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DN
5 tháng 9 2018
\(a,x^2-2x=0\)
\(\Rightarrow x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy ...
\(b,\left(5-2x\right)^2-16=0\)
\(\Rightarrow\left(5-2x\right)^2=16\)
\(\Rightarrow\left(5-2x\right)^2=4^2\)
\(\Rightarrow5-2x=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{9}\end{matrix}\right.\)
Vậy ...
\(c,x\left(x+3\right)-x^2-11=0\)
\(\Rightarrow x^2+3x-x^2-11=0\)
\(\Rightarrow3x-11=0\)
\(\Rightarrow3x=11\)
\(\Rightarrow x=\dfrac{11}{3}\)
Vậy ...
SS
4
LC
16 tháng 10 2018
a,\(x^3-x=0\Rightarrow x\left(x^2-1\right)=0\Rightarrow x\left(x+1\right)\left(x-1\right)=0\)
b,\(x^2-2x+x-2=0\Rightarrow x\left(x-2\right)+\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+1\right)=0\)
c,\(x^2-6x+8=x^2-4x-2x+8=x\left(x-4\right)-2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)
16 tháng 10 2018
\(x^3-x=0\)
\(\Leftrightarrow x\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
x=0 hoặc x-1=0=> x=1 hoặc x+1=0 => x=-1
\(x^2-2x+x-2=0\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
\(x^2-6x+8=0\)
\(\Leftrightarrow x^2-2x-4x+8=0\)
\(\Leftrightarrow x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)
17 tháng 7 2018
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
11 tháng 7 2018
\(x^3+x=0\)
\(\Rightarrow x.\left(x^2+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varnothing\end{cases}}\)
\(x^2-2x-3=0\)
\(\Rightarrow x.\left(x-2\right)=3\)
Vì \(x>x-2\)và \(x\inƯ\left(3\right)=\left\{3;-3\right\}\)
Các phần sau tương tự
KT
11 tháng 7 2018
\(x^3+x=0\)
\(\Leftrightarrow\)\(x\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x=0\)
\(x^2-2x-3=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
\(2x^2+5x-3=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+3=0\\2x-1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
\(x+5x^2=0\)
\(\Leftrightarrow\)\(x\left(5x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\5x+1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}\)
Vậy...
31 tháng 7 2018
1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3
31 tháng 10 2018
x2 - 5x = 0
=> x(x - 5) = 0
=> \(\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
31 tháng 10 2018
b) (3x - 5)2 - 4 = 0
=> (3x - 5)2 = 0 + 4
=> (3x - 5)2 = 4
=> (3x - 5)2 = 22
=> \(\orbr{\begin{cases}3x-5=2\\3x-5=-2\end{cases}}\)
=> \(\orbr{\begin{cases}3x=7\\3x=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x=1\end{cases}}\)
DA
3
SC
18 tháng 9 2018
a,x2+6x-7=0
=>x2+7x-x-7=0
=>(x^2+7x)-(x+7)=0
=>x(x+7)-(x+7)=0 =>(x+7)(x-1)=0
=>\(\orbr{\begin{cases}x+7=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=1\end{cases}}}\)
b, x^3-2x^2-5x+6=0
=>x(x^2-2x-5+6)=0
=>x(x^2-2x+1)=0\(^{\orbr{\begin{cases}x=0\\\left(x-1^2\right)=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
c, 2x^2-5x+3=0
=>2x^2-2x-3x+3=0
D
18 tháng 9 2018
\(x^3-19x-30=0\)
\(\Rightarrow x^3+5x^2+6x-5x^2-25x-30=0\)
\(\Rightarrow\left(x-5\right)\left(x^2+5x+6\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x^2+2x+3x+6\right)=0\)
\(\Rightarrow\left(x-5\right)[x\left(x+2\right)+3\left(x+2\right)]=0\)
\(\Rightarrow\left(x-5\right)\left(x+3\right)\left(x+2\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-5=0\\x+3=0\\x+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=-3\\x=-2\end{cases}}\)
\(x^3+x^2-2x-8=0\)
\(\Leftrightarrow\left(x^3+3x^2+4x\right)-\left(2x^2+6x+8\right)=0\)
\(\Leftrightarrow x\left(x^2+3x+4\right)-2\left(x^2+3x+4\right)=0\)
\(\Leftrightarrow\left(x^2+3x+4\right)\left(x-2\right)=0\)(1)
Ta thấy \(x^2+3x+4\)
\(=x^2+2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+4\)
\(=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0;\forall x\)
\(\Rightarrow\left(1\right)\)xảy ra \(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy x=2
\(\Leftrightarrow\left(x^3-8\right)+\left(x^2-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x+\frac{3}{2}\right)^2+\frac{7}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x\in\varnothing\end{cases}\Rightarrow x=2.}\)
Vậy ........