Find the minimum value of the expression .
Answer: The minimum value is
Find the minimum value of the expression \(\frac{2}{-4x^2+8x-5}\)
Find the minimum value of the expression .
A=(x+y+1)(x+y+1)+4
A=(x+y+1)2+4
Vậy MinA=4 khi.......... của @Nguyễn Huy Thắng đó mà ghi tiếp
ngu Anh nhưng ko sao dịch dc chữ Find the minimum = tìm GTNN :)
\(A=x^2+y^2+2x+2y+2xy+5\)
\(=\left(x^2+y^2+2x+2y+2xy+1\right)+4\)
\(=\left(x+y+1\right)^2+4\ge4\)
Dấu "=" xảy ra khi \(\left(x+y+1\right)^2=0\)\(\Rightarrow x=-y-1\)
Vậy \(Min_A=4\) khi \(x=-y-1\)
Assume that two numbers x and y satisfy: 2x + y = 6.
Find the minimum value of expression A = 4x2 + y2
\(2x+y=6\)
\(\Rightarrow y=6-2x\)
\(\text{Thế vào phương trình ta dc:}\)
\(4x^2+\left(6-2x\right)^2\)
\(=4x^2+36-24x+4x^2\)
\(=8x^2-24x+36\)
\(\Leftrightarrow4x\left(2x-6\right)+36\)
Rồi sao nữa quên ùi
ta có : \(2x+y=6\Leftrightarrow y=6-2y\)
thay vào A, ta có:
\(A=4x^2+\left(6-2x\right)^2\)
\(A=8\left(x^2-3x+2,25\right)+18\)
\(A=8\left(x-1,5\right)^2+18\)
\(\Rightarrow A\ge18\)
Find the minimum value of A=x(x-3)
We have: \(A=x^2-3x=x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}=\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)
\(\Rightarrow A_{min}=-\frac{9}{4}\) at \(x=\frac{3}{2}\)
For positive real numbers x,y,z so that: x+y+z = 3. Find the minimum value of expression
A = 1/( x^2 + x) + 1/(y^2+ y) +1/( z^2 +z)
Exam number 219:12
Fill in the blank with the suitable number (Note: write decimal number with "the dot" between number part and fraction part. Example: 0.5)
Question 1:
Given .
Find the value of "" such that its degree is equal to 4.
Answer: The value of "" is
Question 2:
The value of with is
Question 3:
Given .
The degree of is
Question 4:
Given .
The degree of is
Question 5:
Given .
The value of is
Question 6:
In this figure, if the length of the line segment is an even number.
Then .
Question 7:
Given .
The value of
Question 8:
The value of with is
Question 9:
Given with .
Then the minimum of is
Question 10:
The minimum value of is
Find the maximum and minimum value of the expression
\(\frac{x+y+z}{3}+\frac{2016}{\sqrt[3]{xyz}}\)if \(x,y,z\in\left[1,2016\right]\)
Đặt \(A=\frac{x+y+z}{3}+\frac{2016}{\sqrt[3]{xyz}}\)
Tìm giá trị nhỏ nhất :Áp dụng bđt Cauchy : \(A=\frac{x+y+z}{3}+\frac{2016}{\sqrt[3]{xyz}}\ge\frac{3.\sqrt[3]{xyz}}{3}+\frac{2016}{\sqrt[3]{xyz}}\)
\(\Rightarrow A\ge\sqrt[3]{xyz}+\frac{2016}{\sqrt[3]{xyz}}\ge2\sqrt{\sqrt[3]{xyz}.\frac{2016}{\sqrt[3]{xyz}}}\)
\(\Rightarrow A\ge2\sqrt{2016}=24\sqrt{14}\) .
Dấu "=" xảy ra khi và chỉ khi \(\begin{cases}x=y=z\\\sqrt[3]{xyz}=\frac{2016}{\sqrt[3]{xyz}}\end{cases}\) \(\Leftrightarrow x=y=z=12\sqrt{14}\)
Vậy A đạt giá trị nhỏ nhất bằng \(24\sqrt{14}\) tại \(x=y=z=12\sqrt{14}\)
Find the Minimum value of this expression
\(\sqrt{\left(3x+1\right)^2+1}+\sqrt{\left(3x-3\right)^2+9}\)
The minimum value of 9x2-6-6x is ...........
Dịch: Tìm giá trị nhỏ nhất của \(9x^2-6-6x\)
Ta có: \(9x^2-6-6x=9x^2-6x-6=\left(3x-1\right)^2-7\ge-7\)
Dấu \(''=''\)xảy ra \(\Leftrightarrow\left(3x-1\right)^2=0\Leftrightarrow3x-1=0\Leftrightarrow x=\frac{1}{3}\)
Vậy, Giá trị nhỏ nhất của đa thức \(9x^2-6-6x\) là \(-7\Leftrightarrow x=\frac{1}{3}\)
=> The minimum value of 9x2-6-6x is -7