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A math problem in senior one.
Forget it for a long time. It is inconvenient to use the computer. Later, I calculated it with manuscript paper and finally worked it out. But let me show you my initial idea, and I can't work it out. Later, I changed my mind and understood.

The answer is: subtract the last two digits in brackets first. Then add it to lg root number 27. Then divide by lg 1.2.

The process is as follows:

Lg radical number 27-3lg radical number/10/0 = LG (three-thirds of 3/three-thirds of kloc-0/0) = three-thirds of lg0.3 =3/2lg0.3.

Add lg8, and lg8=3lg2=3/2lg4.

And 3/2lg0.3+3/2lg4 = 3/2lg1.2.

The outside of the stent is divided by lg 1.2, so it is 3/2lg 1.2 divided by lg 1.2, and the result is 3/2.

The specific steps are as follows:

(LG√27+lg8-3lg√ 10)÷LG 1.2

=((3/2)lg3+3lg 2-(3/2))(LG 12- 1)

=(3/2)(lg3+2lg 2- 1)÷(lg3+lg4- 1)

=(3/2)(lg3+2lg 2- 1)÷(lg3+2lg 2- 1)

=3/2.

Below I also show you the previous ideas. .

Lg radical number 27+lg8=3lg6,

3lg6-3lg radical number 10=3lg(6/ radical number 10),

3lg(6/ root 10) divided by lg 1.2=lg(2 16/ root 10 cubic) *lg 10/lg 12=