gải phương trình sau
5/-x^2+5x-6 + x+3/2-x = 0
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PH
2
28 tháng 2 2020
\(\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)
\(\Leftrightarrow x^3-6x^2+9x-3x^2+18x-27-2x+2=x^3-4x^2+4x-5x^2\)
\(\Leftrightarrow x^3-9x^2+25x-25=x^3-9x^2+4x-5x^2\)
\(\Leftrightarrow x^3-9x^2+25x-25=x^3-9x^2+4x\)
\(\Leftrightarrow-9x^2+25x-25=-9x^2+4x\)
\(\Leftrightarrow25x-25=4x\)
\(\Leftrightarrow-25=4x-25x\)
\(\Leftrightarrow-25=-21x\)
\(\Leftrightarrow x=\frac{21}{25}\)
AH
Akai Haruma
Giáo viên
7 tháng 3 2023
Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) đẻ được hỗ trợ tốt hơn. Viết như thế kia rất khó đọc => khả năng bị bỏ qua bài cao.
9 tháng 3 2023
a: =>3x=3
=>x=1
b: =>12x-2(5x-1)=3(8-3x)
=>12x-10x+2=24-9x
=>2x+2=24-9x
=>11x=22
=>x=2
c: =>2x-3(2x+1)=x-6x
=>-5x=2x-6x-3=-4x-3
=>-x=-3
=>x=3
d: =>2x-5=0 hoặc x+3=0
=>x=5/2 hoặc x=-3
e: =>x+2=0
=>x=-2
12 tháng 8 2021
\(a,\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\left(x\ne2;x\ne3\right)\\ \Leftrightarrow\dfrac{5}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}=0\\\Leftrightarrow\dfrac{5-\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}=0 \\ \Leftrightarrow5-x^2+9=0\\ \Leftrightarrow14-x^2=0\\ \Leftrightarrow x^2=14\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{14}\\x=-\sqrt{14}\end{matrix}\right.\)
\(b,\dfrac{x}{2x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\left(x\ne-1;x\ne3\right)\\ \Leftrightarrow\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{x}{2\left(3-x\right)}\\ \Leftrightarrow\dfrac{x\left(x-3\right)-2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\\ \Leftrightarrow x^2-3x-4x=-x^2-x\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
\(c,\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\left(x\ne1\right)\\ \Leftrightarrow\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ \Leftrightarrow-2x^2+x+1=2x^2-2x\\ \Leftrightarrow4x^2-3x-1=0\\ \Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(d,\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}=\dfrac{5-x}{2x^2+10x}\left(x\ne5;x\ne-5\right)\\ \Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\\ \Leftrightarrow\dfrac{x^2+25x-2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(5-x\right)\left(x-5\right)}{2x\left(x+5\right)\left(x-5\right)}\\ \Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=-\left(x^2-10x+25\right)\\ \Leftrightarrow x^2+25x-2x^2-20x-50=-x^2+10x-25\\ \Leftrightarrow-5x=25\\ \Leftrightarrow x=-5\)
Tick nha
WE
2
H
14 tháng 2 2023
`a,(2x-5)(12+5x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
`b, (x-3)(x-4)-2(x-3)=0`
`<=>(x-3)(x-4-2)=0`
`<=>(x-3)(x-6)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)
`c, x(x-1)(x+1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
`d, (2x)/3 +(2x-1)/6=0`
`<=> (4x)/6 +(2x-1)/6=0`
`<=> (4x+2x-1)/6=0`
`<=> (6x-1)/6=0`
`<=> 6x-1=0`
`<=> 6x=1`
`<=>x=1/6` ( đề là vậy à bạn )
H9
HT.Phong (9A5)
CTVHS
14 tháng 2 2023
a) \(\left(2x-5\right)\left(12+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,5\\x=-2,4\end{matrix}\right.\)
b) \(\left(x-3\right)\left(x-4\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x-4\right)-2\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)
c) \(x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\\x=0\end{matrix}\right.\)
d) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=0\)
\(\Leftrightarrow\dfrac{4x+2x-1}{6}=0\)
\(\Leftrightarrow6x-1=0\)
\(\Leftrightarrow6x=1\Leftrightarrow x=\dfrac{1}{6}\)
19 tháng 1 2024
a: 5-3x=6x+7
=>-3x-6x=7-5
=>-9x=2
=>\(x=-\dfrac{2}{9}\)
b: \(\dfrac{3x-2}{6}-5=3-\dfrac{2\left(x+7\right)}{4}\)
=>\(\dfrac{3x-2}{6}+\dfrac{x+7}{2}=8\)
=>\(\dfrac{3x-2+3\left(x+7\right)}{6}=8\)
=>3x-2+3x+14=48
=>6x+12=48
=>6x=36
=>\(x=\dfrac{36}{6}=6\)
c: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
=>\(\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
=>(x-1)(5x+3-3x+8)=0
=>(x-1)(2x+11)=0
=>\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
=>\(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
11 tháng 3 2020
a) ( 4x - 1 ) (x - 3) - ( x - 3 ) ( 5x + 2 ) = 0
<=> (x - 3)(4x - 1 - 5x - 2) = 0
<=> (x - 3)(-x - 3) = 0
<=> x = 3 hoặc x = -3
b) ( x + 3 ) ( x - 5 ) + ( x + 3 ) ( 3x - 4) = 0
<=> (x + 3)(x - 5 + 3x - 4) = 0
<=> (x + 3)(4x - 9) = 0
<=> x = -3 hoặc x = 9/4
c) ( x + 6 ) ( 3x - 1 )+ x2 - 36 = 0
<=> 3x^2 + 17x - 6 + x^2 - 36 = 0
<=> 4x^2 + 17x - 42 = 0
<=> 4x^2 + 24x - 7x - 42 = 0
<=> 4x(x + 6) - 7(x + 6) = 0
<=> (4x - 7)(x + 6) = 0
<=> x = -6 hoặc x = 7/4
d) ( x + 4 ) ( 5x + 9 ) - x2 + 16 = 0
<=> 5x^2 + 29x + 36 - x^2 + 16 = 0
<=> 4x^2 + 29x + 52 = 0
<=> 4x^2 + 16x + 13x + 42 = 0
<=> 4x(x + 4) + 13(x + 4) = 0
<=> (4x + 13)(x + 4) = 0
<=> x = -13/4 và x = -4
Ta có: \(\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\)
\(\Leftrightarrow\dfrac{-5}{\left(x-2\right)\left(x-3\right)}-\dfrac{x^2-9}{\left(x-2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow-5-x^2+9=0\)
\(\Leftrightarrow x=-2\)