tìm x bt:
x+5x^2=0
(x+1)^2=x+1
5x(x-1)=x-1
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TH
2
T
20 tháng 8 2023
\(a,\left(5x-3\right)\left(3x+1\right)-\left(15x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\left(15x^2-4x-3\right)-\left(15x^2-29x-2\right)=0\)
\(\Rightarrow15x^2-4x-3-15x^2+29x+2=0\)
\(\Rightarrow25x-1=0\)
\(\Rightarrow x=\dfrac{1}{25}\)
\(----------\)
\(b,x^2+\left(x+5\right)\left(x-3\right)-25=0\)
\(\Rightarrow x^2+x^2+2x-15-25=0\)
\(\Rightarrow2x^2+2x=40\)
\(\Rightarrow2x\left(x+1\right)=40\)
\(\Rightarrow x\left(x+1\right)=20\)
\(\Rightarrow x;x+1\) là ước của 20
mà \(x;x+1\) là hai số nguyên liên tiếp \(\left(x\in Z\right)\)
nên \(x\left(x+1\right)=4.5=\left(-5\right).\left(-4\right)=20\)
\(\Rightarrow x\in\left\{4;-5\right\}\)
20 tháng 8 2023
a: =>15x^2+5x-9x-3-15x^2+30x-x+2=0
=>25x-1=0
=>x=1/25
b: =>x^2+x^2+2x-15-25=0
=>2x^2+2x-40=0
=>x^2+x-20=0
=>(x+5)(x-4)=0
=>x=4 hoặc x=-5
24 tháng 8 2020
a, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)
\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)
b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P
Tự làm nốt nhé
NM
Nguyễn Minh Quang
Giáo viên
4 tháng 8 2021
ta có
a. (5x-7)(x-9)-(-x+3)(-5x+2)= 2x(x-4)-(x-1)(2x+3)
\(\Leftrightarrow5x^2-52x+63-\left(5x^2-17x+6\right)=2x^2-8x-\left(2x^2+x-3\right)\)
\(\Leftrightarrow-35x+57=-9x+3\Leftrightarrow26x=54\Leftrightarrow x=\frac{27}{13}\)
b. (x-3)(-x+10)+(x-8)(x+3)= (5x^2-1)(x+3)-5x^3-15x^2
\(\Leftrightarrow-x^2+13x-30+x^2-5x-24=5x^3+15x^2-x-3-5x^3-15x^2\)
\(\Leftrightarrow8x-54=-x-3\Leftrightarrow9x=51\Leftrightarrow x=\frac{17}{3}\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
\(5x\left(x-2\right)-3\left(x-1\right)=20x^2-15x\left(2x+1\right)-24\)
\(\Rightarrow5x^2-10x-3x+3=20x^2-30x^2-15x-24\)
\(\Rightarrow5x^2-13x+3=-10x^2-15x-24\)
\(\Rightarrow5x^2+10x^2-13x+15x+3+24=0\)
\(\Rightarrow15x^2+2x+27=0\)
Ta có:
\(\Delta=2^2-4\cdot15\cdot27==-1616< 0\)
Nên pt vô nghiệm
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
10 tháng 8 2023
\(5x\left(x-2\right)-3\left(x-1\right)=20x^2-15x\left(2x+1\right)-24\\ \Leftrightarrow5x^2-10x-3x+3=20x^2-30x^2-15x-24\\ \Leftrightarrow5x^2-20x^2+30x^2-10x-3x+15x+3+24=0\\ \Leftrightarrow15x^2+2x+27=0\\ \Leftrightarrow15x^2-2.x.\sqrt{15}+\dfrac{2}{15}+\dfrac{403}{15}=0\\ \Leftrightarrow\left(\sqrt{15}x+\dfrac{\sqrt{30}}{15}\right)^2+\dfrac{403}{15}=0\left(Vô.lí\right)\\ Vậy:Không.có.x.thoả\)
KV
7 tháng 1 2023
1) \(\dfrac{15-5x}{5x^2-15x}=\dfrac{5\left(3-x\right)}{5x\left(x-3\right)}=-\dfrac{5\left(x-3\right)}{5x\left(x-3\right)}=-\dfrac{1}{x}\)
