Tìm x biết
(X-2)×(x-3)<0
X×x-2x>0
X×x-5x <0
Tìm x,biết
1) 3x^2 - 4x = 0
2) (x^2 - 5x) + x - 5 = 0
3) x^2 - 5x + 6 = 0
4) 5x(x-3) - x+3 = 0
5) x^2 - 2x + 5 = 0
6) x^2 + x -6 = 0
Answer:
\(3x^2-4x=0\)
\(\Rightarrow x\left(3x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
\(\left(x^2-5x\right)+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
\(x^2-2x+5=0\)
\(\Rightarrow\left(x^2-2x+1\right)+4=0\)
\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)
Vậy không có giá trị \(x\) thoả mãn
\(x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
Tìm x biết: a)x(x-3)+x-3=0 b)(5x-4)^2-16^2=0
tìm x biết : a)x(x-3)-x^2+5=0 b)x^2-6x=0 c)2x^3+5x^2-012x=0
a: Ta có: \(x\left(x-3\right)-x^2+5=0\)
\(\Leftrightarrow-3x+5=0\)
hay \(x=\dfrac{5}{3}\)
b: Ta có: \(x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Tìm x, biết:
a) x(x - 2) + x - 2 = 0;
b) 5x(x - 3) - x + 3 = 0
a. x(x-2)+x-2=0
=> (x-2).(x+1)=0
=> x-2=0 hoặc x+1=0
=> x=2 hoặc x=-1
b. 5x(x-3)-x+3=0
=> 5x(x-3)-(x-3)=0
=> (x-3).(5x-1)=0
=> x-3=0 hoặc 5x-1=0
=> x=3 hoặc x=1/5
a) x(x - 2) + x - 2 = 0;
<=>x.(x-2)+(x-2)=0
<=>(x-2)(x+1)=0
<=>x-2=0 hoặc x+1=0
<=>x=2 hoặc x=-1
b) 5x(x - 3) - x + 3 = 0
<=>5x.(x-3)-(x-3)=0
<=>(x-3)(5x-1)=0
<=>x-3=0 hoặc 5x-1=0
<=>x=3 hoặc x=1/5
Tìm x,biết
A) x(x-2)+x-2=0
B) 5x(x-3)-x+3=0
x(x-2) + x-2= x(x-2) + (x-2).1=(x-2)(x+1)
à như thế này
x (x-2) + 1( x - 2) = 0
(x - 2)( x + 1) = 0
Tìm x biết:
a) x\(^2\) + 5x = 0
b) 3x(x – 1) = 1 – x
c) 2x(x + 2) – 3(x + 2) = 0
\(a,\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow\left(x+2\right)\left(2x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Tìm x, biết : a.,(x-y)^2-(x-3).(x+3)=6; b, x^2-5x=0;c,x^2+6x+5=0
tìm x biết
a) x(x -2 ) + x - 2 = 0
b) 5x (x-3) - x + 3 = 0
a) \(x\left(x-2\right)+x-2=0\)
<=> \(\left(x-2\right)\left(x+1\right)=0\)
<=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
Vậy...
b) \(5x\left(x-3\right)-x+3=0\)
<=> \(\left(x-3\right)\left(5x-1\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)
Vậy...
tìm x biết:
a)x2 + 3x = 0 b) x3 – 4x = 0
c) 5x(x-1) = x-1 d) 2(x+5) - x2-5x = 0
e) 2x(x-5)-x(3+2x)=26 f) 5x.(x – 2012) – x + 2012 = 0
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) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
Tìm x , biết
5x(x-1)=x-1
2(x+5)-x^2-5x=0
(x-3)^2 -(x+5)^2