2^x+2^x+^2=5*16
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TN
1
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1
Đề bài sai, pt này ko giải được
Đề đúng: \(\dfrac{4x^2+16}{x^2+6}=...\)
Mẫu số bên trái thừa mất số 1
HQ
Hà Quang Minh
Giáo viên
9 tháng 9 2023
Ta có: \(\begin{array}{l}2{\rm{x}} + 5 = 16\\2{\rm{x}} = 16 - 5\\2{\rm{x}} = 11\\x = \frac{{11}}{2}\end{array}\)
Như vậy, bạn Vuông giải đúng, bạn Tròn giải sai
LN
1
15 tháng 6 2016
x^20-x=0
x(x^19-1)=0
x= 0
hoặc x ^ 19 =1
x = 0 hoặc x= 1
TD
3
27 tháng 12 2017
a)\(\frac{2x}{x+5}+\frac{10}{x+5}=\frac{2x+10}{x+5}=\frac{2\left(x+5\right)}{x+5}=2\)
b)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{x^2-4}=\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}=\frac{8x+16}{\left(x-2\right)\left(x+2\right)}\)\(=\frac{8\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{8}{x-2}\)
PD
27 tháng 12 2017
a) \(\frac{2x}{x+5}+\frac{10}{x+5}\)=\(\frac{2x+10}{x+5}\)=\(\frac{2\left(x+5\right)}{x+5}\)=\(2\)
b)\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{x^2-4}\)=\(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{\left(x+2\right)^2-\left(x-2\right)^2+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{\left(x+2-x+2\right)\left(x+2+x-2\right)+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4\times2x+16}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{8x+16}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{8\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{8}{x-2}\)
NT
1
30 tháng 8 2023
1: =(x+y-3x)(x+y+3x)
=(-2x+y)(4x+y)
2: =(3x-1-4)(3x-1+4)
=(3x+3)(3x-5)
=3(x+1)(3x-5)
3: =(2x)^2-(x^2+1)^2
=-[(x^2+1)^2-(2x)^2]
=-(x^2+1-2x)(x^2+1+2x)
=-(x-1)^2(x+1)^2
4: =(2x+1+x-1)(2x+1-x+1)
=3x(x+2)
5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]
=(2x^2+2)*4x
=8x(x^2+1)
6: =(5x-5y)^2-(4x+4y)^2
=(5x-5y-4x-4y)(5x-5y+4x+4y)
=(x-9y)(9x-y)
7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)
=(x^2+2xy+y^2)(x^2-y^2)
=(x+y)^3*(x-y)
8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)
=[(x-2y)^2-4][(x+2y)^2-36]
=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)
LV
1
Q
1
DD
29 tháng 6 2023
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.
ai biết đâu cha nội :hihi
=1+21827753873883=?
cha nội ???