Chọn A
2) \(\dfrac{x\left(x-5\right)}{x^2+25}=\dfrac{x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{x}{x+5}\)
\(A=0\Leftrightarrow\dfrac{x}{x+5}=0\Leftrightarrow x=0\)
Chọn B
3) \(\dfrac{2x-5}{5-2x}=-\dfrac{5-2x}{5-2x}=-1\)
Chọn D
2 tháng 8 2021
a, \(x^3+3x^2-\left(x+3\right)=0\Leftrightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\Leftrightarrow x=1;x=-1;x=-3\)
b, \(15x-5+6x^2-2x=0\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(3x-1\right)=0\Leftrightarrow x=-\frac{5}{2};x=\frac{1}{3}\)
c, \(5x-2-25x^2+10x=0\)
\(\Leftrightarrow\left(5x-2\right)-5x\left(5x-2\right)=0\Leftrightarrow\left(1-5x\right)\left(5x-2\right)=0\Leftrightarrow x=\frac{2}{5};x=\frac{1}{5}\)
24 tháng 3 2020
a) 2(x + 5) - x^2 - 5x = 0
<=> 2x + 10 - x^2 - 5x = 0
<=> -3x + 10 - x^2 = 0
<=> x^2 + 3x - 10 = 0
<=> (x - 2)(x + 5) = 0
<=> x - 2 = 0 hoặc x + 5 = 0
<=> x = 2 hoặc x = -5
b) 2(x - 3)(x^2 + 1) + 15x - 5x^2 = 0
<=> 2x^3 + 2x - 6x^2 - 6 + 15x - 5x^2 = 0
<=> 2x^3 + 17x - 11x^2 - 6 = 0
<=> (2x^2 - 7x + 3)(x - 2) = 0
<=> (2x^2 - x - 6x + 3)(x - 2) = 0
<=> [x(2x - 1) - 3(2x - 1)](x - 2) = 0
<=> (x - 3)(2x - 1)(x - 2) = 0
<=> x - 3 = 0 hoặc 2x - 1 = 0 hoặc x - 2 = 0
<=> x = 3 hoặc x = 1/2 hoặc x = 2
c) (x + 2)(3 - 4x) = x^2 + 4x + 2
<=> 3x - 4x^2 + 6 - 8x = x^2 + 4x + 2
<=> -5x - 4x^2 + 6 = x^2 + 4x + 2
<=> 5x + 4x^2 - 6 + x^2 + 4x + 2 = 0
<=> 9x + 5x^2 - 4 = 0
<=> 5x^2 + 10x - x - 4 = 0
<=> 5x(x + 2) - (x + 2) = 0
<=> (5x - 1)(x + 2) = 0
<=> 5x - 1 = 0 hoặc x + 2 = 0
<=> x = 1/5 hoặc x = -2
YN
4 tháng 10 2021
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)
+++++\(x+5x^2=0\) \(\Leftrightarrow x.\left(1+5x\right)=0\) \(\Leftrightarrow\hept{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}}\)
++++\(\left(x+1\right)^2=x+1\) \(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\) \(\Leftrightarrow\left(x+1\right).\left(x+1-1\right)=0\)
\(\Leftrightarrow x.\left(x+1\right)=0\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}}\)
++++\(5x.\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\5x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}}\)
a) x + 5x^2 = 0
=> x(1+5x) = 0
=> \(\orbr{\begin{cases}x=0\\1+5x=0\end{cases}}\) => \(\orbr{\begin{cases}x=0\\5x=-1\end{cases}}\) => \(\orbr{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}\)
Vậy: x=0 hoặc x=-1/5
b) (x+1)^2 = x+1
=> (x+1)(x+1) = x+1
=> x+1 = (x+1) : (x+1)
=> x+1 = 1
=> x = 0
Vậy: x = 0
c) 5x(x-1) = x-1
=> 5x = (x-1) : (x-1)
=> 5x = 1
=> x = 1/5
Vậy: x = 1/5