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bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz
= xy(X + y + z) + yz(x + y + z) + xz(X + y + z)
= (x + y +z)(xy + yz+ xz)
b) xy(x + y) - yz(y + z) - xz(z - x)
= x2y + xy2 - y2z - yz2 - xz2 + x2z
= x2(y + z) - yz(y + z) + x(y2 - z2)
= x2(y + z) - yz(y + z) + x(y + z)(y - z)
= (y + z)(x2 - yz + xy - xz)
= (y + z)[x(x + y) - z(x + y)]
= (y + z)(x + y)(x - z)
c) x(y2 - z2) + y(z2 - x2) + z(x2 - y2)
= x(y - z)(y + z) + yz2 - yx2 + x2z - y2z
= x(y - z)(y + z) - yz(y - z) - x2(y - z)
= (y - z)((xy + xz - yz - x2)
= (y - z)[x(y - x) - z(y - x)]
= (y - z)(x - z)(y -x)
`a, 4a^2 + 4a + 1 = (2a+1)^2`
`b, -3x^2 + 6xy - 3y^2`
` = -3(x-y)^2`
`c, (x+y)^2 - 2(x+y)z + z^2`
`= (x+y-z)^2`
a: \(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)
\(=a\left(x-y\right)+b\left(x-y\right)+c\left(x-y\right)\)
\(=\left(x-y\right)\left(a+b+c\right)\)
b: \(a^m-a^{m+2}\)
\(=a^m-a^m\cdot a^2\)
\(=a^m\left(1-a^2\right)\)
\(=a^m\left(1-a\right)\left(1+a\right)\)
a. Câu hỏi của nguyễn khánh linh - Toán lớp 8 - Học toán với OnlineMath
a. \(x^2-2xy+y^2-z^2=\\ \left(x^2-2xy+y^2\right)-z^2\\ =\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
b. \(x^2-6x+9-9y\\ =\left(x^2-6x\right)+\left(9-9y\right)\\ =x\left(x-6\right)+9\left(1-y\right)\)
1,
a, \(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2\)
=\(\left(x-y-z\right)\left(x-y+z\right)\)
b, hình như sai đề phải ko bn?
Phải là: \(x^2-6x+9-9y^2\)=\(\left(x-3\right)^2-\left(3y\right)^2\)\(=\left(x-3y-3\right)(x+3y-3)\)
c,\(\left(x+y\right)\left(y+z\right)\left(x+z\right)+xyz\)
=\((x^2+xy+yz+xz)\left(z+y\right)+xyz\)
=\((x^2z+xyz+xz^2)+(x^2y+xy^2+xyz)+\)\(\left(yz^2+y^2z+xyz\right)\)
= xz(x + y + z) + xy(x + y + z) + yz(x + y +z)
=(x+y+z)(xz+xy+yz)
2,a,\(\left(x-2\right)\left(x+1\right)=0\)\(\Rightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b,Đề phải thế này nha:
\(5x\left(x-3\right)-x+3=0\)\(\Rightarrow\)(x - 3)(5x - 1)=0
\(\Rightarrow\)\(\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\)\(\Rightarrow\)\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
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