(x+2)2+(x-3)2-2(x-1)(x+1)=9
1) \(\dfrac{1}{x^2+6x+9}+\dfrac{1}{6x-x^2+9}+\dfrac{x}{x^2-9}\) 2) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\) 3) \(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\)
Bài 2: Tìm x biết:
1,x\(^2\)+4x+4=25
2,(5-2x)\(^2\)-16=0
3,(x-3)\(^3\)-(x-3)(x\(^2\)+3x+9)+9(x+1)\(^2\)=15
4,3(x+2)\(^2\)+(2x-1)\(^2\)-7(x-3)9x+3)=36
5,(x-3)(x\(^2\)+3x+9)+x(x+2)(2-x)=1
6,(2x+1)\(^2\)-4(x+2)\(^2\)=9
7,(x+3)\(^{^{ }2}\)-(x-4)(x+8)=1
1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)
\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
Tìm x :
a) x (3x + 1) + (x -1)2 - (2x + 1)(2x -1) = 0
b) (x + 1)3 + (2 - x)3 - 9(x - 3)(x+3) = 0
c) (x - 1)3 - (x + 3)(x2 - 3x + 9) + 3x2 = 25
d) (x + 2)3 - ( x +1)(x2 - x + 1) - 6(x - 1)2 = 23
e) (x + 3)(x2 - 3x + 9) - x(x - 2)(x+2) + 11 = 0
f) x(x - 3) - x + 3 = 0
Lời giải:
a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$
$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$
$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$
$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$
$\Leftrightarrow -x+2=0$
$\Leftrightarrow x=2$
b.
$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$
$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$
$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$
$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$
$\Leftrightarrow -x+10=0\Leftrightarrow x=10$
c.
$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$
$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$
$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$
$\Leftrightarrow 3x-28=25$
$\Leftrightarrow x=\frac{53}{3}$
d.
$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$
$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$
$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$
$\Leftrgihtarrow 24x=22$
$\Leftrightarrow x=\frac{11}{12}$
e.
$(x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0$
$\Leftrightarrow x^3+3^3-x(x^2-4)+11=0$
$\Leftrightarrow x^3+27-x^3+4x+11=0$
$\Leftrightarrow (x^3-x^3)+4x+(27+11)=0$
$\Leftrightarrow 4x+38=0$
$\Leftrightarrow x=\frac{-19}{2}$
f.
$x(x-3)-x+3=0$
$\Leftrightarrow x(x-3)-(x-3)=0$
$\Leftrightarrow (x-3)(x-1)=0$
$\Leftrightarrow x-3=0$ hoặc $x-1=0$
$\Leftrightarrow x=3$ hoặc $x=1$
a, [(x/x^2-25) - (x-5/X^2+5x)] : (2x-5/x^2+5x) + ( x/ 5-x)
b, [(9/x^3-9x) + (1/x+3)] : [(x-3/x^2+ 3x) - ( x/3x+9)]
c, (1/x-1) - (x^3-x/x^2+1) . [(x/x^2+1-2x) + (1/1-x^2)]
1, ( x+1/3)^3
2, ( 2x+y^2)^3
3, ( 1/2x^2+1/3y)^3
4, ( 3x^2-2y)^3
5, ( 2/3x^2-1/2y)^3
6, ( 2x+1/2)^3
7, ( x-3)^3
8, ( x+1).(X^2+3x+9)
9, ( x-3).( x^2+3x+9)
10, ( x-2).( x^2+2x+4)
11, ( x+4).( x^2-4x+16)
12, ( x-3y).( x^2+3xy+9y^2)
13, ( x^2-1/3). ( x^4+1/3x^2+1/9)
14, ( 1/3x+2y).( 1/9x^2-2/3xy+4y^2)
Đưa về HĐT
x+1/x-1 - x-1/x+1 = x^2+3/1-x^2
x+2/x-3 + x-2/x+3 = -2(x^2+6)/9-x^2
x+3/x-3 + 17/x^2-9 = x-3/x+3
mn giúp mình mình cảm ơn nha :)
a, [(x/x^2-25) - (x-5/X^2+5x)] : (2x-5/x^2+5x) + ( x/ 5-x)
b, [(9/x^3-9x) + (1/x+3)] : [(x-3/x^2+ 3x) - ( x/3x+9)]
c, (1/x-1) - (x^3-x/x^2+1) . [(x/x^2+1-2x) + (1/1-x^2)]
Cần gấp ạ
√x^2-6x+9=3-x
x^2-1/2x+1/16=x+3/2
√x-2√x-1=√x-1-1
√9-4√5-√5=-2
Let's solve each equation step by step:
√(x^2 - 6x + 9) = 3 - xSquaring both sides of the equation, we get:
x^2 - 6x + 9 = (3 - x)^2
x^2 - 6x + 9 = 9 - 6x + x^2
The x^2 terms cancel out, and we are left with:
-6x = -6x
This equation is true for any value of x. Therefore, there are infinitely many solutions.
x^2 - (1/2)x + 1/16 = x + 3/2Moving all terms to one side of the equation, we get:
x^2 - (1/2)x - x + 3/2 - 1/16 = 0
x^2 - (3/2)x + 29/16 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -3/2, and c = 29/16. Plugging in these values, we get:
x = (3/2 ± √((-3/2)^2 - 4(1)(29/16))) / (2(1))
x = (3/2 ± √(9/4 - 29/4)) / 2
x = (3/2 ± √(-20/4)) / 2
x = (3/2 ± √(-5)) / 2
Since the square root of a negative number is not a real number, this equation has no real solutions.
√(x - 2)√(x - 1) = √(x - 1) - 1Squaring both sides of the equation, we get:
(x - 2)(x - 1) = (x - 1) - 2√(x - 1) + 1
x^2 - 3x + 2 = x - 1 - 2√(x - 1) + 1
x^2 - 4x + 2 = -2√(x - 1)
Squaring both sides again, we get:
(x^2 - 4x + 2)^2 = (-2√(x - 1))^2
x^4 - 8x^3 + 20x^2 - 16x + 4 = 4(x - 1)
x^4 - 8x^3 + 20x^2 - 16x + 4 = 4x - 4
Rearranging terms, we have:
x^4 - 8x^3 + 20x^2 - 20x + 8 = 0
This equation does not have a simple solution and requires further calculations or approximation methods to find the solutions.
√9 - 4√5 - √5 = -2Simplifying the left side of the equation, we get:
3 - 4√5 - √5 = -2
-√5 - 5 = -2
-√5 = 3
This equation is not true since the square root of a number cannot be negative.
Therefore, the given equations either have infinitely many solutions or no real solutions.
A= x+3/x-2 + x+2/3-x + x+2/x2-5x+6
B= x/x-1 - 2x/x2-1
C= x/x-5 - 3x/x2-1
D= x+1/x+2 + 3x+2/x2-4
E= 3x2+3x-3/x2+x-2 - x+1/x+2 + x-2/1-x
F= x/x-1 + 3/x+1 - 6x-4/x2-1
G= 2x/x+3 + x/x-3 - 3x2+3/x2-9
H= x+3/x-3 + x/x+3 - x2+9/x2-9
Tương tự mấy phần kia
\(A=\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\)
\(=\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}=\frac{-3+x}{\left(x-2\right)\left(x-3\right)}=\frac{-1}{x-2}\)