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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
a) 4x2 - 20x + 25 - 36y2
= (2x - 5)2 - 36y2
= (2x - 5 - 6y)(2x - 5 + 6y)
b) x3 + x2 - 2x - 8
= (x3 - 8) + (x2 - 2x)
= (x - 2)(x2 + 2x + 4) + x(x - 2)
= (x - 2)(x2 + 2x + 4 + x)
= (x - 2)(x2 + 3x + 4)
d) x4 + 6x3 + 9x2 - 16
= x2(x2 + 6x + 9) - 16
= x2(x + 3)2 - 16
= (x2 + 3x)2 - 16
= (x2 + 3x - 4)(x2 + 3x + 4)
= (x2 + 4x - x - 4)(x2 + 3x + 4)
= [x(x + 4) - (x + 4)](x2 + 3x + 4)
= (x - 1)(x + 4)(x2 + 3x + 4)
a) x2 + x - 12 = x2 - 3x + 4x - 12 = x( x - 3 ) + 4( x - 3 ) = ( x - 3 )( x + 4 )
b) x2 - 4x - 5 = x2 + x - 5x - 5 = x( x + 1 ) - 5( x + 1 ) = ( x + 1 )( x - 5 )
c) x2 - 2x - 3 = x2 + x - 3x - 3 = x( x + 1 ) - 3( x + 1 ) = ( x + 1 )( x - 3 )
d) x2 - 2x - 8 = x2 + 2x - 4x - 8 = x( x + 2 ) - 4( x + 2 ) = ( x + 2 )( x - 4 )
e) x2 - 5x - 6 = x2 + x - 6x - 6 = x( x + 1 ) - 6( x + 1 ) = ( x + 1 )( x - 6 )
f) x2 - 6x + 8 = x2 - 2x - 4x + 8 = x( x - 2 ) - 4( x - 2 ) = ( x - 2 )( x - 4 )
g) x2 + 4x + 3 = x2 + x + 3x + 3 = x( x + 1 ) + 3( x + 1 ) = ( x + 1 )( x + 3 )
h) x2 - 2x - 15 = x2 + 3x - 5x - 15 = x( x + 3 ) - 5( x + 3 ) = ( x + 3 )( x - 5 )
i) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x( x + 3 ) + 4( x + 3 ) = ( x + 3 )( x + 4 )
j) x2 - 5x - 14 = x2 + 2x - 7x - 14 = x( x + 2 ) - 7( x + 2 ) = ( x + 2 )( x - 7 )
a) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
b) Ta có: \(-x^2+5x-6=0\)
\(\Leftrightarrow-\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow-\left(x^2-2x-3x+6\right)=0\)
\(\Leftrightarrow-\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)
\(\Leftrightarrow-\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow-\left[\left(x-2\right)\left(x-3\right)\right]=0\)
\(\Leftrightarrow-\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: x∈{2;3}
c) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
⇔(4x2-10x)-(2x-5)=0
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
d) Ta có: \(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow\left(2x^2+2x\right)+\left(3x+3\right)=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{-3}{2}\right\}\)
e) Ta có: \(x^3+2x^2-x-2=0\)
\(\Leftrightarrow\left(x^3+2x^2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow x^2\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;1;-1\right\}\)
g) Ta có: \(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow9x^2-6x+1-20x^2-20x-5+12x^2-3-x^2+2x-1=0\)
\(\Leftrightarrow-24x-8=0\)
\(\Leftrightarrow-8\left(3x+1\right)=0\)
⇔3x+1=0
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy: \(x=-\frac{1}{3}\)
h) \(2x^3-7x^2+7x-2=0\)
\(\Leftrightarrow2x^3-4x^2-3x^2+6x+x-2=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-2x-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[2x\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy S = {2; 1; \(\frac{1}{2}\)}
i) \(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\end{matrix}\right.\)
Vậy S = {1;-2}
\(x^3-2x+y^3-2y=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
\(x^2-2xy+y^2-16=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-y+4\right)\)
theo mình đề câu c là 6x2
\(x^3+6x^2+9x-xz^2=x\left(x^2-6x+9-z^2\right)\)
\(=\left(x-3-z\right)\left(x-3+z\right)\)
\(x^2-11x+30=x^2-5x-6x+30\)
\(=x\left(x-5\right)-6\left(x-5\right)=\left(x-5\right)\left(x-6\right)\)
\(4x^2-3x-1=4x^2-4x+x-1\)
\(=4x\left(x-1\right)+x-1=\left(4x+1\right)\left(x-1\right)\)
\(9x^2-7x-2=9x^2-9x+2x-2\)
\(=9x\left(x-1\right)+2\left(x-1\right)=\left(9x+2\right)\left(x-1\right)\)
\(\left(x^2+x\right)^2-2\left(x^2+x\right)-5=\left(x^2+x-1\right)^2-4\)
\(=\left(x^2+x-3\right)\left(x^2+x+1\right)\)
còn lại lát mình làm tiếp
Bài 1:
a, \(x^3-2x-y^3-2y=\left(x^3+y^3\right)-\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
b, \(x^2-2xy+y^2-16=\left(x-y\right)^2-4^2=\left(x-y+4\right)\left(x-y-4\right)\)
c, \(x^3+6x^2+9x-xz^2=x\left(x^2+6x+9-z^2\right)\)
\(=x\left[\left(x+3\right)^2-z^2\right]=x\left(x+3+z\right)\left(x+3-z\right)\)
1. a)\(x^2+x-3x-3=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)