4hx^2+2x^2-7x-h+6 trên (-4)x^2+10x-4

\(a,=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}=\dfrac{x+1}{x}\\ b,=\dfrac{-\left(x^2-5x-6\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+1\right)\left(x-6\right)}{\left(x+2\right)^2}\)

Bài 1 : 

\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)

\(=x^2-4x+4-x+9=x^2-5x+13\)

Bài 2 : 

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

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

b, Thay x = -4 ta được : 

\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)

\(b,=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\left(x\ne-y\right)\\ c,=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\left(x\ne-1;x\ne\pm2;x\ne0\right)\)

b: \(\dfrac{x^3-x^2y+xy^2}{x^3+y^3}=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\)

c: \(\dfrac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\)

ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)

Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)

Bài 1.

Ta có : B = ( x + 2 )² + ( x - 2 )² - 2( x + 2 )( x - 2 )

= [ ( x + 2 ) - ( x - 2 ) ]²

= ( x + 2 - x + 2 )²

= 4² = 16

=> B không phụ thuộc vào x

Vậy với x = -4 thì B vẫn bằng 16

Bài 2.

4x² - 4x + 1 = ( 2x )² - 2.2x.1 + 1² = ( 2x - 1 )²

Bài 3.

Ta có : \(A=\frac{3}{2}x^2+2x+3\)

\(=\frac{3}{2}\left(x^2+\frac{4}{3}x+\frac{4}{9}\right)+\frac{7}{3}\)

\(=\frac{3}{2}\left(x+\frac{2}{3}\right)^2+\frac{7}{3}\ge\frac{7}{3}\forall x\)

Dấu "=" xảy ra khi x = -2/3

=> MinA = 7/3 <=> x = -2/3

\(\frac{x^2+4x+3}{2x+6}\)

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

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

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

\(=\frac{x+1}{2}\)

\(\frac{x^2+4x+3}{2x+6}\)

ĐKXĐ : x ≠ -3

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

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

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

\(=\frac{x+1}{2}\)

KO

Để rút gọn biểu thức, ta sẽ thực hiện các phép tính và kết hợp các thành phần tương tự: P(2x-1).4x^2 + 2x + 1 + (x+1)x^2 - x + 1 = P(8x^3 - 4x^2) + 2x + 1 + x^3 + x^2 - x + 1 = P(8x^3) - P(4x^2) + x^3 + (2x-x) +(1+1) = **8Px^3 - 4Px^2**+ x^3 **+ x**+ **2** Vậy biểu thức đã được rút gọn thành: **8Px³ - 4Px²+x³+x+2**

1) ( 4x + 1 )² + ( 4x - 1 )² - 2( 4x + 1 ).( 4x - 1 )

= ( 4x + 1 - 4x - 1 )²

= 2²

= 4

2) 4x² - 9 + ( 2x + 3 )

= ( 2x )² - 3² + ( 2x + 3 )

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

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

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

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

1, (4x+1)^2 + (4x-1)^2 - 2(4x+1)(4x-1)

=[(4x+1)-(4x-1)]^2

=(4x+1-4x+1)^2

=2^2

=4

2, 4x^2 - 9 +(2x+3)

=(4x^2 - 9)+(2x+3)

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

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

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

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

=.= hok tốt!!

1. \(\dfrac{x^3-4x^2+4x}{x^2-4}=\dfrac{x\left(x^2-4x+4\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)}{x+2}\)

vậy ý còn lại thì sao anh? ._.