Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)
a) (x + 2)(x2 - 2x + 4) - -x(x2 + 2) = 15
<=> x3 + 8 - x3 - 2x = 15
<=> 2x = -7
<=> x= -7/2
Vậy S = {-7/2}
b) (x - 2)3 - (x - 3)(x2 + 3x + 9) + 6(x + 1)2 = 49
<=> x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x + 6 = 49
<=> 24x = 24
<=> x = 1
Vậy S = {1}
(x - 2)3 - (x - 3)(x2 + 3x + 9) + 6(x + 1)2 = 49
<=>x3-6x2+12x-8-(x3-27)+6(x2+2x+1)=49
<=>x3-6x2+12x-8-x3+27+6x2+12x+6=49
<=>24x+25=49
<=>24x=24
<=>x=1
x(x + 5)(x - 5) - (x + 2)(x2 - 2x + 4) = 42
<=>x(x2-25)-(x3+8)=42
<=>x3-25x-x3-8=42
<=>-25x-8=42
<=>-25x=50
<=>x=-2
Tìm x, biết:
1) 2x ( x - 5) - x ( 2x - 4 ) = 15
<=> 2x2 - 10x - 2x2 + 4x - 15 = 0
<=> -6x - 15 = 0
<=> -6x = 15
<=> x = -15/6
2) ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6
<=> x2 + 2x + x + 2 - x2 - 3x - 4x - 12 - 6 = 0
<=> -4x = -16
<=> x = 4
3) 4x2 - 4x + 5 - x ( 4x - 3) = 1 - 2x
<=> 4x2 - 4x + 5 - 4x2 + 3x - 1 + 2x = 0
<=> x + 4 = 0
<=> x = -4
4) ( x + 3 ) ( 2x + 1 ) - 2x2 = 4x - 5
<=> 2x2 + x + 6x + 3 - 2x2 - 4x + 5 = 0
<=> 3x + 8 = 0
<=> 3x = -8
<=> x = -8/3
5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0
<=> - 8x + 32 + 8x2 + 6x - 4x - 3 = 0
.......
6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)
<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0
<=> -2x + 40 = 0
<=> -2x = -40
<=> x = 20
Còn lại tương tự ....
1
a) \(\left(3x+1\right)\left(3x-1\right)=9x^2-1\)
\(\left(x+5y\right)\left(x-5y\right)=x^2-25y\)
b) \(\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\)
\(\left(x-5\right)\left(x^2+5x+25\right)=x^3-125\)
Bài 3:
a: \(\Leftrightarrow x^2+8x+16-x^2+1=16\)
=>8x+1=0
=>x=-1/8
b: \(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
=>2x+255=0
=>x=-255/2
c: \(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6x^2+12x+6=49\)
=>24x+62=49
=>24x=-13
=>x=-13/24
d: =>x^3+8-x^3-2x=15
=>-2x=15-8=7
=>x=-7/2
a) (x - 2)3 - (x - 3)(x2 + 3x + 9) + 6(x + 1)2=49
=> x3 - 6x2 + 24x - 8 - x3 + 3x2 - 3x2 + 9x - 9x + 27 + 6x2 + 12x +6 = 49
=> 36x + 25 - 49 = 0 => 36x - 24 = 0 => x = 2/3
b) (x + 2)(x2 - 2x + 4)-x(x2 + 2)=15
=> x3 + 2x2 - 2x2 - 4x + 4x + 8 - x3 - 2x = 15
=> 8 - 2x - 15 = 0 => -2x - 7 = 0 => x = -7/2