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a) \(\left(2x-1\right)^2-25=0\)
\(\left(2x-1\right)^2=0+25=25\)
\(\left(2x-1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-1=5\\2x-1=-5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}2x=6\\2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\end{array}\right.\)
b) \(8x^3-50x=0\)
\(2x\left(4x^2-25\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=0\\4x^2-25=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-\frac{5}{2}\end{array}\right.\end{array}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x biết
a,\(50x-\left(1+2+3+...+100\right)=0\)
\(1+2+3+...+100\) có số số hạng là:
\(100-1+1=100\) (số hạng)
\(\Rightarrow\)\(\dfrac{\left(100+1\right).100}{2}=5050\)
Ta có: \(50x-5050=0\)
\(\Rightarrow50x=5050\)
\(x=5050:50\)
\(\Rightarrow x=101\)
b, \(\left[\left(2x+14\right):2^3-3\right]:2-1=0\)
\(\Rightarrow\left[\left(2x+14\right):8-3\right]:2=1\)
\(\Rightarrow\left(2x+14\right):8-3=2\)
\(\Rightarrow\left(2x-14\right):8=5\)
\(\Rightarrow2x-14=40\)
\(\Rightarrow2x=26\)
\(\Rightarrow x=13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy x = 2 hoặc x = -3
b, \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow2x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\2x-5=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy...
1.
a) \(2\left(x+3\right)-x^2-3x=0\)
\(2\left(x+3\right)-\left(x^2+3x\right)=0\)
\(2\left(x+3\right)-x\left(x+3\right)=0\)
\(\left(2-x\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy x=2 hoặc x=-3
b) \(8x^3-50x=0\)
\(x\left(8x^2-50\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\8x^2-50=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{25}{4}=\left(\pm\dfrac{5}{2}\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{-5}{2}\) hoặc \(x=\dfrac{5}{2}\)
tik mik nhé !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
d) (x - 2)^2 = 1
= x = 2 + 1 = 3
c) (x^2 + 1). (x + 2011) = 0
Tim x:
a) x^2 + 2x = 0
= \(x^2+2x=0\)
= \(x^2=0:2=0\)
b) (x - 3) + 2x^2 - 6x = 0
Rút gọn thừa số chung :
\(2x^2-5x-3=0\)
x = \(\frac{-1}{2}\)x = 3
=\(x^2=0\)
=> x = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x2 - 121) . (2x + 3) = 0
=>x2-121=0 hoặc 2x+3=0
+)Nếu x2-121=0
=>x2=0+121=121
=>x2=(-11)2 hoặc x2=112
=>x=-11 hoặc x=11
+)Nếu 2x+3=0
=>2x=0-3=-3
=>x=(-3):2=\(\frac{-3}{2}\)
Vậy x=-11 hoặc x=11 hoặc x=\(\frac{-3}{2}\)
b) 2x2 - 8x = 0
=>2x(x-4)=0
=>x=0 hoặc x-4=0
Nếu x-4=0
=>x=0+4=4
Vậy x=0 hoặc x=4
c) (3x + 1)5 = (3x + 1)4
=>(3x+1)5:(3x+1)4=(3x+1)4:(3x+1)4
=>3x+1=1
=>3x=1-1=0
=>x=0:3=0
Vậy x=0
a)(x2 - 121) . (2x + 3) = 0
=>x2-121=0 hoặc 2x+3=0
- Với x2-121=0
<=>x2=121 <=>x=±11
- Với 2x+3=0
<=>2x=-3 <=>x=-3/2
b) 2x2 - 8x = 0
=>2x(x-4)=0
=>2x=0 hoặc x-4=0
=>x=0 hoặc x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+......+50+x=1275\)
\(\Rightarrow50x+\left(1+2+3+4+......+50\right)=1275\)
\(\Rightarrow50x+1221=1275\)
\(\Rightarrow50x=54\)
\(\Rightarrow x=\frac{27}{25}\)
b) \(x+2x+3x+.......+100x=10100\)
\(\Rightarrow x\left(1+2+3+.....+100\right)=10100\)
\(\Rightarrow5050x=10100\)
\(\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(2x^2+50x=0\)
\(\Rightarrow x\left(2x+50\right)=0\)
\(\Rightarrow x=0\) hay \(2x+50=0\)
\(\Rightarrow2x=-50\)
\(\Rightarrow x=-25\)
b) \(2x^3-50x=0\)
\(\Rightarrow x\left(2x^2-50\right)=0\)
\(\Rightarrow x=0\) hay \(2x^2-50=0\)
\(2x^2=50\)
\(x^2=25\)
\(\Rightarrow x=5\) hay \(x=-5\)