Câu hỏi của Trần Linh Phương

Question

Giúp mk nhaBài 1 : Phân tích đa thức thành nhân tửa) x2-6x-y2+9b) 25-4x2-4xy -y2c) x2+2xy+y2- xz-yzd) x2-4xy+4y2-z2+4tz-4t2Bài 2 : Phân tích đa thức thành nhân tửa) ax2+cx2-ay+ay2-cy+cy2b) ax^2+ay^2-b...

Trần Anh · Accepted Answer

Bài 1 : a ) $x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3+y\right)\left(x-3-y\right)$b)  $25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)$c)  $x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)$d)   $x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)$$=\left(x-2y\right)^2-\left(z-2t\right)^2=\left(x-2y+z-2t\right).\left(x-2y-z+2t\right)$BÀi 2 : a)   $ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)$$=x^2.\left(a+c\right)-y\left(a+c\right)+y^2.\left(a+c\right)=\left(a+c\right).\left(x^2-y+y^2\right)$b)   $ax^2+ay^2-bx^2-by^2+b-a=\left(ax^2-bx^2\right)+\left(ay^2-by^2\right)-\left(a-b\right)$$=x^2.\left(a-b\right)+y^2.\left(a-b\right)-\left(a-b\right)=\left(a-b\right)\left(x^2+y^2-1\right)$c)  $ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)$$=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)$$=\left(b+c-a\right)\left(d-c^2\right)$BÀi 3 : a)  $x.\left(x-5\right)-4x+20=0$ $\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0$ $\Leftrightarrow\left(x-5\right)\left(x-4\right)=0$$\Leftrightarrow\hept{\begin{cases}x-5=0\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\x=4\end{cases}}}$b)  $x.\left(x+6\right)-7x-42=0$$\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0$ $\Leftrightarrow\left(x+6\right)\left(x-7\right)=0$$\Leftrightarrow\hept{\begin{cases}x+6=0\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\x=7\end{cases}}}$c)   $x^3-5x^2+x-5=0$ $\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0$ $\Leftrightarrow\left(x-5\right)\left(x^2+1\right)$$\Leftrightarrow\hept{\begin{cases}x^2+1=0\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\x=5\end{cases}}}$d)   $x^4-2x^3+10x^2-20x=0$ $\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0$$\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0$  $\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)$$\Leftrightarrow\hept{\begin{cases}x=0\x-2=0\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\x=2\x^2=-10\left(KTM\right)\end{cases}}}$Câu trả lời: Bài 1 : a ) $x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3+y\right)\left(x-3-y\right)$b)  $25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)$c)  $x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)$d)   $x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)$$=\left(x-2y\right)^2-\left(z-2t\right)^2=\left(x-2y+z-2t\right).\left(x-2y-z+2t\right)$BÀi 2 : a)   $ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)$$=x^2.\left(a+c\right)-y\left(a+c\right)+y^2.\left(a+c\right)=\left(a+c\right).\left(x^2-y+y^2\right)$b)   $ax^2+ay^2-bx^2-by^2+b-a=\left(ax^2-bx^2\right)+\left(ay^2-by^2\right)-\left(a-b\right)$$=x^2.\left(a-b\right)+y^2.\left(a-b\right)-\left(a-b\right)=\left(a-b\right)\left(x^2+y^2-1\right)$c)  $ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)$$=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)$$=\left(b+c-a\right)\left(d-c^2\right)$BÀi 3 : a)  $x.\left(x-5\right)-4x+20=0$ $\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0$ $\Leftrightarrow\left(x-5\right)\left(x-4\right)=0$$\Leftrightarrow\hept{\begin{cases}x-5=0\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\x=4\end{cases}}}$b)  $x.\left(x+6\right)-7x-42=0$$\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0$ $\Leftrightarrow\left(x+6\right)\left(x-7\right)=0$$\Leftrightarrow\hept{\begin{cases}x+6=0\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\x=7\end{cases}}}$c)   $x^3-5x^2+x-5=0$ $\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0$ $\Leftrightarrow\left(x-5\right)\left(x^2+1\right)$$\Leftrightarrow\hept{\begin{cases}x^2+1=0\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\x=5\end{cases}}}$d)   $x^4-2x^3+10x^2-20x=0$ $\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0$$\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0$  $\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)$$\Leftrightarrow\hept{\begin{cases}x=0\x-2=0\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\x=2\x^2=-10\left(KTM\right)\end{cases}}}$

a) 5x – 15y	b) 5x²y² + 15x²y + 30xy²
c) x³ – 2x²y + xy² – 9x	d) x(x² – 1) + 3(x² – 1)
e) x² – 10x + 25	g) x² – 64
h) (x + y)² – (x² – y²)	i) 5x² + 5xy – x – y
k) x² – 25 + y² + 2xy	l) 2xy – x² – y² + 16
m) (x – 2)(x – 3) + (x – 2) - 1	n) 3(x – 1) + 5x( 1 – x)
p) 12y(2x – 5) + 6xy(5 – 2x)	q) ax – 2x – a² + 2a

a) a² – b² – 2a + 1	b) x² – 2x – 4y² – 4y
c) x² + 4x – y² + 4	d) x⁴ – 1
e) x⁴ + x³ + x² + x	g) a² + 2ab + b² – ac - bc

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