Hình như ở câu phân tích đa thức đề là (3x+5)(x-2) chứ không phải (x+2)

1..

(x+1)(2-x)-(3x+5)(x+2)

=(x+1)(2-x)+(3x-5)(2-x)

=(2-x)(x+1+3x-5)

=(2-x)(4x-4)

2...

(3x-5)(2x+7)-(x+1)(6x-5)=16

=>(6x²⁺21x-10x-35)-(6x²-5x+6x-5)-16=0

=>6x²⁺21x-10x-35-6x²+5x-6x+5-16=0

=>(6x²-6x²)+(21x-10x+5x-6x)+(-35+5-16)=0

=>10x-46=0

=>10x     =46

=>    x     =4.6

 Vậy x=4.6(bạn nên đổi ra phân số thì sẽ đúng hơn)

6B, 7A, 8D

1)(x^2+3x+1)(x^2+3x+2)-6

Đặt t = x²  + 3x + 1

Khi đó PT có dạng:

t.(t + 1) - 6

= t² + t - 6

= t² - 2t - 3t - 6

= t.(t - 2) + 3.(t - 2)

= (t + 3).(t - 2)

= (x² + 3x + 1 + 3).(x² + 3x + 1 - 2)

= (x² + 3x + 4).(x² + 3x - 1)

\(1\hept{\begin{cases}\left(x^2+3x+2-1\right)\left(x^2+2x+2\right)-6\\\left(t-1\right)\left(t\right)-6\\t^2-t-6\end{cases}}.\) " đặt x^2+3x+2 = t

\(\hept{\begin{cases}t^2-\frac{2t.1}{2}+\frac{1}{4}-\left(\frac{24+1}{4}\right)\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\end{cases}}\)

\(\hept{\begin{cases}\left(t-\frac{1}{2}-\frac{5}{2}\right)\left(t-\frac{1}{2}+\frac{5}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\end{cases}}\)

2)  \(\hept{\begin{cases}\left\{\left(x+1\right)\left(x+7\right)\right\}\left\{\left(x+5\right)\left(x+3\right)\right\}+15\\\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\\t\left(t+8\right)+15\end{cases}}\)  

\(\hept{\begin{cases}t^2+8t+15\\\left(t^2+8t+16\right)-1\\\left(t+4\right)^2-1\end{cases}}\Leftrightarrow\left(t+5\right)\left(t+4\right)\)

\(\hept{\begin{cases}a^3\left(b-c\right)+b^3\left(c-a+b-b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(-c+a-b+b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(a-b\right)-b^3\left(b-c\right)+c^3\left(a-b\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\\\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+ab+c^2\right)\\\left(a-b\right)\left(b-c\right)\left(a^2+2ab+2b^2+c^2\right)\end{cases}}}\)

Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Bài 1.

a) x( 8x - 2 ) - 8x² + 12 = 0

<=> 8x² - 2x - 8x² + 12 = 0 

<=> 12 - 2x = 0

<=> 2x = 12

<=> x = 6

b) x( 4x - 5 ) - ( 2x + 1 )² = 0

<=> 4x² - 5x - ( 4x² + 4x + 1 ) = 0

<=> 4x² - 5x - 4x² - 4x - 1 = 0

<=> -9x - 1 = 0

<=> -9x = 1

<=> x = -1/9

c) ( 5 - 2x )( 2x + 7 ) = ( 2x - 5 )( 2x + 5 )

<=> -4x² - 4x + 35 = 4x² - 25

<=> -4x² - 4x + 35 - 4x² + 25 = 0

<=> -8x² - 4x + 60 = 0

<=> -8x² + 20x - 24x + 60 = 0

<=> -4x( 2x - 5 ) - 12( 2x - 5 ) = 0

<=> ( 2x - 5 )( -4x - 12 ) = 0

<=> \(\orbr{\begin{cases}2x-5=0\\-4x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)

d) 64x² - 49 = 0

<=> ( 8x )² - 7² = 0

<=> ( 8x - 7 )( 8x + 7 ) = 0

<=> \(\orbr{\begin{cases}8x-7=0\\8x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{7}{8}\end{cases}}\)

e) ( x² + 6x + 9 )( x² + 8x + 7 ) = 0

<=> ( x + 3 )²( x² + x + 7x + 7 ) = 0

<=> ( x + 3 )² [ x( x + 1 ) + 7( x + 1 ) ] = 0

<=> ( x + 3 )²( x + 1 )( x + 7 ) = 0

<=> x = -3 hoặc x = -1 hoặc x = -7

g) ( x² + 1 )( x² - 8x + 7 ) = 0

Vì x² + 1 ≥ 1 > 0 với mọi x

=> x² - 8x + 7 = 0

=> x² - x - 7x + 7 = 0

=> x( x - 1 ) - 7( x - 1 ) = 0

=> ( x - 1 )( x - 7 ) = 0

=> \(\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}\)

Bài 2.

a) ( x - 1 )² - ( x - 2 )( x + 2 )

= x² - 2x + 1 - ( x² - 4 )

= x² - 2x + 1 - x² + 4

= -2x + 5

b) ( 3x + 5 )² + ( 26x + 10 )( 2 - 3x ) + ( 2 - 3x )²

= 9x² + 30x + 25 - 78x² + 22x + 20 + 9x² - 12x + 4

= ( 9x² - 78x² + 9x² ) + ( 30x + 22x - 12x ) + ( 25 + 20 + 4 )

= -60x² + 40x² + 49

d) ( x + y )² - ( x + y - 2 )²

= [ x + y - ( x + y - 2 ) ][ x + y + ( x + y - 2 ) ]

= ( x + y - x - y + 2 )( x + y + x + y - 2 )

= 2( 2x + 2y - 2 )

= 4x + 4y - 4

Bài 3.

 A = 3x² + 18x + 33

= 3( x² + 6x + 9 ) + 6 

= 3( x + 3 )² + 6 ≥ 6 ∀ x

Đẳng thức xảy ra <=> x + 3 = 0 => x = -3

=> MinA = 6 <=> x = -3

B = x² - 6x + 10 + y²

= ( x² - 6x + 9 ) + y² + 1

= ( x - 3 )² + y² + 1 ≥ 1 ∀ x,y

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-3=0\\y^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=0\end{cases}}\)

=> MinB = 1 <=> x = 3 ; y = 0

C = ( 2x - 1 )² + ( x + 2 )²

= 4x² - 4x + 1 + x² + 4x + 4

= 5x² + 5 ≥ 5 ∀ x

Đẳng thức xảy ra <=> 5x² = 0 => x = 0

=> MinC = 5 <=> x = 0

D = -2/7x² - 8x + 7 ( sửa thành tìm Max )

Để D đạt GTLN => 7x² - 8x + 7 đạt GTNN

7x² - 8x + 7 

= 7( x² - 8/7x + 16/49 ) + 33/7

= 7( x - 4/7 )² + 33/7 ≥ 33/7 ∀ x

Đẳng thức xảy ra <=> x - 4/7 = 0 => x = 4/7

=> MaxC = \(\frac{-2}{\frac{33}{7}}=-\frac{14}{33}\)<=> x = 4/7