(x - 5)/(x + 5) - (2x)/(x - 5) = (x(x + 10))/(25 - x ^ 2)
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11 tháng 7 2021
9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)
\(\Leftrightarrow-4x=9\)
hay \(x=-\dfrac{9}{4}\)
11 tháng 7 2021
10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)
Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)
\(\Leftrightarrow5x^2-7x=0\)
\(\Leftrightarrow x\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)
12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)
\(\Leftrightarrow2x^2+x-3=0\)
\(\Leftrightarrow2x^2+3x-2x-3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
22 tháng 5 2022
\(\dfrac{3}{2x+10}+\dfrac{3}{x-5}-\dfrac{2x}{x^2-25}\)
\(=\dfrac{3\left(x-5\right)}{2\left(x+5\right)\left(x-5\right)}+\dfrac{6\left(x+5\right)}{2\left(x+5\right)\left(x-5\right)}-\dfrac{4x}{2\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{3x-15+6x+30-4x}{2\left(x+5\right)\left(x-5\right)}\)
\(=\dfrac{5x+15}{2\left(x+5\right)\left(x-5\right)}\)
TH
1
DH
15 tháng 2 2020
\(Đkxđ:\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\end{matrix}\right.\)
\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)
\(\Leftrightarrow5x=-25\)
\(\Leftrightarrow x=-5\left(ktmđk\right)\)
Vậy pt vô nghiệm
Y
17 tháng 1 2023
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
4 tháng 12 2023
a: \(\left[\left(10-x\right)\cdot2+51\right]:3-2=3\)
=>\(\left[2\left(10-x\right)+51\right]:3=5\)
=>\(\left[2\left(10-x\right)+51\right]=15\)
=>\(2\left(10-x\right)=15-51=-36\)
=>10-x=-36/2=-18
=>\(x=10-\left(-18\right)=10+18=28\)
b: \(\left(x-12\right)-15=20-\left(17+x\right)\)
=>\(x-12-15=20-17-x\)
=>\(x-27=3-x\)
=>\(2x=30\)
=>\(x=\dfrac{30}{2}=15\)
c: \(720-\left[41-\left(2x-5\right)\right]=2^3\cdot5\)
=>\(720-\left[41-2x+5\right]=8\cdot5=40\)
=>\(\left[41-2x+5\right]=720-40=680\)
=>-2x+46=680
=>-2x=680-46=634
=>\(x=\dfrac{634}{-2}=-317\)
O
2
17 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
`(2x - 3)^2`
`= 4x^2 - 12x + 9`
`b,`
`(x + 1)^2`
`= x^2 + 2x + 1`
`c,`
`(2x + 5)(2x - 5)`
`= 4x^2 - 25`
`d,`
`(a + b - c)(a - b + c)`
`= a^2 - b^2 + bc - c^2 + cb`
`e,`
\((x + 1)^2 - 10(x + 1) + 25\)
`= x^2 + 2x + 1 - 10x - 10 + 25`
`= x^2 - 8x +16`
`@` `\text {Kaizuu lv uuu}`
17 tháng 7 2023
`@` CT:
Bình phương của `1` tổng: `(A + B)^2 = A^2 + 2AB + B^2`
Bình phương của `1` hiệu: `(A - B)^2 = A^2 - 2AB + B^2`
`A^2 - B^2 = (A-B)(A+B)`
31 tháng 7 2023
a: =>5-x=-23
=>x=5+23=28
b: =>x-3-x+7-25+x=54
=>x-21=54
=>x=75
c: =>7-9x-2x+4=-5x-35+27-25=-5x-37
=>-11x+3=-5x-37
=>-6x=-40
=>x=20/3
T
31 tháng 7 2023
a.
