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P
1
28 tháng 6 2021
`(x+5)/(x^2-5x)-(x-5)/(2x^2+10x)=(x+25)/(2x^2-50)`
ĐK:`x ne 0,x ne 5,x ne -5`
Nhân 2 vế với `2x(x+5)(x-5)` ta có phương trình:
`2(x+5)(x+5)-(x-5)(x-5)=x(x+25)`
`<=>2(x^2+10x+25)-(x^2-10x+25)=x^2+25x`
`<=>x^2+30x+25=x^2+25x`
`<=>5x+25=0`
`<=>5x=-25`
`<=>x=-5(l)`
Vậy pt vô nghiệm
5 tháng 8 2016
1) \(\left(x+1\right)\left(x^2-2x+1\right)=\left(x+1\right)\left(x-1\right)^2\)
2) \(\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)
4) \(\left(\frac{1}{2}-x\right)\left(\frac{1}{4}+\frac{1}{2}x+x^2\right)=\left(\frac{1}{2}\right)^3-x^3=\frac{1}{8}-x^3\)
5) \(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=\left(2x\right)^3-\left(3y\right)^3=8x^3-27y^3\)
6) \(\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)
3) Đề thiếu
5 tháng 8 2016
\(1,\left(x+1\right)\left(x^2-2x+1\right)=\left(x+1\right)\left(x-1\right)^2\)
\(2,\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)
\(3,\left(x+2\right)^2=x^2+2.x.2+2^2=x^2+4x+2\)
\(4,\left(\frac{1}{2}-x\right)\left(\frac{1}{4}+\frac{1}{2}x+x^2\right)=\left(\frac{1}{2}\right)^3-x^3=\frac{1}{8}-x^3\)
\(5,\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=\left(2x\right)^3-\left(3y\right)^3=8x^3-27y^3\)
\(6,\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)
28 tháng 7 2021
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
28 tháng 7 2021
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
HC
2
19 tháng 7 2021
\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)
\(\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+5}{2\left(x-5\right)\left(x+5\right)}\)
dkxd : x ≠ 0
x ≠ 5
x ≠ -5
MTC : 2x(x - 5)(x + 5)
Quy đồng mẫu thức hai vế của phương trình :
⇒ \(\dfrac{2\left(x-5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}\) = \(\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)
Suy ra : 2(x - 5)(x + 5) - (x - 5)(x + 5) = x(x + 25)
\(\Leftrightarrow\) 2(x2 - 25) - (x2 - 25) = x2 + 25x
\(\Leftrightarrow\) 2x2 - 50 - x2 + 25 - x2 - 25x = 0
\(\Leftrightarrow\) -25 - 25x = 0
\(\Leftrightarrow\) -25x = 25
\(\Leftrightarrow\) x = \(\dfrac{25}{-25}=-1\) (thỏa mãn)
Vậy S = \(\left\{-1\right\}\)
Chúc bạn học tốt
19 tháng 7 2021
Ta có: \(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)^2}{2x\left(x+5\right)\left(x-5\right)}-\dfrac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=\dfrac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=x^2+25x\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)
\(\Leftrightarrow15x+25=0\)
\(\Leftrightarrow15x=-25\)
hay \(x=-\dfrac{5}{3}\)(thỏa ĐK)
KT
2
TT
26 tháng 5 2021
\(x\ne0;x\ne\pm5\)
PT \(\Leftrightarrow\dfrac{x+25}{2\left(x+5\right)\left(x-5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}=0\)
\(\Rightarrow x^2+25x-2x^2-20x-50+x^2-10x+25=0\)
\(\Leftrightarrow-5x-25=0\)
\(\Leftrightarrow x=-5\) (ktm)
Vậy pt vô nghiệm.
26 tháng 5 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\end{matrix}\right.\).
\(PT\Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(5-x\right)\left(x-5\right)}{2x\left(x-5\right)\left(x+5\right)}\)
\(\Rightarrow x\left(x+25\right)-2\left(x+5\right)^2=\left(5-x\right)\left(x-5\right)\)
\(\Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=10x-x^2-25\)
\(\Leftrightarrow-5x=25\Leftrightarrow x=-5\) (loại)
Vậy PT vô nghiệm
15 tháng 5 2020
ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)
Ta có: \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow\frac{x+5}{x\left(x-5\right)}-\frac{x-5}{2x\left(x+5\right)}=\frac{x+25}{2\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow\frac{2\left(x+5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\frac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)
\(\Leftrightarrow2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=x^2+25x\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)
\(\Leftrightarrow5x+25=0\)
\(\Leftrightarrow5x=-25\)
hay x=-5(ktm)
Vậy: Tập nghiệm \(S=\varnothing\)
15 tháng 8 2020
ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)
Ta có: \(\frac{x+25}{2x^2-50}-\frac{x+5}{x^2-5x}=\frac{5-x}{2x^2+10x}\)
\(\Leftrightarrow\frac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}-\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=0\)
Suy ra: \(x^2+25x-2\left(x^2+10x+25\right)+x^2-10x+25=0\)
\(\Leftrightarrow2x^2+15x+25-2x^2-20x-50=0\)
\(\Leftrightarrow-5x-25=0\)
\(\Leftrightarrow-5x=25\)
hay x=-5(loại)
Vậy: \(S=\varnothing\)
R
1
28 tháng 10 2020
( 2x + 1 )3 - ( 3x + 2 )2 = ( 2x - 5 )( 4x2 + 10x + 25 ) + 6x( 2x + 1 ) - 9x2
⇔ 8x3 + 12x2 + 6x + 1 - ( 9x2 + 12x + 4 ) = 8x3 - 125 + 12x2 + 6x - 9x2
⇔ 8x3 + 12x2 + 6x + 1 - 9x2 - 12x - 4 = 8x3 + 3x2 + 6x - 125
⇔ 8x3 + 3x2 - 6x - 3 = 8x3 + 3x2 + 6x - 125
⇔ 8x3 + 3x2 - 6x - 3 - 8x3 - 3x2 - 6x + 125 = 0
⇔ -12x + 122 = 0
⇔ -12x = -122
⇔ x = 61/6
\(\Leftrightarrow2x^2-11x+5-2x^2+10x=25\Leftrightarrow-x=20\Leftrightarrow x=-20\)