Tìm giá trị nhỏ nhất của biểu thức:

a) Ta có: 

\(M=2x^2+4x+7\)

\(M=2\cdot\left(x^2+2x+\dfrac{7}{2}\right)\)

\(M=2\cdot\left(x^2+2x+1+\dfrac{5}{2}\right)\)

\(M=2\cdot\left[\left(x+1\right)^2+2,5\right]\)

\(M=2\left(x+1\right)^2+5\)

Mà: \(2\left(x+1\right)^2\ge0\forall x\) nên:

\(M=2\left(x+1\right)^2+5\ge5\forall x\)

Dấu "=" xảy ra:

\(2\left(x+1\right)^2+5=5\Leftrightarrow2\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy: \(M_{min}=5\) khi \(x=-1\)

b) Ta có:

\(N=x^2-x+1\)

\(N=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

\(N=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Mà: \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\) nên \(N=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=" xảy ra: 

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(N_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2}\)

Tìm giá trị lớn nhất của biểu thức

a) Ta có: 

\(E=-4x^2+x-1\)

\(E=-\left(4x^2-x+1\right)\)

\(E=-\left[\left(2x\right)^2-2\cdot2x\cdot\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{15}{16}\right]\)

\(E=-\left[\left(2x-\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]\)

Mà: \(\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\ge\dfrac{15}{16}\forall x\) nên 

\(\Rightarrow E=-\left[\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]\le-\dfrac{15}{16}\forall x\)

Dấu "=" xảy ra:

\(-\left[\left(2x+\dfrac{1}{4}\right)^2+\dfrac{15}{16}\right]=-\dfrac{15}{16}\Leftrightarrow-\left(2x+\dfrac{1}{4}\right)^2-\dfrac{15}{16}=-\dfrac{15}{16}\)

\(\Leftrightarrow-\left(2x+\dfrac{1}{4}\right)^2=0\Leftrightarrow2x-\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{1}{16}\)

Vậy: \(E_{max}=-\dfrac{15}{16}\) khi \(x=\dfrac{1}{16}\)

b) Ta có:

\(F=5x-3x^2+6\)

\(F=-3x^2+5x-6\)

\(F=-\left(3x^2-5x-6\right)\)

\(F=-3\left(x^2-\dfrac{5}{3}x-2\right)\)

\(F=-3\left[\left(x-\dfrac{5}{6}\right)^2-\dfrac{97}{36}\right]\)

\(F=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}\)

Mà: \(-3\left(x-\dfrac{5}{6}\right)^2\le0\forall x\) nên:

\(F=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}\le\dfrac{97}{36}\forall x\)

Dấu "=" xảy ra:

\(-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{36}=\dfrac{97}{36}\Leftrightarrow-3\left(x-\dfrac{5}{6}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{5}{6}\)

Vậy: \(F_{max}=\dfrac{97}{36}\) khi \(x=\dfrac{5}{6}\)

\(M=2x^2+4x+7\)

\(=2\left(x^2+2x+\dfrac{7}{2}\right)\)

\(=2\left(x^2+2x+1+\dfrac{5}{2}\right)\)

\(=2\left[\left(x+1\right)^2+\dfrac{5}{2}\right]\)

\(=2\left(x+1\right)^2+5\)

Vì \(2\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x+1\right)^2+5\ge5\forall x\)

\(\Rightarrow M_{min}=5\Leftrightarrow2\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Tương tự: \(N=x^2-x+1\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

\(\Rightarrow N_{min}=\dfrac{3}{4}\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

\(E=-4x^2+x-1\)

\(=-4\left(x^2-\dfrac{1}{4}x+\dfrac{1}{4}\right)\)

\(=-4\left[x^2-2.x.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2-\left(\dfrac{1}{8}\right)^2+\dfrac{1}{4}\right]\)

\(=-4\left[\left(x-\dfrac{1}{8}\right)^2+\dfrac{15}{64}\right]\)

\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{15}{16}\)

Vì \(-4\left(x-\dfrac{1}{8}\right)^2\le0\forall x\)

\(\Rightarrow-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{15}{16}\le-\dfrac{15}{16}\forall x\)

\(\Rightarrow E_{max}=-\dfrac{15}{16}\Leftrightarrow-4\left(x-\dfrac{1}{8}\right)^2=0\Leftrightarrow x=\dfrac{1}{8}\)

Tương tự: \(F=5x-3x^2+6\)

\(=-3x^2+5x+6\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)

\(\Rightarrow F_{max}=\dfrac{97}{12}\Leftrightarrow-3\left(x-\dfrac{5}{6}\right)^2=0\Leftrightarrow x=\dfrac{5}{6}\)

Bài 1:

Ta thấy: $(x+\frac{1}{2})^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow (x+\frac{1}{2})^2+\frac{5}{4}\geq \frac{5}{4}$

Vậy gtnn của biểu thức là $\frac{5}{4}$

Giá trị này đạt tại $x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}$

Bài 2:

$x+y-3=0\Rightarrow x+y=3$
\(M=x^2(x+y)-(x+y)x^2-y(x+y)+4y+x+2019\)

