Bài 1: Tìm x
f) x3 - 6x2 + 11x - 6 = 0
g) x + / 2x-1/ = 5
h) 2x3 + 3x2 - 32x= 48
Chú ý: /......./ là dấu giá trị tuyệt đối nhá.........
Bài 1: Tìm x
f) x3 - 6x2 + 11x - 6 = 0
g) x + / 2x-1/ = 5
h) 2x3 + 3x2 - 32x= 48
Chú ý: /......./ là dấu giá trị tuyệt đối nhá.........
Giải:
a) \(x^3-6x^2+11x-6=0\)
\(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-x-2^3+2=0\)
\(\Leftrightarrow x^3-3.x^2.2+3.x.2^2-2^3+2-x=0\)
\(\Leftrightarrow\left(x-2\right)^3+2-x=0\)
\(\Leftrightarrow\left(x-2\right)^3-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\left(x-2\right)^2-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)
Vậy ...
b) \(x+\left|2x-1\right|=5\)
\(\Leftrightarrow\left|2x-1\right|=5-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5-x\\2x-1=x-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=6\\x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy ...
tìm x biêt
/x/+2x=6
thực ra dấu / là dấu giá trị tuyệt đối á. chỉ cho cách bấm dấu giá trị tuyêt đối lun nhá
bài 1: Thực hiện phép tính
a/ (4x-3) (2x+5)
B/ (14X5y - 7x2y3 + 3X4y) :7x2y
c/ (2x3-3x2-11x +6):(x-3)
bài 2: Phân thức đa thức thành nhân tử
a/ x3-25x
b/ x2-2xy+3x-6y
c/ 8x3+4x2-6x-27
Bài 2:
a: =x(x^2-25)
=x(x-5)(x+5)
b: =x(x-2y)+3(x-2y)
=(x-2y)(x+3)
c: =(2x-3)(4x^2+6x+9)+2x(2x-3)
=(2x-3)(4x^2+8x+9)
1) (1-x)(5x+3)=(3x-7)(x-1)
2) (x-2)(x+1)=x2-4
3) 2x3+3x2-32x=48
4) x2+2x-15=0
5) 2x(2x-3)=(3-2x)(2-5x)
6) x3-5x2+6x=0
7) (x2-5)(x+3)=0
8) (x+7)(3x-1)=49-x2
\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)
\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)
\(< =>\left(1-x\right)\left(8x-4\right)=0\)
\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
\(\left(x-2\right)\left(x+1\right)=x^2-4\)
\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)
\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)
\(< =>-1\left(x-2\right)=0\)
\(< =>2-x=0< =>x=2\)
\(2x^3+3x^2-32x=48\)
\(< =>x^2\left(2x+3\right)-16\left(2x+3\right)=0\)
\(< =>\left(x^2-16\right)\left(2x+3\right)=0\)
\(< =>\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)
\(< =>\hept{\begin{cases}x=4\\x=-4\\x=-\frac{3}{2}\end{cases}}\)
a) x3 + x2 + x + 1 = 0
b) x3 - 6x2 + 11x - 6 = 0
c) x3 - x2 - 21x + 45 = 0
d) x4 + 2x3 - 4x2 - 5x - 6 = 0
a) Ta có: \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
mà \(x^2+1>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
b) Ta có: \(x^3-6x^2+11x-6=0\)
\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)
\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
Vậy: S={1;2;3}
c) Ta có: \(x^3-x^2-21x+45=0\)
\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)
\(\Leftrightarrow x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+5x-3x-15\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2\cdot\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy: S={3;-5}
d) Ta có: \(x^4+2x^3-4x^2-5x-6=0\)
\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+4x^2\cdot\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+4x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+3x^2+x^2+4x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+3\right)+\left(x+1\right)\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên (x-2)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy: S={2;-3}
Bài 5; Tìm x
a) x2-4=0
b) 2x(x+5)-3(5+x)=0
c) x3-6x2+11x-6=0
a) x² - 4 = 0
x² = 4
x = 2 hoặc x = -2
b) 2x(x + 5) - 3(5 + x) = 0
(x + 5)(2x - 3) = 0
X + 5 = 0 hoặc 2x - 3 = 0
*) x + 5 = 0
x = -5
*) 2x - 3 = 0
2x = 3
x = 3/2
c) x³ - 6x² + 11x - 6 = 0
x³ - x² - 5x² + 5x + 6x - 6 = 0
(x³ - x²) - (5x² - 5x) + (6x - 6) = 0
x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0
(x - 1)(x² - 5x + 6) = 0
(x - 1)(x² - 2x - 3x + 6) = 0
(x - 1)[(x² - 2x) - (3x - 6)] = 0
(x - 1)[x(x - 2) - 3(x - 2)] = 0
(x - 1)(x - 2)(x - 3) = 0
x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0
*) x - 1 = 0
x = 1
*) x - 2 = 0
x = 2
*) x - 3 = 0
x = 3
Vậy x = 1; x = 2; x = 3
Bài 1 : phân tích đa thức thành nhân tử.
3x2 + 2x – 1
x3 + 6x2 + 11x + 6
x4 + 2x2 – 3
ab + ac +b2 + 2bc + c2
a3 – b3 + c3 + 3abc
cho biểu thức A = {2x-1} -{x-5}
tìm x biết A =5
lưu ý dấu ngoặc của 2x-1 là giá trị tuyệt đối