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Ik mk nha, hôm nay ngày mai, ngày kia mk ik 3 lần lại cho bạn (thành 9 lần)
Nhớ kb với mìn lun nha!! Mk rất vui đc làm quen vs bạn, cảm ơn mn nhìu lắm
Bài 1:(Theo mình câu a nên sửa lại như thế này nhé)
a, a2-5a-14 b,x4+x2-2
=a2+2a-7a-14 =x4-x3+x3-x2+2x2-2x+2x-2
=(a2+2a)-(7a+14) =(x4-x3)+(x3-x2)+(2x2-2x)+(2x-2)
=a(a+2)-7(a+2) =x3(x-1)+x2(x-1)+2x(x-1)+2(x-1)
=(a+2)(a-7) =(x-1)(x3+x2+2x+2)
=(x-1)[(x3+x2)+(2x+2)]
=(x-1)[x2(x+1)+2(x+1)]
=(x-1)(x+1)(x2+2)
Bài 2:
a, x3+x2+x+1=0
<=>(x3+x2)+(x+1)=0
<=>x2(x+1)+(x+1)=0
<=>(x+1)(x2+1)=0
<=>\(\orbr{\begin{cases}x+1=0\\x^2+1=0\left(loại\right)\end{cases}}\)(x2 luôn lớn hơn hoặc bằng 0 =>x2+1 luôn lớn hơn hoặc bằng 1 nên x2+1=0 loại nhé)
<=>x= -1
b, x(2x-7)-4x+14=0
<=>x(2x-7)-(4x-14)=0
<=>x(2x-7)-2(2x-7)=0
<=>(2x-7)(x-2)=0
<=>\(\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
1.a) 2x4-4x3+2x2
=2x2(x2-2x+1)
=2x2(x-1)2
b) 2x2-2xy+5x-5y
=2x(x-y)+5(x-y)
=(2x+5)(x-y)
2.
a) 4x(x-3)-x+3=0
=>4x(x-3)-(x-3)=0
=>(4x-1)(x-3)=0
=> 2 TH:
*4x-1=0 *x-3=0
=>4x=0+1 =>x=0+3
=>4x=1 =>x=3
=>x=1/4
vậy x=1/4 hoặc x=3
b) (2x-3)^2-(x+1)^2=0
=> (2x-3-x-1).(2x-3+x+1)=0
=>(x-4).(3x-2)=0
=> 2 TH
*x-4=0
=> x=0+4
=> x=4
*3x-2=0
=>3x=0-2
=>3x=-2
=>x=-2/3
vậy x=4 hoặc x=-2/3
1,
a, = 2x.(x-2)
b, = (x^2+y^2+2xy)-(2x+2y)
= (x+y)^2-2.(x+y)
= (x+y).(x+y-2)
2,
a,<=> x^2-1-x^2-2x = 3
<=> -2x-1=3
<=> -2x=4
<=> x=4 : (-2) = -2
b, <=>(x^2-4x+4)-7=0
<=>(x-2)^2-7=0
<=> (x-2)^2=7
=> x-2=+-\(\sqrt{7}\)
<=> x=2+-\(\sqrt{7}\)
k mk nha
a, \(2x-4x\)
\(=-2x\)
b, \(x^2+y^2+2xy-2x-2y\)
\(=\left(x+y\right)^2-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-2\right)\)
a, \(\left(x+1\right)\left(x-1\right)-x\left(x+2\right)=3\)
\(\Leftrightarrow x^2-1-x^2-2x=3\)
\(\Leftrightarrow-2x=4\)
\(\Leftrightarrow x=-2\)
b,\(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
f) \(4x\left(x+1\right)=8\left(x+1\right)\)
\(\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
h) \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
i) \(2x\left(x-2\right)-\left(2-x\right)^2=0\)
\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow x=\pm2\)
k) \(\left(1-x\right)^2-1+x=0\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
l) \(\left(x-3\right)^3+3-x=0\)
\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=1\Leftrightarrow x=4\end{matrix}\right.\)
m) \(x+6x^2=0\)
\(\Leftrightarrow x\left(1+6x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{6}\end{matrix}\right.\)
n) \(\left(x+1\right)=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
f ) \(4x\left(x+1\right)=8\left(x+1\right)\)
\(\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy ...
h ) \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy ...
I ) \(2x\left(x-2\right)-\left(2-x\right)^2=0\)
\(\Leftrightarrow-2x\left(2-x\right)-\left(2-x\right)^2=0\)
\(\Leftrightarrow\left(-2x-2+x\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left(-2-x\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2-x=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
Vậy ...
K ) \(\left(1-x\right)^2-1+x=0\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow\left(1-x\right)x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
Vậy ...
i ) \(\left(x-3\right)^3+3-x=0\)
\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x-3=1\\x-3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)
Vậy ...
m ) \(x+6x^2=0\)
\(\Leftrightarrow x\left(1+6x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1+6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{6}\end{matrix}\right.\)
Vậy ...
n ) \(x+1=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
Vậy ...