x^2+20x-x-20=0 tim x
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JH
0
KH
1
18 tháng 8 2021
a, \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left(3x+3\right)^2=0\Leftrightarrow\left(4x-3x-3\right)\left(4x+2x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)
b, \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow x=-2;x=\frac{1}{3}\)
c, \(5x^3-20x=0\Leftrightarrow5x\left(x^2-4\right)=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow x=0;x=\pm2\)
JH
1
18 tháng 8 2021
1: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
2: Ta có: \(\left(5x-4\right)^2-49x^2=0\)
\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)
\(\Leftrightarrow\left(2x+4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3: Ta có: \(5x^3-20x=0\)
\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
A4
1
MA
17 tháng 9 2016
a) \(\frac{45-\left(3x+9\right)}{13}=42\)
\(\Leftrightarrow45-\left(3x+9\right)=546\)
\(\Leftrightarrow3x+9=-501\)
\(\Leftrightarrow3x=-510\)
\(\Leftrightarrow x=-170\)
b) \(\left(x+1\right)+\left(2x+2\right)+...+\left(20x+20\right)=1050\)
\(\Leftrightarrow x+1+2x+2+...+20x+20=1050\)
\(\Leftrightarrow\left(x+1\right)+2\left(x+1\right)+....+20\left(x+1\right)=1050\)
\(\Leftrightarrow\left(x+1\right)\left(1+2+3+...+20\right)=1050\)
\(\Leftrightarrow\left(x+1\right)\left[\frac{\left(20+1\right).20}{2}\right]=1050\)
\(\Leftrightarrow\left(x+1\right).210=1050\)
\(\Leftrightarrow x+1=5\)
\(\Leftrightarrow x=4\)
MR
1
KT
22 tháng 7 2018
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...
LV
1
24 tháng 10 2021
c: ta có: \(7x^2-2x-5=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{7}\end{matrix}\right.\)
K
3
15 tháng 6 2018
a. x(x-5) -4x+20=0
<=> x(x-5) - 4(x-5)=0
<=> (x-5)(x-4)=0
<=>(x-5)=0 hoặc x-4=0
<=> x=5 hoặc x=4
Vậy x={4;5}
b.tương tự
c. x3-5x2+x-5 =0
<=> x2(x-5) + (x-5) = 0
<=> (x-5) (x2+1) = 0
<=> x-5=0 hoặc x2+1=0(loại vì x2=-1)
<=> x=5
vậy x=5
d. bạn kiểm tra lại đề
15 tháng 6 2018
Tìm x :
a) \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x^2-5x-4x+20=0\)
\(\Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
b) \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x^2+6x-7x-42=0\)
\(\Leftrightarrow\left(x^2+6x\right)-\left(7x+42\right)=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-6\end{matrix}\right.\)
c) \(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\left(x^3-5x^2\right)+\left(x-5\right)=0\)
\(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vôlí\right)\\x=5\end{matrix}\right.\)
HT
6 tháng 8 2016
1, x(x - 5) - 4x + 20 = 0
=> x(x - 5) - 4(x - 5) = 0
=> (x - 4)(x - 5) = 0
=> x - 4 = 0 hoặc x - 5 = 0
=> x = 4 hoặc x = 5
=> x thuộc {4; 5}
2, 3(x + 1) + x(x + 1)
= (3 + x)(x + 1)
3, 2x3 + x = 0
=> x(2x2 + 1) = 0
=> x = 0 hoặc 2x2 + 1 = 0
=> x = 0 hoặc 2x2 = -1
=> x = 0 hoặc x2 = -1/2 (vô lí vì x2 > hoặc = 0 với mọi x)
=> x = 0
4, x3 - 16x = 0
=> x(x2 - 16) = 0
=> x = 0 hoặc x2 - 16 = 0
=> x = 0 hoặc x2 = 16
=> x = 0 hoặc x = 4 hoặc x = -4
=> x thuộc {-4; 0; 4}
5, x2 + 6x = -9
=> x2 + 6x + 9 = 0
=> x2 + 2.3.x + 32 = 0
=> (x + 3)2 = 0
=> x + 3 = 0
=> x = -3
6, x4 - 2x3 + 10x2 - 20x = 0
=> x2(x2 + 10) - 2x(x2 + 10) = 0
=> (x2 + 2x)(x2 + 10) = 0
=> x(x +2)(x2 + 10) = 0
-TH1: x = 0
-TH2: x + 2 = 0 => x = -2
-TH3: x2 + 10 = 0 => x2 = -10 (vô lí vì x2 > hoặc = 0 với mọi x)
=> x thuộc {0; -2}
7, (2x - 3)2 = (x + 5)2
-TH1: 2x - 3 = x + 5
=> x = 8
- TH2: - 2x + 3 = x + 5
=> -3x = 2
=> x = \(\frac{-2}{3}\)
- TH3: 2x - 3 = - x - 5
=> 3x = -2
=> x = \(\frac{-2}{3}\)
- TH4: - 2x + 3 = - x - 5
=> -x = -8
=> x = 8`
=> x thuộc {\(\frac{-2}{3}\); 8}
\(x^2+20x-x-20=0\Leftrightarrow x^2+19x-20=0\Leftrightarrow\left(x^2-x\right)+\left(20x-20\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left(x+20\right)=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+20=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-20\end{cases}}}\)
Ta có:
x^2+20x-x-20 = 0
(x^2-x)+(20x-20)= 0
x.(x-1)+20.(x-1)= 0
(x-1).(x+20) = 0
=> x-1 = 0 hoặc x+20 = 0
x = 0+1 x = 0-20
x =1 x = -20
VẬY x=1 hoặc x=-20