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1 tháng 8 2017
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...
N
1 tháng 8 2017
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 !!!
16 tháng 10 2015
x2+2x2+3x2+.......+50x2=10200
=> x2(1+2+3+4+....+50)=10200
=> x2.1275=10200
=> x2=10200:1275
=> x2=8
??????????
NT
1
10 tháng 11 2018
\(2x-8x^2=0\Rightarrow2x\left(1-4x\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\1-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}}\)
\(x-x^2=0\Rightarrow x\left(1-x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Cn lại lm tương tự nha e!
=.= hok tốt!!
AN
0
15 tháng 10 2017
a) \(2x^3-32x=0\)
\(2x\left(x^2-16\right)=0\)
\(2x\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow2x=0\)hoặc \(\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(x=0\) hoặc \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
b) \(\left(3x-2\right)^2-\left(x+5\right)^2=0\)
\(\left(3x-2-x-5\right)\left(3x-2+x+5\right)=0\)
\(\left(2x-7\right)\left(4x+3\right)=0\)
\(\orbr{\begin{cases}2x-7=0\\4x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-3}{4}\end{cases}}\)
c) \(2\left(x+3\right)-x^2-3x=0\)
\(2\left(x+3\right)-\left(x^2+3x\right)=0\)
\(2\left(x+3\right)-x\left(x+3\right)=0\)
\(\left(2-x\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
vậy \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
d) \(4x^2-25-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(4x^2-25\right)-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5\right)-\left(2x+5\right)\left(x+7\right)=0\)
\(\left(2x+5\right)\left(2x-5-x-7\right)=0\)
\(\left(2x+5\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+5=0\\x-12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{-5}{2}\\x=12\end{cases}}\)
LL
1
TD
20 tháng 7 2017
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, \(\left[\left(2x+14\right):2^3-3\right]:2-1=0\)
\(\Rightarrow\left[\left(2x+14\right):8-3\right]:2=1\)
\(\Rightarrow\left(2x+14\right):8-3=2\)
\(\Rightarrow\left(2x-14\right):8=5\)
\(\Rightarrow2x-14=40\)
\(\Rightarrow2x=26\)
\(\Rightarrow x=13\)
4 tháng 2 2018
1/ \(\left\{{}\begin{matrix}\left(x-2\right)^{72}\ge0\\\left(y+1\right)^{70}\ge0\end{matrix}\right.\)
Mà \(\left(x-2\right)^{72}+\left(y+1\right)^{70}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^{72}=0\\\left(y+1\right)^{70}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy ...
2/ \(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|y-3\right|\ge0\end{matrix}\right.\)
Mà \(\left|x+1\right|+\left|y-3\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x+1\right|=0\\\left|y-3\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\)
Vậy ...
3/ \(\left\{{}\begin{matrix}\left(2x-10\right)^{100}\ge0\\\left(x-y\right)^{102}\ge0\end{matrix}\right.\)
Mà \(\left(2x-10\right)^{100}+\left(x-y\right)^{102}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-10\right)^{100}=0\\\left(x-y\right)^{102}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-10=0\\x-y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5\end{matrix}\right.\)
Vậy ....
4/ \(\left\{{}\begin{matrix}\left|2x+8\right|\ge0\\\left|y+x\right|\ge0\end{matrix}\right.\)
Mà \(\left|2x+8\right|+\left|y+x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|2x+8\right|=0\\\left|y+x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+8=0\\y+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=8\end{matrix}\right.\)
Vậy ..
<=> 2x(25x2 - 1) = 0
TH1: x = 0
TH2: 25x2-1 = 0
<=> 25x2 = 1
<=> x = 1/5 hoặc -1/5
Vậy x = 0 hoặc x = 1/5 hoặc x = -1/5
=x3x(50-2)=0
=x3x48=0
=x3=0
=x=0
Vậy x =0