min a nếu x = 0

=>0 + 0 - 0 + 2038

=> A = 2038

Hội con 🐄 chúc bạn học tốt!!! 

\(A=x^4+6x^2+3^2+x^2-4x+2^2+2025.\)

\(A=\left(x^2+3\right)^2+\left(x-2\right)^2+2025\)

Vì \(\hept{\begin{cases}\left(x^2+3\right)^2\ge0\forall x\\\left(x-2\right)^2\ge0\forall x\end{cases}}\Rightarrow\left(x^2+3\right)^2+\left(x-2\right)^2+2025\ge2025\forall x\)

Dấu '' = " xảy ra khi 

\(\left(x^2+3\right)^2=0\)                                    hoặc                                              \(\left(x-2\right)^2=0\)

\(\Rightarrow x=\pm\sqrt{3}\)                                                                                             \(\Rightarrow x=2\)

Vậy \(Min_A=2025\Leftrightarrow x=\pm\sqrt{3};x=2\)

Study well 

A = x² -4x - 9 

A = ( x - 2 ) ² -13 

Vì ( x - 2 ) ² >= 0

=> A >= -13

Dấu "=" xảy ra khi x -2 =0

          <=> x =2

 Vậy A min = -13 khi x =2

Ta có : \(P=2x^2-8x+1=2\left(x^2-4x\right)+1=2\left(x^2-4x+4-4\right)+1=2\left(x-2\right)^2-7\)

Vì \(2\left(x-2\right)^2\ge0\forall x\) 

Nên : \(P=2\left(x-2\right)^2-7\ge-7\forall x\in R\)

Vậy \(P_{min}=-7\) khi x = 2

1.a) Không tồn tại\(\)

   b) 1997 tại x=4

   c) 4 tại x=1;y=2

   d) 164 tại x=8

2.a) x>3 và x<-1

   b) Không tốn tại x

a)  -4x(x - 7) + 4x(x² - 5) = 28x² - 13

=> -4x² + 28x + 4x² - 20x = 28x² - 13

=> (-4x² + 4x²) + (28x - 20x) = 28x² - 13

=> 8x = 28x² - 13

=> 8x - 28x² + 13 = 0

=> phương trình vô nghiệm

b) (4x² - 5x)(3x + 2) - 7x(x + 5) = (-4 + x)(-2x - 3) + 12x² + 2x²

=> 4x²(3x + 2) - 5x(3x + 2) - 7x² - 35x = -4(-2x - 3) + x(-2x - 3) + 14x²

=> 12x³ + 8x² - 15x² - 10x - 7x² - 35x = 8x + 12 - 2x² - 3x + 14x²

=> 12x³ + (8x² - 15x² - 7x²) + (-10x - 35x) = (8x - 3x) + 12 + (-2x² + 14x²)

=> 12x³ - 14x² - 45x = 5x + 12 + 12x²

=> 12x³ - 14x² - 45x - 5x - 12 - 12x² = 0

=> 12x³ + (-14x² - 12x²) + (-45x - 5x) - 12 = 0

=> 12x³ - 26x² - 50x - 12 = 0

Làm nốt

Cái câu b sửa cái đề lại nhé dấu " = " ở chỗ (-2x = 3) là gì vậy?

a, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)

\(\Leftrightarrow-32x^2+8x+4x^3+13=0\)( vô nghiệm ) 

Ta có A= x^3 + 2x^2 + 5x + 10/ x^2 + 4x+4 

A= x^2(x+2)+5(x+2)/ (x+2)^2

A= (x^2)(x^2+5)/ (x+2)(x+2)

