It is known that x ＜ 0, (X & # 178; + 9) △ X has a maximum value of

y=f(x)=(x²+9)/x=x+(9/x)
Ling X0
f(-x)=-x-(9/x)≥2√(-x)*(-9/x)=6
F (x) is an odd function
f(x)≤-6
If and only if x = 9 / x, that is, x = - 3, the equal sign holds

2(x²+1/x²)-3(x+1/x)=1
solve equations

2(x^2+1/x^2)-3(x+1/x)=1
2(x^2+1/x^2+2-2)-3(x+1/x)=1
2(x+1/x)^2-4-3(x-1/x)=1
Consider (x + 1 / x) as Z
We can get 2Z ^ 2-4-3z = 1
To find Z is to find (x + 1 / x) and then X
(x ^ 2 is the square of x)

What triangle inequality of vector: | a | - | B | = | a ± B | = | a | + | B |

The vector inequality you give can be proved according to the triangle inequality
Triangle inequality: in △ ABC, three sides satisfy the following inequality:
　　|a - b| < c < a + b；　　　　　　　　　　　　　①
That is: any side of a triangle is greater than the difference between the other sides and less than the sum of the two sides
(Note: A, B and C above are side lengths and numbers; a, B and C in the following discussion are vectors.)
For the inequality you give, it can be divided into two parts
　　|a| - |b| ≤ |a+b| ≤ |a| + |b|；　　　　　　　　　　②
　　|a| - |b| ≤ |a- b| ≤ |a| + |b|；　　　　　　　　　　③
1. Proof formula 2:
It needs to be discussed in two situations
(1) The two vectors are not collinear;
The sum of their vectors: a + B is exactly the third side of the triangle, denoted as: C
　　｜|a| - |b|｜ < |c| < |a| + |b|；　　　　　　　　　　④
Where, C = a + B; and according to "the absolute value of a number is certainly not less than the number itself", we can see that:
　　|a| - |b| ≤ ｜|a| - |b|｜
Substituting the above formula into formula 4, we can see that:
　　|a| - |b| < |a+b| < |a| + |b|；
Obviously, the formula satisfies inequality 2;
(2) Two vectors are collinear;
In this case, if a and B are connected end to end, the length of resultant vector is only two kinds of values
1）|a+b| = |a| + |b|；
2）|a+b| = ｜|a| - |b|｜；
And these two formulas also satisfy (2)
Synthesis (1) and (2) shows that inequality 2 holds for any two vectors
2. The formula of proof 3 is as follows
For Formula 3, we can change it into formula 2, and then prove it
Let the vector x = - B, then formula 3 becomes:
　　|a| - |-x| ≤ |a+x| ≤ |a| + |-x|；
Because the absolute value of a negative vector is equal to a positive vector, the above formula is equivalent to:
　　|a| - |x| ≤ |a+x| ≤ |a| + |x|；　　　　　　　　　　　　⑤
Obviously, formula 5 conforms to formula 2, so it holds; and formula 5 is equivalent to formula 3, so formula 3 also holds
Synthesizing 1 and 2, we can prove that the inequality you give holds for any two vectors

It is known that s △ doc = 15 square centimeter, Bo = 23bd

Let H be the height of the trapezoid, which is also the height of △ DBC, because ob = 23bd, BD = Bo + OD, so Bo = 2od, and because in △ AOD and △ DBC, ad ‖ BC, Bo = 2od, so ad = 12bc, because △ doc is equal to △ BOC, Bo = 2od, s △ doc = 15 square centimeter, so s △ BOC = 2 △ doc = 2 × 15 = 30 (square centimeter), because s △ DBC = s △ doc + s △ BOC, so s △ DBC = 15 + 30 = 45 (square centimeter), and Because s △ DBC = 12 × BC × h, so 12bch = 45, because the area of trapezoid ABCD = 12 (AD + BC) h, so the area of trapezoid ABCD = 12 (12bc + BC) h, = 32 × 12bch, = 32 × 45, = 67.5 (square centimeter), a: the area of trapezoid is 67.5 square centimeter

The definition field of F (x) = LG (1-lgx) is and the value field is
What is the range?

First, x > 0
in addition
1-lgx > 0, so
0

A cone has a circumference of 12.56 cm and a volume of 4 cm
emergency

What is the radius of the bottom
12.56 ﹣ 3.14 ﹣ 2 = 2cm
The volume is
2 × 2 × 3.14 × 2.4 × 1 / 3 = 10.048 cm3

If 3a2bn and 4amb4 are similar, then M= ，n= ．

If the solution ∵ 3a2bn and 4amb4 are of the same kind, then M = 2, n = 4, so the answer is: 2, 4

Triangle ABC is a right triangle. The area of shadow a is 57 square centimeters larger than that of shadow B. find the length of AC

The area of shadow a is 57 square centimeters larger than that of shadow B
The area of semi circle is 57cm ^ 2 larger than that of RT △ ABC
It is known that the diameter of the semicircle is 20cm,
Its area is 50 π
14, s (semicircle) = 157cm ^ 2
∴S（RT△ABC）=100cm^2=1/2*AB*AC
∴ AC=10cm

What are the numbers that are both even and prime numbers within 20?

Within 20, the number that is both even and prime is 2
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It is known that the area of a right triangle is 2 √ 3 square centimeter and the perimeter is (6 + 2 √ 3) cm. The circumcircle radius of the right triangle is calculated

The two right sides of the right triangle are 2 and 2 √ 3 long respectively, and the hypotenuse is 4 long. The triangle is a right triangle, so the radius of the circumscribed circle is half of the length of the hypotenuse, that is, 2 cm

It is known that x ＜ 0, (X & # 178; + 9) △ X has a maximum value of