A B C D E K H I F

a,  xét tam giác FIC và tam giác EBC có : ^ECB chung

^FIC = ^EBC = 90

=> tam giác FIC đồng dạng với tam giác EBC (g-g)

b, có ^IFC + ^ICF = 90

^ICD + ^ICF = 90

=> ^IFC = ^ICD

xét tam giác FIC và tg CID có : ^FIC = ^CID = 90

=> tam giác FIC đồng dạng với tg CID (g-g)

=> IC/IF = ID/IC

=> IC^2 = IF.ID

a) 2(x-1)² - 4(x+3)² + 2x(x-5)

= 2(x² -2x +1)- 4(x² + 6x +9) + 2x² -10x

= 2x² - 4x + 2 -4x² - 24x - 36 + 2x² - 10x

= (2x² + 2x² - 4x²) - (4x + 24x+10x) +(2-36)

= -38x-34

b) 2(2x+5)²  -3(4x+1)(1-4x)

= 2(4x² + 20x + 25) + 3(4x+1)(4x-1)

= 8x² +40x + 50 + 3(16x² -1)

= 8x² + 40x + 50 + 48x² - 3

=56x² +40x + 47

a, \(2\left(x-1\right)^2-4\left(x+3\right)^2+2x\left(x-5\right)\)

\(=2\left(x^2-2x+1\right)-4\left(x^2+6x+9\right)+2x\left(x-5\right)\)

\(=2x^2-4x+2-4x^2-24x-36+2x^2-10=-28x-44\)

b, \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)

\(=2\left(4x^2+20x+25\right)-3\left(1-16x^2\right)\)

\(=8x^2+40x+50-3+48x^2=56x^2+40x+47\)

Tại x = 16 => x +1 = 17

Thay vào A ta được:

A = x⁴ - (x+1)x³ + (x+1)x² - (x+1)x + 20

A= x⁴ -(x⁴ + x³)  + (x³ + x²)  -(x² + x) +20

A= x⁴ - x⁴ - x³ + x³ + x² - x² -x + 20

A= - x+20

Mà  x = 16

=> A= -16 + 20 = 4

Vậy A= 4 khi x =16

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

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

= x^2 + 2x + 1 - x^2 + 2x -1 - 3x^2 + 3 

= -3x^2 + 4x + 3 

( x + y ) ^3 - ( x - y ) ^3 - 6x^2y 

= x^3 + 3x^2y + 3xy^2 + y^3 - ( x^3 - 3x^2y + 3xy^2 - y^3 ) - 6x^2y 

= x^3 + 3x^2y + 3xy^2 + y^3 - x^3 + 3x^2y - 3xy^2 + y^3 - 6x^2y 

= 2y^3 

Ta có x⁵ + x⁴ - 2x³ - 3x² - x + 1 

= x⁵ + x⁴ - 3x³ - 3x² + (x³ - x) + 1 

= x⁴(x + 1) - 3x²(x + 1) + 2 + 1 

= (x + 1)(x⁴ - 3x²) + 3 

= x(x + 1)(x³ - 3x) + 3 

= x(x + 1)(x³ - x - 2x) + 3 

= x(x + 1)(2 - 2x) + 3 

 = -2x(x + 1)(x - 1) + 3 

= -2(x³ - x) + 3 = -2.2 + 3 = -1

Khi đó P(x) = (x⁵ + x⁴ - 2x³ - 3x² - x + 1)²⁰¹⁸ = (-1)²⁰¹⁸ = 1

12,ĐK : x >= 0

 \(\left(\sqrt{2x}-y\right)^2=2x-2\sqrt{2x}y+y^2\)

13, \(\left(\frac{3}{2}x+3y\right)^2=\frac{9}{4}x+\frac{2.3x}{2}.3y+9y^2=\frac{9}{4}x+9xy+9y^2\)

14, ĐK : x ; y >= 0 

 \(\left(\sqrt{2x}+\sqrt{8y}\right)^2=2x+2\sqrt{16xy}+8y=2x+8\sqrt{xy}+8y\)

15, \(\left(x+\frac{1}{6}y+3\right)^2=x^2+\frac{1}{36}y^2+9-\frac{1}{3}xy-y-6x\)

sửa 13 = \(\frac{9}{4}x^2+9xy+9y^2\)

16, \(\left(\frac{1}{2}x-4y\right)^2=\frac{1}{4}x^2-4xy+16y^2\)

17, \(\left(\frac{x}{2}+2y^2\right)\left(\frac{x}{2}-2y^2\right)=\frac{x^2}{4}-4y^4\)

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

19, \(\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2-2\left(x+y\right)\left(x-y\right)\)

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

20, \(\left(2x+3\right)^2-\left(x+1\right)^2=\left(2x+3-x-1\right)\left(2x+3+x+1\right)=\left(x+2\right)\left(3x+4\right)\)