A= x^2+5/ x+2 

Để A= x^2+5/ x+2 bé nhất thì x^2+5 phải bé nhất

MÀ x^2 lớn hơn hoặc = 0 vs mọi x => x^2=0 => x^2 + 5 = 5 vs x=0

Thay x=0 vào A có 0^2 + 5/ 0+2 = 5/2

Vậy MinA=5/2 vs x=0

a) x³ - 7x + 6 = 0

x³ - x - 6x + 6 = 0

(x³ - x) - (6x - 6) = 0

x(x² - 1) - 6(x - 1) = 0

x(x - 1)(x + 1) - 6(x - 1) = 0

(x - 1)[x(x + 1) - 6] = 0

(x - 1)(x² + x - 6) = 0

(x - 1)(x² - 2x + 3x - 6) = 0

(x - 1)[(x² - 2x) + (3x - 6)] = 0

(x - 1)[x(x - 2) + 3(x - 2)] = 0

(x - 1)(x - 2)(x + 3) = 0

x - 1 = 0 hoặc x - 2 = 0 hoăkc x + 3 = 0

*) x - 1 = 0

x = 1

*) x - 2 = 0

x = 2

*) x + 3 = 0

x = -3

Vậy x = -3; x = 1; x = 2

a: \(x^3-7x+6=0\)

=>\(x^3-x-6x+6=0\)

=>\(x\left(x^2-1\right)-6\left(x-1\right)=0\)

=>x(x-1)(x+1)-6(x-1)=0

=>(x-1)(x^2+x-6)=0

=>(x-1)(x+3)(x-2)=0

=>\(\left[\begin{array}{l}x-1=0\\ x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-3\\ x=2\end{array}\right.\)

b: \(x^4+4x^2-5=0\)

=>\(x^4+5x^2-x^2-5=0\)

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

=>\(x^2-1=0\)

=>\(x^2=1\)

=>\(\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)

c: \(x^4+x^3-x^2-x=0\)

=>\(x^3\left(x+1\right)-x\left(x+1\right)=0\)

=>\(\left(x+1\right)\left(x^3-x\right)=0\)

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

=>\(\left[\begin{array}{l}x=0\\ x+1=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-1\\ x=1\end{array}\right.\)

d: \(x^2+6x-x-6=0\)

=>x(x+6)-(x+6)=0

=>(x+6)(x-1)=0

=>\(\left[\begin{array}{l}x+6=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-6\\ x=1\end{array}\right.\)

e: \(x^2-4x+5x-20=0\)

=>x(x-4)+5(x-4)=0

=>(x-4)(x+5)=0

=>\(\left[\begin{array}{l}x-4=0\\ x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-5\end{array}\right.\)

f: \(x^2-10x+2x-20=0\)

=>x(x-10)+2(x-10)=0

=>(x-10)(x+2)=0

=>\(\left[\begin{array}{l}x-10=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=10\\ x=-2\end{array}\right.\)

g: \(x^4-x^3-x^2+1=0\)

=>\(x^3\left(x-1\right)-\left(x^2-1\right)=0\)

=>\(x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)

=>\(\left(x-1\right)\left(x^3-x-1\right)=0\)

TH1: x-1=0

=>x=1

TH2: \(x^3-x-1=0\)

=>x≃1,32

h: \(x^5+x^4+x^3+x^2+x+1=0\)

=>\(x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)=0\)

=>\(\left(x^2+x+1\right)\left(x^3+1\right)=0\)

mà \(x^2+x+1=\left(x+\frac12\right)^2+\frac34\ge\frac34>0\forall x\)

nên \(x^3+1=0\)

=>\(x^3=-1\)

=>x=-1

i: \(x^2-9+\left(x+3\right)\left(3x-5\right)=0\)

=>(x-3)(x+3)+(x+3)(3x-5)=0

=>(x+3)(x-3+3x-5)=0

=>(x+3)(4x-8)=0

=>4(x+3)(x-2)=0

=>(x+3)(x-2)=0

=>\(\left[\begin{array}{l}x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=2\end{array}\right.\)

j: \(64x^2-9+8x+3=0\)

=>(8x+3)(8x-3)+(8x+3)=0

=>(8x+3)(8x-3+1)=0

=>(8x+3)(8x-2)=0

=>\(\left[\begin{array}{l}8x+3=0\\ 8x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac38\\ x=\frac28=\frac14\end{array}\right.\)

a)

\(A=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}\)

\(A-2=-\dfrac{3}{x^2-8x+22}=-\dfrac{3}{\left(x-4\right)^2+6}\ge-\dfrac{3}{6}=-\dfrac{1}{2}\)

\(A\ge\dfrac{3}{2}\) khi x =4