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a) 5x - 15y = 5(x - 3y)
b) \(\dfrac{3}{5}\)x2 + 5x4 - x2 - y
= \(\dfrac{3}{5}\)x2 + 5x2.x2 - x2 - y
= x2(\(\dfrac{3}{5}\) + 5x2 -1) - y
c) 14x2y2 - 21xy2 + 28x2y
= 7xy.xy - 7xy.3y + 7xy.4x
= 7xy(xy - 3y + 4x)
= 7xy[(xy - 3y) + 4x]
= 7xy[y(x - 3) +4x]
d) \(\dfrac{2}{7}x\)(3y - 1) - \(\dfrac{2}{7}y\)(3y - 1)
= (3y - 1).(\(\dfrac{2}{7}x\) - \(\dfrac{2}{7}y\) )
= (3y - 1).[\(\dfrac{2}{7}\)(x - y)]
e) x3 - 3x2 + 3x - 1
= x2.x - 3x.x + 3.x - 1
= x(x2-3x+3) - 1
g) 27x3 + \(\dfrac{1}{8}\)
= (3x)3 + \(\left(\dfrac{1}{2}\right)^3\)
= (3x + \(\dfrac{1}{2}\)).(9x2 - \(\dfrac{3}{2}\)x + \(\dfrac{1}{4}\))
h) (x+y)3 - (x-y)3
= 2(3x2y) + 2y3
f) (x+y)2 - 4x2
= -3x2 + y(2x + y)
Trả lời tội ghê đó bạn nhưng mk gửi một bài mà sao bạn trả lời một câu vậy bạn nhưng dù sao vẫn cảm on nha
Bài 1 :
a ) \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)
\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)
\(\)\(=2y^2-10xy\)
Câu b tương tự
Bài 2 :
a ) \(x^2-9+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3+x-3\right)\)
\(=2x\left(x-3\right)\)
b ) \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
c ) \(x^3-4x^2+12x-27\)
\(=x^3-9x^2+5x^2+27x-15x-3^3\)
\(=\left(x^3-9x^2+27x-3^3\right)+\left(5x-15x\right)\)
\(=\left(x-3\right)^3+5\left(x-3\right)\)
\(=\left(x-3\right)\left[\left(x-3\right)^2+5\right]\)
\(=\left(x-3\right)\left(x^2-6x+14\right)\)
d ) \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(3x\left(x+1\right)-10x\left(x+1\right)\)
\(=-7x\left(x+1\right)\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
b1:
câu a,f áp dụng a2-b2=(a-b)(a+b)
câu b,c áp dụng a3-b3=(a-b)(a2+ab+b2)
câu d: \(x^2+2xy+x+2y=x\left(x+2y\right)+\left(x+2y\right)=\left(x+1\right)\left(x+2y\right)\)
câu e: \(7x^2-7xy-5x+5y=7x\left(x-y\right)-5\left(x-y\right)=\left(7x-5\right)\left(x-y\right)\)
câu g xem lại đề
a) \(=2xy^2\left(x^2+8x+15\right)\)
\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)
\(=2xy^2\left[\left(x+4\right)^2-1\right]\)
\(=2xy^2\left(x+4+1\right)\left(x+4-1\right)\)
\(=2xy^2\left(x+5\right)\left(x-3\right)\)
mấy câu sau tự làm nha :*
b,=(x^2-10x+25)-4
=(x-5)^2-2^2
=(x-5-2)(x-5+2)
=(x-7)(x-3)
a) 5x-15y=5x-3.5.y=5(x-3y)
c) 14xy(xy+28x)
d) \(\dfrac{2}{7}\left(3x-1\right)\left(x-1\right)\)
e) (x-1)3
f) (x+y-2x)(x+y+2x)=(y-x)(3x+y)
g) (3x+\(\dfrac{1}{2}\))(9x2+\(\dfrac{3}{2}x\)+\(\dfrac{1}{4}\))
h) (x+y-x+y)\(\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
2a)
(x+1)(x2+2x)=0
(x+1)x(x+2)=0
\(\left[{}\begin{matrix}x+1=0\\x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\)
2b)
\(\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=0\)
\(\Leftrightarrow\)(\(\dfrac{2}{3}-5x\))(\(\dfrac{2}{3}+5x\))=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}\dfrac{2}{3}-5x=0\\\dfrac{2}{3}+5x=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\dfrac{2}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.\)
2c)
x(3x-2)+5(3x-2)=0
\(\Leftrightarrow\)(3x-2)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
2d)
x2-2.x.\(\dfrac{1}{2}\)+\(\left(\dfrac{1}{2}\right)^2\)=0
\(\Leftrightarrow\)(x-\(\dfrac{1}{2}\))2=0
\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)
Bổ sung bài 2 :
a ) \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\\x=-1\\x=-2\end{matrix}\right.\)
Vậy \(x=0;x=-1;x=-2\)
b ) \(\dfrac{4}{9}-25x^2=0\)
\(\Leftrightarrow\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=0\)
\(\Leftrightarrow\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}-5x=0\\\dfrac{2}{3}+5x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.\)
Vậy \(x=\dfrac{2}{15};x=-\dfrac{2}{15}\)
c ) \(x\left(3x-2\right)-5\left(2-3x\right)=0\)
\(\Leftrightarrow x\left(3x-2\right)+5\left(3x-2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
Vậy \(x=\dfrac{2}{3}\) và \(x=-5\)
d ) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
cảm ơn bạn
cảm ơn bạn