sai đề rồi, nếu là 3vs 7 thì tổng đâu chia hết 3

( x² - 1 )( x² + 4x + 3 ) = 192

<=> ( x - 1 )( x + 1 )( x + 3 )( x + 1 ) - 192 = 0 [ phân tích thì quá dễ rồi mình k làm bước trung gian ]

<=> [ ( x - 1 )( x + 3 ) ]( x + 1 )² - 192 = 0

<=> ( x² + 2x - 3 )( x² + 2x + 1 ) - 192 = 0

<=> ( x² + 2x - 1 - 2 )( x² - 2x - 1 + 2 ) - 192 = 0

<=> ( x² + 2x - 1 )² - 4 - 192 = 0

<=> ( x² + 2x - 1 )² - 14² = 0

<=> ( x² + 2x - 15 )( x² + 2x + 13 ) = 0

Vì x² + 2x + 13 = ( x + 1 )² + 12 ≥ 12 > 0  ∀ x

nên pt <=> x² + 2x - 15 = 0

<=> x² - 3x + 5x - 15 = 0

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

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

<=> x = 3 hoặc x = -5

Vậy phương trình có tập nghiệm S = { 3 ; -5 }

\(\left(x^2-1\right)\left(x^2+4x+3\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+x+3x+3\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left[\left(x+1\right)x+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x+1\right)\left(x+3\right)=0\)
  ~Còn lại bạn tự giải nốt :3 ~

a) ĐKXĐ : x ≠ -1 ; x ≠ -2

\(Q=\left[\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{6x+3}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right]\times\frac{1}{x+2}\)

\(=\frac{x^2-x+1+6x+3-2x-2}{\left(x+1\right)\left(x^2-x+1\right)}\times\frac{1}{x+2}\)

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

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

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

b) Ta có : x² - x + 1 = ( x² - x + 1/4 ) + 3/4 = ( x - 1/2 )² + 3/4 ≥ 3/4 ∀ x

hay x² - x + 1 ≥ 3/4 ∀ x

=> \(\frac{1}{x^2-x+1}\le\frac{4}{3}\)hay Q ≤ 4/3 ∀ x

Dấu "=" xảy ra <=> x = 1/2(tm) . Vậy MaxQ = 4/3

\(A=\frac{xyz}{x+y}\Rightarrow\frac{1}{A}=\frac{x+y}{xyz}\)

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

 \(\frac{x+y}{xyz}=\frac{x}{xyz}+\frac{y}{xyz}=\frac{1}{yz}+\frac{1}{xz}\ge\frac{\left(1+1\right)^2}{yz+xz}=\frac{4}{z\left(x+y\right)}\)(1)

Lại có \(z\left(x+y\right)\le\frac{\left(x+y+z\right)^2}{4}=\frac{9}{4}\)(theo AM-GM) => \(\frac{4}{z\left(x+y\right)}\ge\frac{16}{9}\)(2)

Từ (1) và (2) => \(\frac{x+y}{xyz}\ge\frac{4}{z\left(x+y\right)}\ge\frac{16}{9}\)=> \(\frac{x+y}{xyz}\ge\frac{16}{9}\)hay \(\frac{1}{A}\ge\frac{16}{9}\)

=> A ≤ 9/16. Đẳng thức xảy ra <=> z = 3/2 ; x = y = 3/4

Vậy MaxA = 9/16 <=> x = y = 3/4 ; z = 3/2

\(9=3^2=\left(x+y+z\right)^2\ge4\left(x+y\right)z\)

\(\rightarrow9.\frac{x+y}{xyz}\ge4.\frac{\left(x+y\right)^2}{xy}\ge4.\frac{4xy}{xy}=16\)

\(\rightarrow\frac{x+y}{xyz}\ge\frac{16}{9}\rightarrow\frac{xyz}{x+y}\le\frac{9}{16}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{3}{4};z=\frac{3}{2}\)

1 + 1 + 1 = 3