( 4x - 1 )³ + ( 3 - 4x )( 9 + 12x + 16x² ) = ( 8x - 1 )( 8x + 1 ) - ( 3x - 5 )

<=> 64x³ - 48x² + 12x - 1 + [ 3³ - ( 4x )³ ] = ( 8x )² - 1 - 3x + 5

<=> 64x³ - 48x² + 12x - 1 + 27 - 64x³ = 64x² - 3x + 4

<=> -48x² + 12x + 26 = 64x² - 3x + 4

<=> -48x² + 12x + 26 - 64x² + 3x - 4 = 0

<=> -112x² + 15x + 22 = 0 (*)

\(\Delta=b^2-4ac=15^2-4\cdot\left(-112\right)\cdot22=225+9856=10081\)

\(\Delta>0\)nên (*) có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{\sqrt{10081}-15}{-224}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-15-\sqrt{10081}}{-224}\end{cases}}\)

Lớp 8 sao nghiệm xấu thế -..-

Dạ em cảm ơn

\(B=\left(x^2-8x\right)\left(x^2-8x+24\right)\)

Đặt \(x^2-8x+12=t\) ta có:
\(B=\left(t-12\right)\left(t+12\right)=t^2-144\ge-144\)

Dấu "=" xảy ra khi \(t^2=0\Leftrightarrow t=0\Leftrightarrow x^2-8x+12=0\)

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

\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\Leftrightarrow x=2;x=6\)

\(C=5x^2+9y^2-6xy-12x+13\)

\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-12x+9\right)+4\)

\(=\left(x-3y\right)^2+\left(2x-3\right)^2+4\ge4\)

Dấu "=" xảy ra tại \(x=\frac{3}{2};y=\frac{1}{2}\)

\(C=4x^2-4xy+y^2+4x^2-16x+16+1\)

    \(=\left(2x-y\right)^2+(2x-4)^2+1\ge1\forall x;y\in R\)

Dấu "=" xảy ra<=> 2x-y=0  và   2x-4=0

                   <=>2x-y=0 và  x=2   <=>y=4 và x=

Vậy....

\(B=3x^2-12x+16\) 

   \(=x^2-12x+36+2x^2-20\) 

   \(=\left(x-6\right)^2+2x^2-20\ge-20\forall x\in R\) 

Dấu "=" xảy ra <=> \(\left(x-6\right)^2=0\)và \(2x^2=0\) 

                    <=>x₁ =6 và x₂ =0

Vậy....

a,A=(2x)²-2.2x.2+2²+11=(2x-2)²+11

Vì (2x-2)²luôn lớn hơn hoặc bằng 0

=>A>hoặc =0+11 hay a>hoặc =11

vậy GTNN của A là 11 khi x=1

a: \(A=x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)

Dấu '=' xảy ra khi x=3/2

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

=>\(\dfrac{3}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}< =3:\dfrac{7}{4}=\dfrac{12}{7}\)

=>C>=-12/7

Dấu '=' xảy ra khi x=1/2

\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\)ĐKXĐ là \(x\ne-2;x\ne-8;x\ne-4;x\ne-6\)

\(\dfrac{x^2+4x+4+2}{x+2}+\dfrac{x^2+16x+64+8}{x+8}=\dfrac{x^2+8x+16+4}{x+4}+\dfrac{x^2+12x+36+6}{x+6}\)\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)

\(\Leftrightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)

\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)

\(\Leftrightarrow\left(\dfrac{2}{x+2}-1\right)+\left(\dfrac{8}{x+8}-1\right)=\left(\dfrac{4}{x+4}-1\right)+\left(\dfrac{6}{x+6}-1\right)\)\(\Leftrightarrow\dfrac{-x}{x+2}+\dfrac{-x}{x+8}=\dfrac{-x}{x+4}+\dfrac{-x}{x+6}\)

\(\Leftrightarrow\dfrac{x}{x+2}+\dfrac{x}{x+8}-\dfrac{x}{x+4}-\dfrac{x}{x+6}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)=0\)

Do \(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\ne0\)

=> x=0

Vậy ....

thiếu nghiệm r bạn

P= 9x^2 + 12x -5

  = (3x)^2 + 2.3.2x + 4 -4 -5

  =(9x^2 + 2.3.2x + 4) -9

  = (3x+2)^2 -9 

min p = -9 => (3x+2)^2 = 0

                => x= -2/3

max p = -9 => x= -2/3