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Given the complete set s = R, set a = {x ∧ 2-x-6 ＜ 0}, set B = {x ∩ x + 4 / X-2 ＞ 0}, given C = {x ∩ x ∧ 2-4ax + 3a ＜ 0} to find a ∩ B, a ∩ CSB
✅ prompt: you can see the detailed answer of the other party's setting separately, and you can view it after adopting more content
The solution set of inequality 2lg x-lgx-1 > 0 is
2 LG square x-lgx-1 > 0
(2lgx+1)(lgx-1)>0
lgx>1,x>10
lgx
How to write 2lg & # 178; x-lgx Λ 4 + 1 = 0
Let lgx = t, lgx ^ 4 = 4lgx. The original formula is 2T ^ 2-4t + 1 = 0, and the solution is 1 / 2 under t = 1 + radical or 1 / 2 under t = 1-radical. Then x = e ^ t, just bring in
What is the value of a when LG2 + lgx = 2lg (x + a) has one solution and two solutions
The original equation is
LG (2x) = LG (x + a) & sup2;, i.e
2X = (x + a) & sup2; (x > 0, and x > - a)
Namely
Let f (x) = 2x (x > 0), G (x) = (x + a) & sup2; (x > - a)
There is one intersection, two intersections and no intersection between the straight line f (x) and the conic g (x)
The following provides the method to solve the problem of Vader's theorem
F (x) = g (x), i.e
X & sup2; + (2a-2) x + A & sup2; = 0 (x > 0, and x > - a),
△=(2a-2)²-4a²=4-8a
(1) A solution, then
Δ = 0, i.e
4-8a=0,a=1/2
(2) Second solution, then
Δ > 0, i.e
4-8a>0,a
LG (x + 1) LG (x-1) or lgx · lgx? Why
(1) LG (x + 1) LG (x-1) = lgx · lgx because LG (x + 1) LG (x-1) = LG (x + 1) + (x-1) = LG (2x) because lgx · lgx = LG (x + x) = LG (2x), LG (x + 1) LG (x-1) = lgx · X or (2) LG (x + 1) LG (x-1) > lgx · lgx because LG (x + 1) LG (x-1) = lgx * LG1 * lgx / LG1 = 0 LG1
Find the value of 2lg & # 178; 2 + LG & # 178; 5 + 3lg2lg5-lg2
2lg²2+lg²5+3lg2lg5-lg2
=lg²2+2lg2lg5+lg²5+lg²2+lg2lg5-lg2
=(lg2+lg5)²+lg2(lg2+lg5-1)
=1+0
=1
2lg^2 2+2lg^2 5+3lg2lg5-lg2
The original question is:
2lg^2(2)+lg^2(5)+3lg2lg5-lg2=?
2lg^2 2+2lg^2 5+3lg2lg5-lg2
=2lg^2 2+2lg^2 5+4lg2lg5-lg2lg5-lg2
=2(lg^2 2+lg^2 5+2lg2lg5)-lg2(lg5+1)
=2(lg2+lg5)^2-lg2lg50
=2(lg10)^2-lg2lg50
=2-lg2lg50
Excuse me, 2lg ^ 2 + LG ^ 25 + 3lg2lg5 - LG2 =?
Original formula = (LG2) ^ 2 + (lg5) ^ 2 + 2lg2lg5 + (LG2) ^ 2 + lg2lg5-lg2
=（lg2+lg5）^2+lg2（lg2+lg5）-lg2
=1
2(lg√2)²+lg√2Xlg5+√(lg√2)²-lg2+1
ji
2(lg√2)²+lg√2Xlg5+√(lg√2)²-lg2+1
=2 (LG radical 2) ^ 2 + LG2 * (1-lg2) + LG radical 2-lg2-1
=2(1/2lg2) ^2+lg2-(lg2)^2+1/2lg2-lg2-1
=1/2(lg2)^2+1/2lg2-(lg2)^2-1
=-1/2(lg2)^2+1/2lg2-1
=-1/2lg2[lg2-1]-1
=-1/2lg2*(-lg5)-(lg2+lg5)
=1/2lg2*lg5-lg2-lg5
Calculate LG & sup2; 5 + LG2 · LG25 + LG & sup2; 2
lg²5+lg2·lg25+lg²2=lg²5+lg2·2lg5+lg²2=(lg5+lg2)^2=(lg10)^2=1
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