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a) \(100x^2-\left(x^2+25\right)^2=\left(10x\right)^2-\left(x^2+25\right)^2=\left(10x-x^2-25\right)\left(10x+x^2+25\right)\)
\(=-\left(x-5\right)^2\left(x+5\right)^2\)
b) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1=\left(x-y+5-1\right)^2=\left(x-y+4\right)^2\)
c) \(\left(x^2+4y^2-5\right)^2-16\left(x^2+y^2+2xy+1\right)\)
Có lẽ bạn ghi sai đề rồi nha.
a)ta co: 125x^3+y^6=(5x)^3+(y^2)^3=(5x+y^2)(5x-5xy^2+y^2) b)ta co 5xy^2-10xyz+5xz^2=5x(y^2-2yz+z^2)=5x(y-z)^2 (may cau sau gan giong ban tu lam nha)
b) \(5xy^2-10xyz+5xz^2\)
\(=5xy^2-5xyz-5xyz+5xz^2\)
\(=5xy\left(y-z\right)-5xz\left(y-z\right)\)
\(=\left(y-z\right)\left(5xy-5xz\right)\)
\(=5x\left(y-z\right)\left(y-z\right)\)
\(=5x\left(y-z\right)^2\)
a) \(4x^3\left(x^2+x\right)-\left(x^2+x\right)=\left(x^2+x\right)\left(4x^3-1\right)\)
b)\(\left(1-2a+a^2\right)-\left(b^2-2bc+c^2\right)=\left(1-a\right)^2-\left(b-c\right)^2=\)\(\left(1-a+b-c\right)\left(1-a-b+c\right)\)
lm tiếp câu c
c) \(C=\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)
\(=\left[\left(x-7\right)\left(x-2\right)\right]\left[\left(x-5\right)\left(x-4\right)\right]-72\)
\(=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\)
Đặt \(x^2-9x+17=a\) ta có:
\(C=\left(a-3\right)\left(a+3\right)-72\)
\(=a^2-9-72\)
\(=a^2-81=\left(a-9\right)\left(a+9\right)\)
Thay trở lại ta được: \(C=\left(x^2-9x++8\right)\left(x^2-9x+26\right)\)
1/
a, x2+36=12x
<=>x2-12x+36=0
<=>(x-6)2=0
<=>x-6=0
<=>x=6
b, 5x(x-3)+3-x=0
<=>5x(x-3)-(x-3)=0
<=>(5x-1)(x-3)=0
<=>\(\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}}\)
2/ Sửa đề x2z2 = y2z2
Đặt \(A=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2=4x\left(x+y+z\right)\left(x+y\right)\left(x+z\right)+y^2z^2\)
\(=4\left(x^2+xy+xz\right)\left(x^2+xz+xy+yz\right)+y^2z^2\)
Đặt x2+xy+xz=t, ta có
\(A=4t\left(t+yz\right)+y^2z^2=4t^2+4tyz+y^2z^2=\left(2t+yz\right)^2=\left(2x^2+2xy+2xz+y^2z^2\right)^2\ge0\)
1a) (x - 2y) (x2 - 2xy + y2)
= (x - 2y) (x - y)2
= x2 - xy - 2xy + 2y2
= (x2 - xy) - (2xy - 2y2)
= x (x - y) - 2y (x - y)
= (x - y) (x - 2y)
2a) x (x - 3) - y (3 - x)
= x (x - 3) + y (x - 3)
= (x - 3) (x + y)
b) 3x2 - 5x - 3xy + 5y
= (3x2 - 3xy) - (5x - 5y)
= 3x (x - y) - 5 (x - y)
= (x - y) (3x - 5)
3) 12x (3 - 4x) + 7 (4x - 3) = 0
12x (3 - 4x) - 7 (3 - 4x) = 0
(3 - 4x) (12x - 7) = 0
=> 3 - 4x = 0 hoặc 12x - 7 = 0
* 3 - 4x = 0 => x = \(\frac{3}{4}\)
* 12x - 7 = 0 => x = \(\frac{7}{12}\)
Vậy x =\(\frac{3}{4}\)hoặc x =\(\frac{7}{12}\)
Bai 1
\(x^2+x-30=x^2+6x-5x-30=\left(x-5\right)\left(x+6\right)\)
Bai 2
a, \(\left(x-2\right)^2-x\left(x-5\right)=13\)
\(\Leftrightarrow x^2-4x+4-x^2+5x=13\)
\(\Leftrightarrow x+4=13\Leftrightarrow x=9\)
b, \(4x^3-100x=0\Leftrightarrow x\left(4x^2-100\right)=0\)
\(\Leftrightarrow x\left(2x-10\right)\left(2x+10\right)=0\Leftrightarrow x=0;\pm5\)