Những câu hỏi liên quan
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
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)
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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 \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
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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.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
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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.\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(3-2x\right)\)
\(\Leftrightarrow x+2=3-2x\)
\(\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)
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Xem thêm câu trả lời
Bài 1 Rút gọn biểu thứca, [(3x - 2)(x + 1) - (2x + 5)(x2 - 1)] : (x + 1)b, (2x + 1)2 - 2(2x + 1)(3 - x) + (3 - x)2c, (x - 1)2 - (x + 1) (x2 - x + 1) - (3x + 1)(1 - 3x)d, (x2 + 1)(x - 3) - (x - 3)(x2 + 3x + 9)e, (3x +2)2 + (3x - 2)2 - 2(3x + 2)(3x - 2) + xBài 2 Phân tích các đa thức sau thành nhân tử1, 3(x + 4) - x2 - 4x2, x2 - xy + x - y3, 4x2 -25 + (2x + 7)(5 - 2x)4, x2 + 4x - y2 + 45, x3 - x2 - x + 16, x3 + x2y - 4x - 4y7, x3 - 3x2 + 1 - 3x8, 2x2 + 3x - 59, x2 - 7xy + 10y210, x3 - 2x2 + x - xy...
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Bài 1 Rút gọn biểu thức
a, [(3x - 2)(x + 1) - (2x + 5)(x2 - 1)] : (x + 1)
b, (2x + 1)2 - 2(2x + 1)(3 - x) + (3 - x)2
c, (x - 1)2 - (x + 1) (x2 - x + 1) - (3x + 1)(1 - 3x)
d, (x2 + 1)(x - 3) - (x - 3)(x2 + 3x + 9)
e, (3x +2)2 + (3x - 2)2 - 2(3x + 2)(3x - 2) + x
Bài 2 Phân tích các đa thức sau thành nhân tử
1, 3(x + 4) - x2 - 4x
2, x2 - xy + x - y
3, 4x2 -25 + (2x + 7)(5 - 2x)
4, x2 + 4x - y2 + 4
5, x3 - x2 - x + 1
6, x3 + x2y - 4x - 4y
7, x3 - 3x2 + 1 - 3x
8, 2x2 + 3x - 5
9, x2 - 7xy + 10y2
10, x3 - 2x2 + x - xy2
Giải các phương trình sau:a, (9x2 - 4)(x + 1) (3x +2)(x2 - 1)b, (x - 1)2 - 1 + x2 (1 - x)(x + 3)c, (x2 - 1)(x + 2)(x - 3) (x - 1)(x2 - 4)(x + 5)d, x4 + x3 + x + 1 0e, x3 - 7x + 6 0f, x4 - 4x3 + 12x - 9 0g, x5- 5x3 + 4x 0h, x4 - 4x3 + 3x2 + 4x - 4 0
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Giải các phương trình sau:
a, (9x2 - 4)(x + 1) = (3x +2)(x2 - 1)
b, (x - 1)2 - 1 + x2 = (1 - x)(x + 3)
c, (x2 - 1)(x + 2)(x - 3) = (x - 1)(x2 - 4)(x + 5)
d, x4 + x3 + x + 1 = 0
e, x3 - 7x + 6 = 0
f, x4 - 4x3 + 12x - 9 = 0
g, x5- 5x3 + 4x = 0
h, x4 - 4x3 + 3x2 + 4x - 4 = 0
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
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b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
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Bài 3: Rút gọn các biểu thức sau:
1) ( x+ 3)(x2 -3x + 9) - (x3 + 54)
2) (2x + y)(4x2 + 2xy + y2 ) - (2x – y)(4x2 + 2xy + y2 )
3) (x – 1)3 – (x + 2)(x2 -2x +4) +3(x +4)(x – 4)
4) x(x + 1)(x - 1) – (x + 1)(x2 – x +1)
5) 8x3 - 5 (2x + 1)(4x2 – 4x + 1)
6) 27 + (x – 3)(x2 +3x + 9)
7) (x – 1)3 – (x +2)(x2 -2x + 4) +3(x +4)(x -4)
8) (x – 2)3 +6( x – 1)2 –(x +1)(x2 -x +1) +3x
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
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\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
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a)A=3x(2/3x2-3x4)+(3x2)(x3-1)+(-2+9).x2-12
b)B=x(2x3+x+2)-2x2(x2+1)+x2-2x+1
c)C=x.(2x+1)-x2(x+2)+x3-x+3
a, \(A=2x^3-9x^5+3x^5-3x^2+7x^2-12=-6x^5+2x^3+4x^2-12\)
b, \(B=2x^4+x^2+2x-2x^3-2x^2+x^2-2x+1=2x^4-2x^3+1\)
c, \(C=2x^2+x-x^3-2x^2+x^3-x+3=3\)
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a) x2(x - 5) + 5 - x = 0; b) 3x4 - 9x3 = -9x2 + 27x;
c) x2(x + 8) + x2 = -8x; d) (x + 3)(x2 -3x + 5) = x2 + 3x.
e) 3x(x - 1) + x - 1 = 0;
f) (x - 2)(x2 + 2x + 7) + 2(x2 - 4) - 5(x - 2) = 0;
g) (2x - 1)2 - 25 = 0;
h) x3 + 27 + (x + 3)(x - 9) = 0.
i)8x3 - 50x = 0; k) 2(x + 3)-x2 - 3x = 0;
m)6x2 - 15x - (2x - 5)(2x + 5) =
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
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Mng giúp em với ạ :((
GIẢI PHƯƠNG TRÌNH
a) A= (3x - x2/ 9 - x2 - 1 ) : ( 9-x2 / x2 + x + 6 - x-3/ 2-x - x+2/ x+3
b) B= (x+2/ x-1 - x+1/ x+2 + x+5/x2 + 2 - 2) . ( x3-1)
Ui em k rõ là giải phương trình hay rút gọn nữa :((
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Tìm x, biết:a) 3x(x - 1) + x - 1 0;b) (x - 2)(
x
2
+ 2x + 7) + 2(
x
2
- 4) - 5(x - 2) 0;c)
(
2
x
-
1
)
2
- 25 0;d)
x
3
+ 27 + (x + 3)(x - 9) 0.
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Tìm x, biết:
a) 3x(x - 1) + x - 1 = 0;
b) (x - 2)( x 2 + 2x + 7) + 2( x 2 - 4) - 5(x - 2) = 0;
c) ( 2 x - 1 ) 2 - 25 = 0;
d) x 3 + 27 + (x + 3)(x - 9) = 0.
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
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