p=(2x+1)(x-2)+3(x2-1)-9x
18
Những câu hỏi liên quan
Bài 1: Giải các phương trình dưới đây1) x2 - 9 (x - 3)(5x +2)2) x3 - 1 (x - 1)(x2 - 2x +16)3) 4x2 (x - 1) - x + 1 04) x3 + 4x2 - 9x - 36 05) (3x + 5)2 (x - 1)26) 9 (2x + 1)2 4 (x - 5)27) x2 + 2x 158) x4 + 5x3 + 4x2 09) (x2 - 4) - (x - 2)(3 - 2x) 010) (3x + 2)(x2 - 1) (9x2 - 4) (x + 1)11) (3x - 1)(x2 + 2) (3x - 1)(7x - 10)12) (2x2 + 1) (4x - 3) (x - 12)(2x2 + 1)
Đọc tiếp
Bài 1: Giải các phương trình dưới đây
1) x2 - 9 = (x - 3)(5x +2)
2) x3 - 1 = (x - 1)(x2 - 2x +16)
3) 4x2 (x - 1) - x + 1 = 0
4) x3 + 4x2 - 9x - 36 = 0
5) (3x + 5)2 = (x - 1)2
6) 9 (2x + 1)2 = 4 (x - 5)2
7) x2 + 2x = 15
8) x4 + 5x3 + 4x2 = 0
9) (x2 - 4) - (x - 2)(3 - 2x) = 0
10) (3x + 2)(x2 - 1) = (9x2 - 4) (x + 1)
11) (3x - 1)(x2 + 2) = (3x - 1)(7x - 10)
12) (2x2 + 1) (4x - 3) = (x - 12)(2x2 + 1)
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
Đúng 1
Bình luận (0)
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
Đúng 1
Bình luận (1)
7.
\(\Leftrightarrow x^2+2x-15=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
8.\(\Leftrightarrow x^4+x^3+4x^3+4x^2=0\)
\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0;x=-4\end{matrix}\right.\)
9.\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(3-2x\right)\)
\(\Leftrightarrow x+2=3-2x\)
\(\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)
Đúng 1
Bình luận (0)
Xem thêm câu trả lời
1/ Thực hiện phép nhân :
a) x2 ( 5x3 - x - \(\dfrac{1}{2}\))
b) ( 3xy - x2 + y ) \(\dfrac{2}{3}\)x2y
c) x2 ( 4x3 - 5xy + 2x ) ( -\(\dfrac{1}{2}\) xy )
2/ Tìm x, biết
a) 3x( 12x - 4 ) - 9x (4x - 3 ) = 30
b ) x( 5 - 2x ) + 2x ( x - 1 )= 15
2.
a. 3x(12x - 4) - 9x(4x - 3) = 30
<=> 36x2 - 12x - 36x2 + 27x = 30
<=> 36x2 - 36x2 - 12x + 27x = 30
<=> 15x = 30
<=> x = 2
b. x(5 - 2x) + 2x(x - 1) = 15
<=> 5x - 2x2 + 2x2 - 2x = 15
<=> -2x2 + 2x2 + 5x - 2x = 15
<=> 3x = 15
<=> x = 5
Đúng 1
Bình luận (0)
a) x2 ( 5x3 - x - 2323x2y= 6969x3y2- 2323x4y+ 2323x2y2
c) x2 ( 4x3 - 5xy + 2x ) ( -
Đúng 0
Bình luận (0)
Thực hiện phép chia:
1. (-3x3 + 5x2 - 9x + 15) : ( 3x + 5)
2. ( 5x4 + 9x3 - 2x2 - 4x - 8) : ( x-1)
3. ( 5x3 + 14x2 + 12x + 8 ) : (x + 2)
4. ( x4 - 2x3 + 2x -1 ) : ( x2 - 1)
5. ( 5x2 - 3x3 + 15 - 9x ) : ( 5 - 3x)
6. ( -x2 + 6x3 - 26x + 21) : ( 3 -2x )
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
Đúng 3
Bình luận (0)
B) (2x+3)2-(5x-4) (5x+4)=(x+5)2-(3x-1) (7x+2)-(x2-x+1)
C) (1-3x)2-(x-2) (9x+1)=(3x-4) (3x+4)-9(x+3)2
D) (3x+4) (3x-4) - (2x+5)2=(x-5)2+(2x+1)2-(x2-2x)+(x-1)2 cần gắp
Thực hiện phép tính:
a,4.(x+3)/3x2-x : x2+3x/1-3x
b, x+1/x2-2x-8 . 4-x/x2+x
