 = 3 A + B = C 4a-3a + 4b-3b=
4c-3c 4A + 4b-4c = 3A + 3b-3c (transfer)
4 (a + B-C) = 3 (a + B-C)
4-3!

Because a + B = C, so a + B - C = 0, we can't get rid of the similar term
Actually, it's not that complicated, because 4 = 3, so 4A = 3A, 4b = 3B,
So 4a-3a = 0, 4b-3b = 0,
So 4a-3a + 4b-3b = (4a-3a) + (4b-3b) = 0 + 0 = 0
"A + B = C" is an invalid condition to confuse you

The solution of inequality ax & # 178; + 1 & gt; 0 (a ≠ 0) about X

When a > 0 & nbsp; no matter what the value of X is
Ax ^ 2 + 1 is always greater than 1 & nbsp;
When a < 0 & nbsp;
ax^2＞-1
x^2＜1/-a  
-√1/-a＜x＜√1/-a

On the inequality X & # 178; - (2 + a) x + 2A & gt; 0 of X
The answer given by our teacher is x ∈ (negative infinity, 0) ∪ (2, positive infinity). I don't know why

∵ x ^ 2 - (2 + a) x + 2A ＞ 0, ∪ (X-2) (x-a) ＞ 0. Then: 1. When a ＜ 2, we need: X ＜ a, or X ＞ 2, ∪ (2, + ∞) the solution set of the original inequality. 2. When a = 2, we get: 0 ＞ 0, which is naturally unreasonable. At this time, the original inequality has no solution. 3. When a ＞ 2

The inequality x ^ 2-2mx + (m ^ 2-1) about X expressed by interval

△=（-2m）^2-4(m^2-1)=4>0
x^2-2mx+(m^2-1)=0
（x-m-1）（x-m+1)=0
x1=m+1 x2=m-1
x1>x2
x^2-2mx+(m^2-1)

Known: inequality x square + ax + b less than 0 solution set is (1,2), find a, b value

The solution set is (1,2) (x-1) (X-2) = x-3x + 2

Solving inequality: log ∨ 3 (3 + 2x-x & # 178;) > log ∨ 3 (3x + 1)

Because the logarithmic function with base 3 is an increasing function,
So the original inequality is equivalent to
3+2x-x²>3x+1>0
That is X & # 178; + x-2-1 / 3
-2

Solving inequality log (2x-3) ^ (x ^ 2-3) > 0

A:
log(2x-3)(x^2-3)>0
The base is 2x-3 and the true number is x ^ 2-3
1）0

Solving inequality log (2x-3) (x ^ 2-2x-2) > 0
Don't copy

log(2x-3) (x²-2x-2) >log(2x-3) 1
(1) If 2x-3 > 1, then x & # 178; - 2x-2 > 1
That is, x > 2 and X & # 178; - 2x-3 > 0
That is, x > 2 and x > 3 or X3
（2） 0

∫ ((log2x) ^ 2) DX the base number is 2 and the true number is X

Seeking indefinite integral ∫ [(log &; x) &;] DX
Let Log &; X = u, then x = 2 ^ u, DX = D (2 ^ u) = (2 ^ u) ln2du, then
The original formula = ∫ U & # 178; D (2 ^ u) = u & # 178; (2 ^ u) - 2 ∫ U (2 ^ u) Du
=u²(2^u)-(2/ln2)∫ud(2^u)
=u²(2^u)-(2/ln2)[u(2^u)-∫(2^u)du]
=u²(2^u)-(2/ln2)[u(2^u)-2^u/ln2]+C
=x(log₂x)²-(2/ln2)[x(log₂x)-x/ln2]+C