<=> x^2+4x+4-x^2+4x-4=8x
<=>8x=8x
<=>x thuộc R
Ta có: (x+2)(x+2)-(x-2)(x-2)=8x
\(\Leftrightarrow x^2+4x+4-x^2+4x-4-8x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
Tìm x, biết:
a) (x-2)(x-1) = x(2x+1)+2
b) (x+2)(x+2) - (x-2)(x-2) = 8x
c) (2x-1)(x2-x+1) = 2x3-3x2+2
d) (x+1)(x2+2x+4) - x3 - 3x2 + 16 = 0
e) (x+1)(x+2)(x+5) - x3- 8x2 = 27
(x+1)(x+2)(x+5)-(x^3-8x^2)
giải các bất phương trình sau:
a) 3x-5 > 2(x-1)+x
b)\(\left(x+2\right)^2-\left(x-2\right)^2>8x-2\)
c)3(4x+1) - 2(5x+2)≥ 8x-2
d)\(2x^2+2x+1-\dfrac{15\left(x+1\right)}{2}\le2x\left(x+1\right)\)
1. (2x +3) (x- 5) +(X-4 ) + (x-5) ( x-2 0 = (3x -5) ( x-4)
2. (8x-3) (3x +2) - (4x+ 7) (x+4) = (2x + 1) (5x-1)
3. 2x^2 + (x -1) (x+1)= 5x (x+10
4. 4- ( x-1) (x+5) - (x+2) (x+5) = 3(x-1) (x+2)
1) (x2-4x+16) (x+4)-x(x+1) (x+2)+3x2=0
2) (8x+2) (1-3x)+(6x-1) (4x-10)=-50
3) (x2+2x+4) (2-x)+x(x-3) (x+4)-x2+24=0
4) (\(\dfrac{x}{2}\)x2+3) (5-6x)+(12x-2) (\(\dfrac{x}{4}\)x4+3)=0
Ai giúp e với ah!
a)(8-5x).(x+2)+4.(x-2).(x+1)+2.(x-2).(x+2)=0
b)(8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1)-33
1) Thực hiện phép tính
a) (2x – 1)2 – 4 (x – 1)(x + 1) – (x – 2)(x + 3)
b) (2x + 3)(4x2 – 6x + 9) – 8x(x – 3)(x + 3)
1) 5(x-3) (x-7)-(5x+1) (x-2)= -8
2) x(x+1) (x+2)-(x+4) (3x-5)= 84-5x
3) (9x2-5) (x+3)-3x2(3x+9)=(x-5) (x+4)-x(x-11)
4) (x2-4x+16) (x+4)-x(x+1) (x+2)+3x2=0
5) (8x+2) (1-3x)+(6x-1) (4x-10)=-50
6) (x2+2x+4) (2-x)+x(x-3) (x+4)-x2+24=0
7) (\(\dfrac{x}{2}\)+3) (5-6x)+(12x-2) (\(\dfrac{x}{4}\)+3)=0
Ai giúp mình giải bài này với ạ!
Tìm x: (8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)-33