c, 9x+5/2(x-1)(x+3)2- 5x-7/2(x-1)(x+3)2
d, 18/(x-3)(x2-9)-3/x^2-6x+9-x/x^2-9
e, 1/x2-x+1+1/1-x2+2/x3+1
Chủ đề 1: Thực hiện phép tính1) (2x+3).(2x-3)-4x.(x+5) 2) 6/x2 - 9 + 5/x-3 + 1/x+3 3)5x.(x-3)+(x-2)24) 4x/x+2 - 3x/x-2 + 12x/ x2 - 45) x(x+2) - ( x-3)(x+3)6) 1/3x-2 + -4/3+2 + 6-3x/9x2 - 47)2x.(3x-1)+(x+2)2 8) 6/x+3 - 6/x-3 + 9x+9/x2 - 99) (2x - 5)2 - x(4x-13)10) x-1/x + 4/x+8 + 8/x2 + 8x11) (2x+1)2 + (x-5)(x+5)-x(5x+7)12) 6/x2-9 + 5/x-3 + 1/x+313) 6x(5x-2)+(2x+3)214) x/x-2 + -2/x-3 + x(1-x)/x2-915) (x-2)2-x(x+5)16) 2/x+3 + 3/x-3 + -6/x2-917) 3x(x-3) + (3x-1)2
Đọc tiếp
Chủ đề 1: Thực hiện phép tính
1) (2x+3).(2x-3)-4x.(x+5)
2) 6/x2 - 9 + 5/x-3 + 1/x+3
3)5x.(x-3)+(x-2)2
4) 4x/x+2 - 3x/x-2 + 12x/ x2 - 4
5) x(x+2) - ( x-3)(x+3)
6) 1/3x-2 + -4/3+2 + 6-3x/9x2 - 4
7)2x.(3x-1)+(x+2)2
8) 6/x+3 - 6/x-3 + 9x+9/x2 - 9
9) (2x - 5)2 - x(4x-13)
10) x-1/x + 4/x+8 + 8/x2 + 8x
11) (2x+1)2 + (x-5)(x+5)-x(5x+7)
12) 6/x2-9 + 5/x-3 + 1/x+3
13) 6x(5x-2)+(2x+3)2
14) x/x-2 + -2/x-3 + x(1-x)/x2-9
15) (x-2)2-x(x+5)
16) 2/x+3 + 3/x-3 + -6/x2-9
17) 3x(x-3) + (3x-1)2
\(\left(2x+3\right)\left(2x-3\right)-4x\left(x+5\right)=4x^2-9-4x^2-20x=-20x-9\)
\(5x\left(x-3\right)+\left(x-2\right)^2=5x^2-15x+x^2-4x+4=6x^2-19x+4\)
\(x\left(x+2\right)-\left(x-3\right)\left(x+3\right)=x^2+2x-\left(x^2-9\right)=x^2+2x-x^2+9=2x+9\)
Đúng 0
Bình luận (0)
1) (\(\dfrac{1}{2}\)x + 3)*(x2- 4x- 6)
2) (6x2 -9x +15)*(\(\dfrac{2}{3}\)x+1)
3) (3x2 -x+5)*(x3+5x-1)
4) (x-1)*(x+1)*(x-2)
5) D=(x-7)*(x+5)-(x-4)*(x+3)
6) E= 4x*(x2-x-1)-(x+3)*(x2-2)
7) F= 5x*(x-3)*(x-1)-4x*(x2-2x)
1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)
\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)
\(=\dfrac{1}{2}x^3+x^2-15x-18\)
2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)
\(=4x^3+6x^2-6x^2-9x+10x+15\)
\(=4x^3+x+15\)
3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)
\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)
\(=3x^5-x^4+5x^3+10x^2+26x-5\)
4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)
Đúng 2
Bình luận (0)
1Giải các phương trình sau bằng cách đưa về phương trình tích:
a) 3x(2x – 3) = 5(3 – 2x)
b) (x2 + 1)(2x + 5) = (x – 1)(x2 + 1)
c) 3x3 = x2 + 3x - 1
d) x2 – 9x + 20 = 0
2Giải các phương trình sau bằng cách đưa về phương trình tích:
a) 3x(2x – 3) = 5(3 – 2x)
b) (x2 + 1)(2x + 5) = (x – 1)(x2 + 1)
c) 3x3 = x2 + 3x - 1
d) x2 – 9x + 20 = 0
a/ \(3x(2x-3)=5(3-2x) \Leftrightarrow 3x(2x-3)+5(2x-3)=0 \\\ \Leftrightarrow (2x-3)(3x+5)=0 \)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{5}{3}\end{matrix}\right.\)
KL: .............
