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

`=6x^2-2x-2x^2+6x-x+3`

`=4x^2+3x+3`

2) `3(x^2-3x)-(4x+2)(x-1)`

`=3x^2-9x-4x^2+4x-2x+2`

`=-x^2-7x+2`

3) `3x(x-5)-(x-2)^2-(2x+3)(2x-3)`

`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`

`=3x^2-15x-x^2+4x-4-4x^2+9`

`=-2x^2-11x+5`

4) `(2x-3)^2+(2x-1)(x+4)`

`=4x^2-12x+9+2x^2+8x-x-4`

`=6x^2-5x+5`

Lời giải:
a.

$5x-[2x+1-(2x-3)-(4x+1)]=5x-(2x+1-2x+3-4x-1)$

$=5x-(-4x+3)=5x+4x-3=9x-3$

b.

$(-3x^2+2x-1)+(4x^2-2x+3)$

$=-3x^2+2x-1+4x^2-2x+3=x^2+2$

Ta có: \(2x\left(3x-1\right)-\left(2x+1\right)\left(x-3\right)\)

\(=6x^2-2x-\left(2x^2-6x+x-3\right)\)

\(=6x^2-2x-2x^2+5x+3\)

\(=4x^2+3x+3\)

Ta có: \(3\left(x^2-2x\right)-\left(4x+2\right)\left(x-1\right)\)

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

\(=3x^2-6x-4x^2+2x+2\)

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

\(2x\left(3x-1\right)-\left(2x+1\right)\left(x-3\right)=6x^2-2x-2x^2+5x+3=4x^2+3x+3\)

\(3\left(x^2-2x\right)-\left(4x+2\right)\left(x-1\right)=3x^2-6x-4x^2+2x-2=-x^2-4x-2\)

cậu xem câu hỏi của Quỳnh Phạm nha

\(\left(2x-3\right)\left(2x+3\right)-\left(2x-1\right)^2\)

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

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

\(=4x^2-9-4x^2+4x-1\)

\(=4x-10\)

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

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

=2^2=4

b)\(\left(2x^3-3x^2+6x-9\right)\left(2x-3\right)\)

\(=\left[x^2\left(2x-3\right)+x\left(2x-3\right)\right]\left(2x-3\right)\)

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

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

tự làm tiếp đi nha

tiếp câu b)

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

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

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

\(=\left[2x+1-\left(2x-1\right)\right]^2\)

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

\(A=\dfrac{2tan^2a+\dfrac{5}{cos^2a}}{4-\dfrac{3}{cos^2a}}=\dfrac{2tan^2a+5\left(1+tan^2a\right)}{4-3\left(1+tan^2a\right)}=...\) (bạn tự thay số bấm máy nhé)

\(B=\dfrac{3cot^2a-1}{cot^2a+2}=...\)

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

<=> (2x)^2 - 9 - (2x)^2 + 4x + 1
<=> 4x - 8
nếu x = 1/2
=> 4*1/2 - 8
<=> 2 - 8
<=> -6

give me a like or die noob

\(\left(2x+3\right)\left(2x-3\right)-\left(2x+1\right)^2\)

\(=4x^2-9-4x^2-4x-1\)

\(=-4x-10\)

Thay \(x=\dfrac{1}{2}\) vào \(-4x-10\) ta được:

\(\left(-4\right)\cdot\dfrac{1}{2}-10\)

\(=-12\)

Vậy: Giá trị của \(-4x-10\) tại \(x=\dfrac{1}{2}\) là \(-12\)

bằng 2^2 =4