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Bài 1:
a) 25x2 - 10xy + y2 = (5x - y)2
b) 81x2 - 64y2 = (9x)2 - (8y)2 = (9x - 8y)(9x + 8y)
c) 8x3 + 36x2y + 54xy2 + 27y3
= 8x3 + 27y3 + 36x2y + 54xy2
= (2x + 3y)(4x2 - 6xy + 9y2) + 18xy(2x + 3y)
= (2x + 3y)(4x2 - 6xy + 18xy + 9y2)
= (2x + 3y)(4x2 + 12xy + 9y2)
= (2x + 3y)(2x + 3y)2 = (2x + 3y)3
c) (a2 + b2 - 5)2 - 4(ab + 2)2 = (a2 + b2 - 5)2 - 22(ab + 2)2
= (a2 + b2 - 5)2 - (2ab + 4)2
= (a2 + b2 - 5 - 2ab - 4)(a2 + b2 - 5 + 2ab + 4)
= (a2 - 2ab + b2 - 9)(a2 + 2ab + b2 - 1)
= \(\left [ (a - b)^{2} - 3^{2} \right ]\)\(\left [ (a + b)^{2} - 1\right ]\)
= (a - b - 3)(a - b + 3)(a + b - 1)(a + b + 1)
pn đăng mỗi lần vài bài thôi chứ đăng nhìn ngán lắm
Bài 2:
a) 2x3 + 3x2 + 2x + 3
= 2x3 + 2x + 3x2 + 3
= 2x(x2 + 1) + 3(x2 + 1)
= (x2 + 1)(2x + 3)
b)x3z + x2yz - x2z2 - xyz2
= xz(x2 + xy - xz - yz)
= \(xz\left [ x(x + y) - z(x + y) \right ]\)
= xz(x + y)(x - z)
c) x2y + xy2 - x - y
= xy(x + y) - (x + y)
= (x + y)(xy - 1)
d) 8xy3 - 5xyz - 24y2 + 15z
= 8xy3 - 24y2 - 5xyz + 15z
= 8y2(xy - 3) - 5z(xy - 3)
= (xy - 3)(8y2 - 5z)
e) x3 + y(1 - 3x2) + x(3y2 - 1) - y3
= x3 - y3 + y - 3x2y + 3xy2 - x
= (x - y)(x2 + xy + y2) - 3xy(x - y) - (x - y)
= (x - y)(x2 + xy + y2 - 3xy - 1)
= (x - y)(x2 - 2xy + y2 - 1)
= \((x - y)\left [ (x - y)^{2} - 1 \right ]\)
= (x - y)(x - y - 1)(x - y + 1)
câu f tương tự
a) \(1+2xy-x^2-y^2=\left[-\left(y-x-1\right)\right]\left(y-x+1\right)\)
b) \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)=\left(x-y\right)\left(z-x\right)\left(z-y\right)\)
c) \(x^4-64=\left(x^2\right)^2-8^2=\left(x^2+8\right)\left(x^2-8\right)\)
d) \(x^2-15x+36=\left(x-3\right)\left(x-12\right)\)
e) \(\left(x^2-8\right)^2-784=\left(x-6\right)\left(x+6\right)\left(x^2+20\right)\)
1, \(25x^2-10xy+y^2=\left(5x-y\right)^2\)
2, \(8x^3+36x^2y+54xy^2+27y^3=\left(2x+3y\right)^3\)
4, \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
5, \(2x^3+3x^2+2x+3\)
\(=x^2\left(2x+3\right)+2x+3\)
\(=\left(x^2+1\right)\left(2x+3\right)\)
6, \(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xy^2\)
\(=xz\left(x^2-xz\right)+xz\left(xy-yz\right)\)
\(=xz\left[x\left(x-z\right)+y\left(x-z\right)\right]\)
\(=xz\left(x+y\right)\left(x-z\right)\)
8, \(x^3+3x^2y+3xy^2+y+y^3\)\(=\left(x+y\right)^3+y\)
9, \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-2\right)\left(x-4\right)\)
10, \(x^2-8x+12\)
\(=x^2-6x-2x+12\)
\(=x\left(x-6\right)-2\left(x-6\right)\)
\(=\left(x-2\right)\left(x-6\right)\)
Chỗ còn lại mai làm nốt nha.