\(=-3y+4y+x+2019=x+y+2019=3+2019=2022\)

Chọn đáp án A

Chọn A

Chắc đề đúng là số dương, vì ko tồn tại x;y nguyên dương thỏa mãn x+y=1

\(A=\dfrac{y^2}{xy+y}+\dfrac{x^2}{xy+x}\ge\dfrac{\left(x+y\right)^2}{x+y+2xy}\ge\dfrac{\left(x+y\right)^2}{x+y+\dfrac{1}{2}\left(x+y\right)^2}=\dfrac{2}{3}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Bài 1:

a: \(M=x^2-10x+3\)

\(=x^2-10x+25-22\)

\(=\left(x^2-10x+25\right)-22\)

\(=\left(x-5\right)^2-22>=-22\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

b: \(N=x^2-x+2\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x-1/2=0

=>x=1/2

c: \(P=3x^2-12x\)

\(=3\left(x^2-4x\right)\)

\(=3\left(x^2-4x+4-4\right)\)

\(=3\left(x-2\right)^2-12>=-12\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

\(a,A=x^2+y^2\\=x^2-2xy+y^2+2xy\\=(x-y)^2+2xy\\=2^2+2\cdot1\\=4+2\\=6\)

\(b,x+y=1\\\Leftrightarrow (x+y)^3=1^3\\\Leftrightarrow x^3+3x^2y+3xy^2+y^3=1\\\Leftrightarrow x^3+3xy(x+y)+y^3=1\\\Leftrightarrow x^3+3xy\cdot1+y^3=1\\\Rightarrow A=1\)

a) Ta có:

\(x-y=2\)

\(\Rightarrow\left(x-y\right)^2=2^2\)

\(\Rightarrow x^2-2xy+y^2=4\)

Mà: \(xy=1\)

\(\Rightarrow\left(x^2+y^2\right)-2\cdot1=4\)

\(\Rightarrow x^2+y^2=4+2\)

\(\Rightarrow x^2+y^2=6\)

b) Ta có: 

\(x+y=1\)

\(\Rightarrow\left(x+y\right)^3=1^3\)

\(\Rightarrow x^3+3x^2y+3xy+y^3=1\)

\(\Rightarrow x^3+3xy\left(x+y\right)+y^3=1\) 

Mà: x + y = 1

\(\Rightarrow x^3+3xy\cdot1+y^3=1\)

\(\Rightarrow x^3+3xy+y^3=1\)

Đáp án B

Ta có

Đáp án B

Ta có:

3 x 2 + y 2 − 2 . log 2 x − y = 1 2 1 + log 2 1 − x y ⇔ 3 x 2 + y 2 − 2 . log 2 x − y 2 = log 2 2 − 2 x y  

⇔ 3 x 2 + 2 x y + y 2 − 2 + 2 x y . log 2 x − y 2 = log 2 2 − 2 x y ⇔ 3 x − y 2 . log 2 x − y = 3 2 − 2 x y . log 2 2 − 2 x y  

Xét hàm số f t = 3 t . log 2 t trên khoảng 0 ; + ∞ , có  f ' t = 3 t ln 3. log 2 t + 3 t t . ln 2 > 0 ; ∀ t > 0

Suy ra f t là hàm số đồng biến trên 0 ; + ∞ mà

f x − y 2 = f 2 − 2 x y ⇒ x 2 + y 2 = 2

Khi đó:

M = 2 x 3 + y 3 − 3 x y = 2 x + y x + y 2 − 3 x y − 3 x y ⇔ 2 M = 2 x + y 2 x + y 2 − 3.2 x y − 3.2 x y 2 x + y 2 x + y 2 − 3 x + y 2 + 6 − 3 x + y 2 + 6 = 2 x + y 6 − x + y 2 − 3 x + y 2 + 6 = − 2 a 3 − 3 a 2 + 12 a + 6 ,

Với  a = x + y ∈ 0 ; 4

Xét hàm số f a = − 2 a 3 − 3 a 2 + 12 a + 6 trên 0 ; 4 ,

suy ra m ax 0 ; 4 f a = 13.  

Vậy giá trị lớn nhất của biểu thức M là 13 2

Đáp án B

Ta có 

3 x 2 + y 2 − 2 . log 2 x − y = 1 2 1 + log 2 1 − x y ⇔ 3 x 2 + y 2 − 2 . log 2 x − y 2 = log 2 2 − 2 x y

⇔ 3 x 2 + 2 x y + y 2 − 2 + 2 x y . log 2 x − y 2 = log 2 2 − 2 x y ⇔ 3 x − y 2 . log 2 x − y = 3 2 − 2 x y . log 2 2 − 2 x y

⇔ 2 M = 2 x + y 2 x + y 2 − 3.2. x y − 3.2 x y                  = 2 x + y 2 x + y 2 − 3 x + y 2 + 6 − 3 x + y 2 + 6

                  = 2 x + y 6 − x + y 2 − 3 x + y 2 + 6 = − 2 a 3 − 3 a 2 + 12 a + 6 ,

Vậy giá trị lớn nhất của biểu thức M là  13 2