10-x-5 = (-5) - 7 -11
=>5-x = 0
=>x=5
b
(x-3) - (x+17-24) - (25-x) = 24 - (-30)
=>x - 3 - x - 17 + 24 - 25 - x = 24 + 30
=>-x - 21 = 54
=>-x = 75
=>x = -75
c
(7 - 9x) - (2x - 4) = - (5x + 35) - (-27) - 25
=>7-9x - 2x + 4 = -5x - 35 + 27 - 35
=>11 - 11x + 5x = -43
=>16x = 11 + 43
=>16x = 54
=>x=4
7 tháng 3 2021
a, \(\dfrac{10-2x}{2}=\dfrac{25-5x}{5}\)
\(\Leftrightarrow\dfrac{2\left(5-x\right)}{2}=\dfrac{5\left(5-x\right)}{5}\)
\(\Leftrightarrow5-x=5-x\)
\(\Leftrightarrow0x=0\)
⇒ Có vô số giá trị của x thỏa mãn.
Vậy...
b, ĐKXĐ: \(x\ne\pm1\)
\(\dfrac{x-3}{x-1}-\dfrac{2x+1}{x+1}=\dfrac{x-x^2}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+1\right)-\left(2x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-x^2}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow x^2-2x-3-2x^2+x+1=x-x^2\)
\(\Leftrightarrow-2x=2\)
\(\Leftrightarrow x=-1\left(ktm\right)\)
Vậy...
7 tháng 3 2021
a) Ta có: \(\dfrac{10-2x}{2}=\dfrac{25-5x}{5}\)
\(\Leftrightarrow5\left(10-2x\right)=2\left(25-5x\right)\)
\(\Leftrightarrow50-10x=50-10x\)
\(\Leftrightarrow0x=0\)(phương trình có vô số nghiệm)
Vậy: S={x|\(x\in R\)}
K
3
10 tháng 9 2015
2346 : (25 + x) = 23
=> 25 + x = 2346 : 23 = 102
=> x = 102 - 25 = 77
Vậy x = 77
2(x-3) + 5(x+4) = 49
=> 2x - 6 + 5x + 20 = 49
=> 7x + (20 - 6) = 49
=> 7x + 14 = 49
=> 7x = 49 - 14 = 35
=> x = 35 : 7 = 5
Vậy x = 5
(2x-10)3 . 251007 = \(5^2.5^{2015}\)
\(\left(2x-10\right)^3.5^{2014}=5^{2014}.5^3\)
\(\left(2x-10\right)^3=5^3\)
2x - 10 = 5
=> 2x = 5 + 10 = 15
=> x = 15 2 = 7,5
Vậy x = 7,5
MH
10 tháng 9 2015
1. a. 2346:(25+x)=23
=> 25+x=2346:23
=> 25+x=102
=> x=102-25
=> x=77
b. 2(x-3)+5(x+4)=49
=> 2x-6+5x+20=49
=> 7x+14=49
=> 7x=49-14
=> 7x=35
=> x=35:7
=> x=5
c. (2x-10)3.251007=52.52015
=> (2x-10)3.(52)1007=52017
=> (2x-10)3.52014=52017
=> (2x-10)3=52017:52014
=> (2x-10)3=53
=> 2x-10=5
=> 2x=5+10
=> 2x=15
=> x=15:2
=> x=7,5
\(\dfrac{\left(x-5\right)}{x+5}-\dfrac{2x}{x-5}=\dfrac{x\left(x+10\right)}{25-x^2}\)
\(\dfrac{\left(x-5\right)}{5+x}+\dfrac{2x}{5-x}=\dfrac{x\left(x+10\right)}{25-x^2}\)
\(\dfrac{\left(x-5\right)^2}{25-x^2}+\dfrac{2x\left(x+5\right)}{25-x^2}=\dfrac{x^2+10x}{25-x^2}\)
\(\dfrac{x^2-10x+25}{25-x^2}+\dfrac{2x^2+10x}{25-x^2}-\dfrac{x^2+10x}{25-x^2}=0\)
\(\dfrac{x^2-10x+25+2x^2+10x-x^2-10x}{25-x^2}=0\)
\(\dfrac{25-10x}{25-x^2}=0\)
\(25-10x=0\)
\(10x=25\)
\(x=\dfrac{25}{10}=\dfrac{5}{2}\)