x-5=x2-25
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Sửa đề: \(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{5-x}{2x^2+10x}\)
ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)
Ta có: \(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{5-x}{2x^2+10x}\)
\(\Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x+25}{2\left(x^2-25\right)}=\dfrac{5-x}{2x\left(x+5\right)}\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(2\left(x+5\right)^2-x\left(x+25\right)=-\left(x-5\right)^2\)
\(\Leftrightarrow2\left(x^2+10x+25\right)-x^2-25x=-\left(x^2-10x+25\right)\)
\(\Leftrightarrow2x^2+20x+50-x^2-25x=-x^2+10x-25\)
\(\Leftrightarrow x^2-5x+50+x^2-10x+25=0\)
\(\Leftrightarrow2x^2-15x+75=0\)
\(\Leftrightarrow2\left(x^2-\dfrac{15}{2}x+\dfrac{75}{2}\right)=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{15}{4}+\dfrac{225}{16}+\dfrac{375}{16}=0\)
\(\Leftrightarrow\left(x-\dfrac{15}{4}\right)^2+\dfrac{375}{16}=0\)(vô lý)
Vậy: \(S=\varnothing\)
\(C=\dfrac{2x}{x-3}-\dfrac{3x+9}{x^2-9}\)
\(C=\dfrac{2x}{x-3}-\dfrac{3\left(x+3\right)}{x^2-3^2}\)
\(C=\dfrac{2x}{x-3}-\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(C=\dfrac{2x}{x-3}-\dfrac{3}{x-3}\)
\(C=\dfrac{2x-3}{x-3}\)
============================
\(D=\left(\dfrac{15-x}{x^2-25}+\dfrac{2}{x+5}\right):\dfrac{x+1}{x-5}\)
\(D=\left(\dfrac{15-x}{\left(x+5\right)\left(x-5\right)}+\dfrac{2\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}\right):\dfrac{x+1}{x-5}\)
\(D=\left(\dfrac{15-x+2x-10}{\left(x+5\right)\left(x-5\right)}\right):\dfrac{x+1}{x-5}\)
\(D=\left(\dfrac{x+5}{\left(x+5\right)\left(x-5\right)}\right):\dfrac{x+1}{x-5}\)
\(D=\dfrac{1}{x-5}:\dfrac{x+1}{x-5}\)
\(D=\dfrac{1}{x-5}\cdot\dfrac{x-5}{x+1}\)
\(D=\dfrac{1}{x+1}\)
\(1,\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\left(B\right)\\ 2,Giống.1\\ 3,=5x^2\left(B\right)\\ 4,x^3=x\Leftrightarrow x^3-x=0\\ \Leftrightarrow x\left(x^2-1\right)=0\\ \Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\\ \Leftrightarrow A\)
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
3: \(\left(x+5\right)\left(x^2-5x+25\right)-x\left(x-4\right)^2+16x\)
\(=x^3+125-x^3+8x^2-16x+16x\)
\(=8x^2+125\)
a. 5x + 3(x2 - x - 1)
= 5x + 3x2 - 3x - 3
= 3x2 + 5x - 3x - 3
= 3x2 + 2x - 3
b. (5 - x)(5 + x) - (2x - 1)2
25 - x2 - (4x2 - 4x + 1)
= 25 - x2 - 4x2 + 4x - 1
= 25 - 1 - x2 - 4x2 + 4x
= 24 - 5x2 + 4x
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)
\(\Leftrightarrow-11x=-22\)
hay x=2
b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)
\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)
\(\Leftrightarrow x=-5\)
https://olm.vn/bai-viet/-15903
x-5=x2-25
x-5-(x2-25)=0
x-5-(x-5)(x+5)=0
(x-5)(x+5-1)=0
(x-5)(x+4)=0
*x-5=0 <=> x=5
*x+4=0 <=> x=-4
Vậy x=5 hoặc x=-4