b/ \(\left(x^2+1\right)\left(2x+5\right)=\left(x-1\right)\left(x^2+1\right)\Leftrightarrow\left(x^2+1\right)\left(2x+5\right)-\left(x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+5-x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x+6=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-6\end{matrix}\right.\)
KL: .............
c/ \(3x^3=x^2+3x-1\Leftrightarrow3x^3-x^2-3x+1=0\Leftrightarrow x^2\left(3x-1\right)-\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=1\\x=-1\end{matrix}\right.\)
KL: ..........
d/ \(x^2-9x+20=0\Leftrightarrow x^2-5x-4x+20=0\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
KL: .............
Đúng 0
Bình luận (0)
\(a,3x\left(2x-3\right)=5\left(3-2x\right)\Leftrightarrow6x^2-9x=15-10x\Leftrightarrow6x^2-9x-15+10x=0\Leftrightarrow6x^2+x-15=0\Leftrightarrow\left(3x+5\right)\left(2x-3\right)=0\)
\(\left(3x+5\right)\left(2x-3\right)=0\)
\(\left[{}\begin{matrix}x=-\frac{5}{3}\\x=\frac{3}{2}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tìm x biết.
a)(x+2)3-x2(x+6)=0
b) (2x+3)3-8x(x-1)(x+1)=9x(4x-3)
c)(2-x)3+(2+x)3-12x(x+1)=0
a) \(\left(x+2\right)^3-x^2\left(x+6\right)=0\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=0\)
\(\Leftrightarrow12x+8=0\)
\(\Leftrightarrow12x=-8\)
\(\Leftrightarrow x=-\dfrac{8}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
b) \(\left(2x+3\right)^3-8x\left(x+1\right)\left(x-1\right)=9x\left(4x-3\right)\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x\left(x^2-1\right)=36x^2-27x\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3+8x=36x^2-27x\)
\(\Leftrightarrow8x^3-8x^3+36x^2-36x^2+54x+27x+8x+27=0\)
\(\Leftrightarrow89x+27=0\)
\(\Leftrightarrow x=-\dfrac{27}{89}\)
c) \(\left(2-x\right)^3+\left(2+x\right)^3-12x\left(x+1\right)=0\)
\(\Leftrightarrow8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x^2-12x=0\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(6x^2+6x^2-12x^2\right)-\left(12x-12x\right)+12x+\left(8+8\right)=0\)
\(\Leftrightarrow12x+16=0\)
\(\Leftrightarrow x=-\dfrac{16}{12}\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
Đúng 2
Bình luận (2)
`#040911`
`a)`
`(x + 2)^3 - x^2(x + 6) = 0`
`<=> x^3 + 6x^2 + 12x + 8 - x^3 - 6x^2 = 0`
`<=> (x^3 - x^3) + (6x^2 - 6x^2) + 12x = 0`
`<=> 12x = 0`
`<=> x = 0`
Vậy, `x = 0.`
`b)`
`(2x + 3)^3 - 8x(x - 1)(x + 1) = 9x(4x - 3)`
`<=> 8x^3 + 36x^2 + 54x + 27 - 8x(x^2 - 1) = 36x^2 - 27x`
`<=> 8x^3 + 36x^2 + 54x + 27 - 8x^3 + 8x - 36x^2 + 27x = 0`
`<=> (8x^3 - 8x^3) + (36x^2 - 36x^2) + (54x + 8x + 27x) + 27 = 0`
`<=> 89x + 27 = 0`
`<=> 89x = -27`
`<=> x = -27/89`
Vậy, `x = -27/89`
`c)`
`(2 - x)^3 + (2 + x)^3 - 12x(x + 1) = 0`
`<=> 8 - 12x + 6x^2 - x^3 + 8 + 12x + 6x^2 + x^3 - 12x^2 - 12x = 0`
`<=> (-x^3 + x^3) + (12x - 12x - 12x) + (6x^2 + 6x^2 - 12x^2) + (8 + 8)=0`
`<=> -12x + 16 = 0`
`<=> -12x = -16`
`<=> 12x = 16`
`<=> x=4/3`
Vậy, `x = 4/3.`
Đúng 3
Bình luận (0)