Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI

C= x^6+27/x^4 - 3x^3 +6x^2 -9x + 9

= (x^2+3)(x^4-3x^2+9)/(x^4+3x^2)-(3x^3+9x)+(3x^2+9)

=(x^2+3)(x^4+6x^2+9-9x^2)/(x^2+3x)(x^2-3x+3)

= (x^2+3+3x)(x^2+3-3x)/x^2+3-3x =x^2+3x+3

=(x^2+3x+9/4) -9/4+3 = (x+3/2)^2 +3/4 >= 3/4

Dấu = xảy ra khi x=-3/2

Vậy Cmin = 3/4 <=> x=-3/2

\(A=\dfrac{3x^2-6x+17}{x^2-2x+5}\)

= \(\dfrac{3x^2-6x+15+2}{x^2-2x+5}\)

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

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

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

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

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

vì (x-1)² ≥ 0 ∀ x

⇔ (x-1)² +4 ≥ 4

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

⇔\(3+\dfrac{2}{\left(x-1\right)^2+4}\le\dfrac{7}{2}\)

⇔ A \(\le\dfrac{7}{2}\)

⇔ Min A =\(\dfrac{7}{2}\)

khi x-1=0

⇔ x=1

vậy ....

Ta có:\(B=\dfrac{2x^2-16x+41}{x^2-8x+22}\)

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

\(B=2-\dfrac{3}{x^2-8x+16+6}\)

\(B=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{5}{2}\)

\(\Rightarrow MINB=\dfrac{5}{2}\Leftrightarrow x=4\)

d)\(D=\dfrac{x^6+512}{x^2+8}\)

\(D=\dfrac{x^6+8x^4-8x^4-64x^2+64x^2+512}{x^2+8}\)

\(D=\dfrac{x^4\left(x^2+8\right)-8x^2\left(x^2+8\right)+64\left(x^2+8\right)}{x^2+8}\)

\(D=\dfrac{\left(x^2+8\right)\left(x^4-8x^2+64\right)}{x^2+8}\)

\(D=x^4-8x^2+64\)

\(D=\left(x^2-4\right)^2+48\ge48\)

\(\Rightarrow MIND=48\Leftrightarrow x=\pm2\)

Bài 1: Tìm x

a) Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16-9=0\)

\(\Leftrightarrow-12x-24=0\)

\(\Leftrightarrow-12x=24\)

hay x=-2

Vậy: x=-2

b) Ta có: \(\left(3x-1\right)^2+2\left(x+3\right)^2+11\left(x+1\right)\left(1-x\right)=6\)

\(\Leftrightarrow9x^2-6x+1+2\left(x^2+6x+9\right)-11\left(x-1\right)\left(x+1\right)-6=0\)

\(\Leftrightarrow9x^2-6x+1+2x^2+12x+18-11\left(x^2-1\right)-6=0\)

\(\Leftrightarrow11x^2+6x+12-11x^2+11=0\)

\(\Leftrightarrow6x+23=0\)

\(\Leftrightarrow6x=-23\)

hay \(x=-\frac{23}{6}\)

Vậy: \(x=-\frac{23}{6}\)

c) Ta có: \(8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)

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

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

hay \(x=\frac{1}{2}\)

Vậy: \(x=\frac{1}{2}\)

d) Ta có: \(x^3+9x^2+27x+27=0\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3=0\)

\(\Leftrightarrow\left(x+3\right)^3=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

Vậy: x=-3

a) (2x + 1)² - 4(x + 2)² = 9

4x² + 4x + 1 - 4(x² + 4x + 4) = 9

4x² + 4x + 1 - 4x² - 16x - 16 = 9

-12x - 15 = 9

-12x = 9 + 15

-12x = 24

x = 12 : (-2)

x = -2

b) (3x - 1)² + 2(x + 3)² + 11(x + 1)(1 - x) = 6

9x² - 6x + 1 + 2(x² + 6x + 9) - 11(x + 1)(x - 1) = 6

9x² - 6x + 1 + 2x² + 12x + 18 - 11(x² - 1) = 6

9x² - 6x + 1 + 2x² + 12x + 18 - 11x² + 11 = 6

6x + 30 = 6

6x = 6 - 30

6x = -24

x = -24 : 6

x = -4

c) 8x³ - 12x² + 6x - 1 = 0

(2x)³ - 3.(2x)².1 + 3.2x.1² - 1³ = 0

(2x - 1)³ = 0

2x - 1 = 0

2x = 1

x = 1/2

d) x³ + 9x² + 27x + 27 = 0

x³ + 3.x².3 + 3.x.3² + 3³ = 0

(x + 3)³ = 0

x + 3 = 0

x = 0 - 3

x = -3

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

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

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

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

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

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

MinC = 3/4 (khi x = -3/2)

\(C=\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{x^4+3x^2-3x^3-9x+3x^2+9}=\dfrac{\left(x^2+3\right)\left(x^4+6x^2+9-9x^2\right)}{\left(x^2+3\right)\left(x^2-3x+3\right)}\\ C=\dfrac{\left(x^2+3\right)^2-9x^2}{x^2-3x+3}=\dfrac{\left(x^2-3x+3\right)\left(x^2+3x+3\right)}{x^2-3x+3}\\ C=x^2+3x+3=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)