\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)

\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)

\(P=x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3\\ P=2x^2+x-x^3-2x^2+x^3-x+3\\ P=3\left(đfcm\right)\)

a) \(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)

\(minA=-3\Leftrightarrow x=2\)

b) \(B=-x^2-8x+5=-\left(x+4\right)^2+21\le21\)

\(maxB=21\Leftrightarrow x=-4\)

c) \(C=2x^2-8x+19=2\left(x-2\right)^2+11\ge11\)

\(minC=11\Leftrightarrow x=2\)

d) \(D=-3x^2-6x+1=-3\left(x+1\right)^2+4\le4\)

\(maxD=4\Leftrightarrow x=-1\)

a) A = (x-2)^2 - 3 >= -3

--> A nhỏ nhất bằng -3

 <=> x = 2

b) B = -(x+4)^2 + 21 <= 21

--> B lớn nhất bằng 21

<=> x = -4

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

Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)

\(b,=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(c,=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)

A = 5x² + 6

Do x² ≥ 0 

⇒ 5x² ≥ 0

⇒ 5x² + 6 ≥ 6

Vậy giá trị nhỏ nhất của A là 6 khi x = 0

--------------------

B = 4(2x - 4)² + 2023

Do (2x - 4)² ≥ 0

⇒ 4(2x - 4)² ≥ 0

⇒ 4(2x - 4)² + 2023 ≥ 2023

Vậy giá trị nhỏ nhất của B là 2023 khi x = 2

Có hai cách trình bày với bài này: một là bạn có thể liệt kê hết các phần tử ra hoặc bạn sắp xếp theo cùng thứ tự và tính như sau:

1: Ta có: \(x^2-2x-5\)

\(=x^2-2x+1-6\)

\(=\left(x-1\right)^2-6\ge-6\forall x\)

Dấu '=' xảy ra khi x=1

2: ta có: \(3x^2+5x-2\)

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

\(=3\left(x^2+2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{49}{36}\right)\)

\(=3\left(x+\dfrac{5}{6}\right)^2-\dfrac{49}{12}\ge-\dfrac{49}{12}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{6}\)

nhanh giùm mình được không

Bài 1: 

a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

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

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

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)

\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)

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: \(F=5x-3x^2+6\)

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

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

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