a, \(A=\left(-5x+4\right)\left(3x-2\right)+\left(-2x+3\right)\left(x-2\right)\)

\(=-15x^2+10x+12x-8=-15x^2+22x-8\)

Thay x = -2 vào biểu thức ta có : \(-15\left(-2\right)^2+22\left(-2\right)-8\)

\(=-15.4-44-8=-112\)

b, \(B=\left(x-9\right)\left(2x+3\right)-2\left(x+7\right)\left(x-5\right)\)

\(=2x^2+3x-18x-27=2x^2-15x-27\)

Thay x = -1/2 vào biểu thức ta có : \(2\left(-\frac{1}{2}\right)^2-15\left(-\frac{1}{2}\right)-27\)

\(=2.\frac{1}{4}+\frac{15}{2}-27=\frac{11}{2}+\frac{15}{2}+27=40\)

Bài làm:

a) \(A=\left(-5x+4\right)\left(3x-2\right)+\left(-2x+3\right)\left(x-2\right)\)

\(A=-15x^2+22x-8-2x^2+7x-6\)

\(A=-17x^2+29x-14\)

Thay x = -2 vào ta được:

\(A=-17.\left(-2\right)^2+29.\left(-2\right)-14\)

\(A=-68-58-14\)

\(A=-140\)

b) \(B=\left(x-9\right)\left(2x+3\right)-2\left(x+7\right)\left(x-5\right)\)

\(B=2x^2-15x-27-2\left(x^2+2x-35\right)\)

\(B=2x^2-15x-27-2x^2-4x+70\)

\(B=-19x+43\)

Thay x = -1/2 vào B ta được:

\(B=-19.\left(-\frac{1}{2}\right)+43=\frac{19}{2}+43=\frac{105}{2}\)

A = ( -5x + 4 )( 3x - 2 ) + ( -2x + 3 )( x - 2 )

A = -15x² + 22x - 8 - 2x² + 7x - 6

A = -17x² + 29x - 14

Thế x = -2 ta được :

A = -17(-2)² + 29(-2) - 14 = -68 - 58 - 14 = -140

b) B = ( x - 9 )( 2x + 3 ) - 2( x + 7 )( x - 5 )

B = 2x² - 15x - 27 - 2( x² + 2x - 35 )

B = 2x² - 15x - 27 - 2x² - 4x + 70

B = -19x + 43

Thế x = -1/2 ta được :

B = -19.(-1/2) + 43 = 19/2 + 43 = 105/2

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

c: \(x^6-3x^5+3x^4-x^3=\left(x^2-x\right)^3\)

a ) \(\left(2x-3\right)^3=\left(2x\right)^3-3.\left(2x\right)^2.3+3.2x.3^2-3^3=8x^3-36x^2+54x-27\)

Có tổng hệ số là \(8-36+54-27=-1\)

b ) \(\left(x^2+2\right)^4=x^8+8x^6+24x^4+32x^2+16\)

Có tổng hệ số là : \(1+8+24+32+16=81\)

c ) \(\left(3x-5\right)^5=243x^5-2025x^4+6750x^3-11250x^2+9375x-3125\)

Có tổng hệ số là : \(243-2025+6750-11250+9375-3125=-32\)

A = (2x - 1)(x + 2) - 3x² + (x - 1)²

= 2x² + 4x - x - 2 - 3x² + x² - 2x + 1

= (2x² - 3x² + x²) + (4x - x - 2x) + (-2 + 1)

= x - 1

B = (x - 2)(x² + 2x + 4) - (x³ + x²) - (3 - x)(3 + x)

= x³ - 8 - x³ - x² - 9 + x²

= (x³ - x³) + (-x² + x²) + (-8 - 9)

= -17

A = (2x - 1)(x + 2) - 3x² + (x - 1)²

= 2x² + 4x - x - 2 - 3x² + x² - 2x + 1

= (2x² - 3x² + x²) + (4x - x - 2x) + (-2 + 1)

= x - 1

B = (x - 2)(x² + 2x + 4) - (x³ + x²) - (3 - x)(3 + x)

= x³ - 8 - x³ - x² - 9 + x²

= (x³ - x³) + (-x² + x²) + (-8 - 9)

= -17

kết quả đây

chúc bạn học tốt

im chưa học\

Ta có : x² + 3x 

= x² + \(2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)

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

Ta có : (2x - 3)(4x² - 6x + 9) 

= (2x -3)(2x + 3)²

= (4x² - 9)(2x + 3)

\(a,=x^2+x+\dfrac{1}{4}\\ b,=4x^2+2x+\dfrac{1}{4}\\ c,=x^2-2+\dfrac{1}{x^2}\\ d,=4x^2+\dfrac{8}{3}x+\dfrac{4}{9}x^2\\ e,=a^2-1\\ f,=25x^4-4\)

\(a,\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)

\(b,\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)

\(c,\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)

\(d,\left(\dfrac{2x+2}{3x}\right)^2=\dfrac{\left(2x+2\right)^2}{9x^2}=\dfrac{4x^2+8x+4}{9x^2}\)

\(e,\left(a-1\right).\left(a+1\right)=a^2-1\)

\(f,\left(5x^2-2\right).\left(5x^2+2\right)=25x^4-4\)

a. \(\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)

b. \(\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)

c. \(\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)

d. \(\left(2x+\dfrac{2}{3x}\right)^2=4x^2+\dfrac{8}{3}+\dfrac{4}{9x^2}\)

e. (a - 1)(a + 1) = a² - 1

f. (5x² - 2)(5x² + 2) = 25x⁴ - 4 

g. (2a - 3)(2a + 3) = 4a² - 9

Bài 3:

a) \(4x^2+4x+1=\left(2x+1\right)^2\)

b) \(9x^2-12x+4=\left(3x-2\right)^2\)

c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)

Phần d bài 3 mk chưa lm dc

Bài 3: 

a: \(4x^2+4x+1=\left(2x+1\right)^2\)

b: \(9x^2-12x+4=\left(3x-2\right)^2\)

c: \(\dfrac{1}{4}a^2b^4+ab^2+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)

d: