tìm x
a \(x+5x^2=0\)
b \(x^3+x=0\)
c \(5x\left(x-1\right)=x-1\)
d \(x^2-10x=-25\)
K
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AH
Akai Haruma
Giáo viên
13 tháng 8 2018
a)
\(3x^2-5x=0\Leftrightarrow x(3x-5)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ 3x-5=0\rightarrow x=\frac{5}{3}\end{matrix}\right.\)
b)
\(x^3-0,36x=0\Leftrightarrow x(x^2-0,36)=0\)
\(\Leftrightarrow x(x-0,6)(x+0,6)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ x-0,6=0\\ x+0,6=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=0\\ x=0,6\\ x=-0,6\end{matrix}\right.\)
c)
\((5x+2)^2-(3x-1)^2=0\)
\(\Leftrightarrow (5x+2-3x+1)(5x+2+3x-1)=0\)
\(\Leftrightarrow (2x+3)(8x+1)=0\)
\(\Rightarrow \left[\begin{matrix} 2x+3=0\\ 8x+1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-3}{2}\\ x=\frac{-1}{8}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
13 tháng 8 2018
d)
\(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow x^2-2.5x+5^2=0\Leftrightarrow (x-5)^2=0\)
\(\Rightarrow x=5\)
e)
\(3(x+5)-x^2-5x=0\)
\(\Leftrightarrow 3(x+5)-x(x+5)=0\)
\(\Leftrightarrow (3-x)(x+5)=0\)
\(\Rightarrow \left[\begin{matrix} 3-x=0\rightarrow x=3\\ x+5=0\rightarrow x=-5\end{matrix}\right.\)
f)
\((x-1)^2-2(x-1)(3x+2)+(3x+2)^2=0\)
\(\Leftrightarrow [(x-1)-(3x+2)]^2=0\)
\(\Leftrightarrow (-2x-3)^2=0\Rightarrow -2x-3=0\Rightarrow x=\frac{-3}{2}\)
10 tháng 7 2023
\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)
\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)
\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)
\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)
10 tháng 7 2023
a: =>(x+2-3)(x+2+3)=0
=>(x-1)(x+5)=0
=>x=1 hoặc x=-5
b: =>(x-1)^2=25
=>x-1=5 hoặc x-1=-5
=>x=-4 hoặc x=6
c: =>25x^2+10x+1-25x^2+9=30
=>10x+10=30
=>x+1=3
=>x=2
d: =>x^3-1-x(x^2-4)=5
=>x^3-1-x^3+4x=5
=>4x=6
=>x=3/2
15 tháng 2 2020
a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42
23 tháng 11 2022
a: =>(x^2+4x-5)(x^2+4x-21)=297
=>(x^2+4x)^2-26(x^2+4x)+105-297=0
=>x^2+4x=32 hoặc x^2+4x=-6(loại)
=>x^2+4x-32=0
=>(x+8)(x-4)=0
=>x=4 hoặc x=-8
b: =>(x^2-x-3)(x^2+x-4)=0
hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)
c: =>(x-1)(x+2)(x^2-6x-2)=0
hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)
NN
2
10 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
`(x - 2)(x - 3) =0`
`<=>`\(\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0+2\\x=0+3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy, `S = {2; 3}`
`b,`
`x^2 - 5x = 0`
`<=> x(x - 5) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=0+5\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy, `S = {0; 5}`
`c,`
`x^2 - 9 = 0`
`<=> x^2 = 0 + 9`
`<=> x^2 = 9`
`<=> x^2 = (+-3)^2`
`<=> x = +-3`
Vậy, `S = {3; -3}`
`d,`
`4x^2 - 25 = 0`
`<=> 4x^2 = 25`
`<=> x^2 = 25/4`
`<=> x^2 = (+-5/2)^2`
`<=> x = +-5/2`
Vậy,` S = {5/2; -5/2}.`
10 tháng 7 2023
a: =>x-2=0 hoặc x-3=0
=>x=2 hoặc x=3
b: =>x(x-5)=0
=>x=0 hoặc x=5
c: =>(x-3)(x+3)=0
=>x=3 hoặc x=-3
d: =>(2x-5)(2x+5)=0
=>x=5/2 hoặc x=-5/2
G
2
BA
21 tháng 10 2019
\(a,x^2-25-x-5=0\)
\(x^2-x-30=0\)
\(x^2+5x-6x-30=0\)
\(x\cdot\left(x+5\right)-6\cdot\left(x+5\right)=0\)
\(\left(x+5\right)\cdot\left(x-6\right)=0\)
\(\orbr{\begin{cases}x+5=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=6\end{cases}}}\)
KN
21 tháng 10 2019
b) \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(\Leftrightarrow\left(10x^2+9x\right)-\left(10x^2+13x-3\right)=8\)
\(\Leftrightarrow-4x+3=8\)
\(\Leftrightarrow-4x=5\Leftrightarrow x=\frac{-5}{4}\)
25 tháng 1 2021
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
2
15 tháng 9 2023
`a)(2x^2-5x+3)(x^2-4x+3)=0`
`<=>[(2x^2-5x+3=0),(x^2-4x+3=0):}<=>[(x=3/2),(x=1),(x=3):}`
`=>A={3/2;1;3}`
`b)(x^2-10x+21)(x^3-x)=0`
`<=>[(x^2-10x+21=0),(x^3-x=0):}<=>[(x=7),(x=3),(x=0),(x=+-1):}`
`=>B={0;+-1;3;7}`
`c)(6x^2-7x+1)(x^2-5x+6)=0`
`<=>[(6x^2-7x+1=0),(x^2-5x+6=0):}<=>[(x=1),(x=1/6),(x=2),(x=3):}`
`=>C={1;1/6;2;3}`
`d)2x^2-5x+3=0<=>[(x=1),(x=3/2):}` Mà `x in Z`
`=>D={1}`
`e){(x+3 < 4+2x),(5x-3 < 4x-1):}<=>{(x > -1),(x < 2):}<=>-1 < x < 2`
Mà `x in N`
`=>E={0;1}`
`f)|x+2| <= 1<=>-1 <= x+2 <= 1<=>-3 <= x <= -1`
Mà `x in Z`
`=>F={-3;-2;-1}`
`g)x < 5` Mà `x in N`
`=>G={0;1;2;3;4}`
`h)x^2+x+3=0` (Vô nghiệm)
`=>H=\emptyset`.
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
a) \(x+5x^2=0\)
\(=>x\left(1+5x\right)=0\)
\(=>\hept{\begin{cases}x=0\\5x+1=0\end{cases}}\)
\(=>\hept{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}\)
b) \(x^3+x=0\)
\(=>x\left(x^2+1\right)=0\)
\(=>\hept{\begin{cases}x=0\\x^2+1=0\end{cases}}\)
\(=>\hept{\begin{cases}x=0\\x\in\phi\end{cases}}\)
c) \(5x\left(x-1\right)=x-1\)
\(=>5x\left(x-1\right)-x+1=0\)
\(=>5x\left(x-1\right)-\left(x-1\right)=0\)
\(=>\left(x-1\right)\left(5x-1\right)=0\)
\(=>\hept{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)
\(=>\hept{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
d) \(x^2-10x=-25\)
\(=>x^2-10x+25=0\)
\(=>\left(x-5\right)^2=0\)
\(=>x-5=0\)
\(=>x=5\)
\(a,x+5x^2=0\)
\(x.\left(1+5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}\)