Gặp chút sự cố đăng nhập nên hơi muộn, xin lỗi nha
11, \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)
\(=a^2b-ab^2+abc-a^2c+b^2c-abc+ac^2-c^2b\)
\(=ab\left(a-b\right)-ac\left(a-b\right)-bc\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
12, \(x^3-7x-6\)
\(=x^3-3x^2+3x^2-9x+2x-6\)
\(=x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+2\right)\)
\(=\left(x-3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x-3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x-3\right)\left(x+2\right)\left(x+1\right)\)
13, \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
14, \(a^4+64\)
\(=a^4+16a^2+64-16a^2\)
\(=\left(a^2+8\right)^2-16a^2\)
\(=\left(a^2-4a+8\right)\left(a^2+4a+8\right)\)
15, \(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)
16, \(x^5+x-1\)
\(=x^5-x^4+x^3+x^4-x^3+x^2-x^2+x-1\)
\(=x^3\left(x^2-x+1\right)-x^2\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x^2-1\right)\)
17, \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-15\)
19, \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\) (*)
Đặt \(x^2+8x+7=a\) ta có:
(*) \(\Leftrightarrow a\left(a+8\right)+15\)
\(\Leftrightarrow a^2+8a+15\)
\(\Leftrightarrow a^2+3a+5a+15\)
\(\Leftrightarrow a\left(a+3\right)+5\left(a+3\right)\)
\(\Leftrightarrow\left(a+3\right)\left(a+5\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
20, \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\) (*)
Đặt \(x^2+3x+1=a\) ta có:
(*) \(\Leftrightarrow a\left(a+1\right)-6\)
\(\Leftrightarrow a^2+a-6\)
\(\Leftrightarrow a^2+3a-2a-6\)
\(\Leftrightarrow a\left(a+3\right)-2\left(a+3\right)\)
\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+3x-1\right)\left(x^2+3x+5\right)\)
a) \(x^2-2x=24\)
\(\Rightarrow x^2-2x-24=0\)
\(\Rightarrow x^2-6x+4x-24=0\)
\(\Rightarrow x\left(x-6\right)+4\left(x-6\right)=0\)
\(\Rightarrow\left(x-6\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b) \(\left(5-2x\right)^2-16=0\)
\(\Rightarrow\left(5-2x\right)^2-4^2=0\)
\(\Rightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\)
\(\Rightarrow\left(1-2x\right)\left(9-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}1-2x=0\\9-2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
c)Sửa đề
\(x^2-4x+4-9x^2+6x-1=0\)
\(\Rightarrow\left(x^2-4x+4\right)-\left(9x^2-6x+1\right)=0\)
\(\Rightarrow\left(x-2\right)^2-\left(3x-1\right)^2=0\)
\(\Rightarrow\left(x-2-3x+1\right)\left(x-2+3x-1\right)=0\)
\(\Rightarrow\left(-2x-1\right)\left(4x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-2x-1=0\\4x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2x=1\\4x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
d) \(2x^2+y^2+2xy-4x+4=0\)
\(\Rightarrow\left(x^2+2xy+y^2\right)+\left(x^2-4x+4\right)=0\)
\(\Rightarrow\left(x+y\right)^2+\left(x-2\right)^2=0\)
Vì \(\left(x+y\right)^2\ge0\) với mọi x và y
\(\left(x-2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+y\right)^2+\left(x-2\right)^2\ge0\) với mọi x và y
Mà \(\left(x+y\right)^2+\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-x\\x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\)
Bài 1:
a) \(\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
b) \(\left(a^2+2a-3\right)\left(a^2+2a+3\right)=\left(a^2+2a\right)^2-9\)
c) \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)=a^2-\left(2a+3\right)^2\)
d) \(\left(a^2-2a+3\right)\left(a^2+2a+3\right)=9-\left(a^2-2a\right)^2\)
e) \(\left(-a^2-2a+3\right)\left(-a^2-2a+3\right)=\left(-a^2-2a+3\right)^2\)
g) \(\left(a^2+2a+3\right)\left(a^2-2a+3\right)=\left(a^2+3\right)^2-4a^2\)
f) \(\left(a^2+2a\right)\left(2a-a^2\right)=4a^2-a^4\)
Bài 2 :
a) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)
b) \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+yx+y^2+yz+zx+zy+z^2=x^2+2xy+2yz+2xz+y^2+z^2\)
c) \(\left(x-y+z\right)^2=\left(x-y+z\right)\left(x-y+z\right)=x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=x^2+y^2+z^2-2xy+2xz-2yz\)d) \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=\left(x-2y\right)^3\)
e) \(\left(x-y-z\right)^2=\left(x-y-z\right)\left(x-y-z\right)=x^2-xy-xz-xy+y^2+yz-xz+yz+z^2=x^2-2xy-2xz+2yz+y^2+z^2\)