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x + x2 = 0
=> x(1 + x) = 0
=> x = 0 hoặc x + 1 = 0
=> x = 0 hoặc x = -1
vậy_
mk biến đổi về pt tích, sau đó bạn tính nốt nhé:
b) \(x+1-\left(x+1\right)^2=0\)
<=> \(\left(x+1\right)\left(1-x-1\right)=0\)
<=> \(-x\left(x+1\right)=0\)
c) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\)
<=> \(3\left(4y-9\right)\left(5y-1\right)=0\)
d) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\)
<=> \(\left(25z+7\right)\left(8-27z\right)=0\)
a) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\)
\(\Leftrightarrow\left(15y-3\right)\left(4y-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}15y-3=0\\4y-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{5}\\y=\frac{9}{4}\end{matrix}\right.\)
Vây \(y\in\left\{\frac{1}{5};\frac{9}{4}\right\}\)
b) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\)
\(\Leftrightarrow\left(8-27z\right)\left(25z+7\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}z=\frac{8}{27}\\z=-\frac{7}{25}\end{matrix}\right.\)
Vậy \(z\in\left\{\frac{8}{27};-\frac{7}{25}\right\}\)
c) \(13y\left(y-8\right)-2y+16=0\)
\(\Leftrightarrow13y\left(y-8\right)-2\left(y-8\right)=0\)
\(\Leftrightarrow\left(13y-2\right)\left(y-8\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{2}{13}\\y=8\end{matrix}\right.\)
Vậy \(y\in\left\{\frac{2}{13};8\right\}\)
d) \(-10y\left(y+2\right)-y-2=0\)
\(\Leftrightarrow\left(-10y-1\right)\left(y+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=-\frac{1}{10}\end{matrix}\right.\)
Vậy \(y\in\left\{-2;-\frac{1}{10}\right\}\)
e) \(x\left(x+19\right)^2-\left(x+19\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+19\right)^2=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-19\end{matrix}\right.\)
Vậy \(x\in\left\{1;-19\right\}\)
Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0
=> 6x2 - 21x - (6x2 + x - 90x - 15) - 2010 = 0
=> 6x2 - 21x - 6x2 + 89x + 15 - 2010 = 0
=> 68x - 1995 = 0
?
b) 2x(x - 2012) - x + 2012 = 0
=> 2x(x - 2012) - (x - 2012) = 0
=> (x - 2012) (2x - 1) = 0
⇔[
| x−2012=0 |
| 2x−1=0 |
⇔[
| x=2012 |
| 2x=1 |
⇔[
| x=2012 |
| x=12 |
Vậy x = {2012;12 }
Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0
=> 6x2 - 21x - (6x2 + x - 90x - 15) - 2010 = 0
=> 6x2 - 21x - 6x2 + 89x + 15 - 2010 = 0
=> 68x - 1995 = 0
?
b) 2x(x - 2012) - x + 2012 = 0
=> 2x(x - 2012) - (x - 2012) = 0
=> (x - 2012) (2x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-2012=0\\2x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2012\\2x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2012\\x=\frac{1}{2}\end{cases}}\)
Vậy x = \(\left\{2012;\frac{1}{2}\right\}\)
Bài 2;
\(a)x^4-16x=0\Rightarrow x^4=16x\Leftrightarrow x^3=16\Leftrightarrow x=\sqrt[3]{16}\)
\(c)4x^2-\frac{1}{4}=0\Leftrightarrow4x^2=\frac{1}{4}\Leftrightarrow x^2=\frac{1}{16}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)
1)\(21x^2y-12xy^2=xy.\left(21x-12y\right)\)
2)\(x^3+x^2-2x=x.\left(x^2+x-2\right)\)
3)\(3x.\left(x-1\right)+7x^2\left(x-1\right)=\left(x-1\right).\left(3x+7x^2\right)=x.\left(x-1\right)\left(3+7x\right)\)
15)\(\left(2a+3\right)^2-\left(2a+1\right)^2=\left(2a+3-2a-1\right)\left(2a+3+2a+1\right)=2.\left(4a+4\right)=8\left(a+1\right)\)
14) \(-4y^2+4y-1=-\left[\left(2y\right)^2-2.2y.1+1^2\right]=-\left(2y-1\right)^2\)
13) \(x^6+1=\left(x^2\right)^3+1=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
12) \(\left(x+1\right)^2-\left(y+6\right)^2=\left(x+1-y-6\right)\left(x+1+y+6\right)=\left(x-y-5\right)\left(x+y+7\right)\)
4) \(3x\left(x-a\right)+4a\left(a-x\right)=3x.\left(x-a\right)-4a\left(x-a\right)=\left(x-a\right)\left(3x-4a\right)\)
Sao nhiều thế!
a,Rút gọn biểu thức sau :2x(x+3)-x(2x-1)
2x(x+3)-x(2x-1) = 2x^2 + 6x - 2x^2 + x = 7x
a: \(=2x^2+6x-2x^2+x=7x\)
b: \(\Leftrightarrow x^3+2x^2-x^2-2x+x+2+a-2⋮x+2\)
=>a-2=0
=>a=2
c: \(36x^2-49=0\)
=>(6x-7)(6x+7)=0
=>x=7/6 hoặc x=-7/6
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
=>x^3+27-x^3+x-27=0
=>x=0
a) \(x+x^2=0\Leftrightarrow x\left(1+x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
b) \(x+1-\left(x+1\right)^2=0\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow-x\left(x+1\right)\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
c) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\Leftrightarrow\left(15y-3\right)\left(4y-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{15}=\dfrac{1}{5}\\x=\dfrac{9}{4}\end{matrix}\right.\)
d) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\Leftrightarrow\left(8-27z\right)\left(25z+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}z=\dfrac{8}{27}\\z=\dfrac{-7}{25}\end{matrix}\right.\)