1. Tìm x
a) x ( x - 6 ) + 10 ( x + 6 ) = 0
b) x3 - x = 0
c) ( x + 2 )2 = x + 2
d) x3 + 4x = 0
e) 3x ( x - 10 ) = x - 10
g) x ( x + 7 ) = 4x + 28.
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20 tháng 10 2021
a) \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)
\(\Leftrightarrow x\in\left\{\frac{1}{5};2000\right\}\)
5 tháng 11 2017
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
30 tháng 9 2018
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
19 tháng 12 2021
a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
1 tháng 11 2021
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
M
1 tháng 4 2020
e, 3x(2-x) =15(x-2)
\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy..
f, (x+5)(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)
Vậy..
g, x(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
,h, (2x -4)(x-2)=0
\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
i, (x+1/5)(2x-3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)
k, x²-4x=0
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
m, 4x²-1=0
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
n, x²-6x+9=0
\(\Leftrightarrow x^2-2.x.3+3^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)
<=> x=3
l, (3x-5)²-(x+4)²=0
\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)
\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy ..
o, 7x(x+2)-5(x+2)=0
\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)
Vậy....
p, 3x(2x-5)-4x+10=0
\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)
\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...
q, (2-2x)-x²+1=0
\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)
\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy ....
r, x(1-3x)=5(1-3x)
\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)
s, 2x-3/4+x+1/6=3
\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)
\(x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\Leftrightarrow\left(x-1\right)x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x=-1\end{matrix}\right..Vậy:x\in\left\{-1;0;-1\right\}\)
\(x^3+4x=0\Leftrightarrow x\left(x^2+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+4=0\end{matrix}\right.mà:x^2+4\ge0+4=4\Rightarrow x=0\)
\(\left(x+2\right)^2=x+2\Leftrightarrow\left(x+2\right)\left(x+2-1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)