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B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)
b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)
\(\dfrac{2\left(x-1\right)}{x^2}.\dfrac{x}{\left(x-1\right)}=\dfrac{2\left(x-1\right)x}{x^2\left(x-1\right)}=\dfrac{2}{x}\)
B2:
a)\(2x^2-4x+2=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)
b)\(x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv
x2-2.x.1/2 +(1/2)2-9/4
=(x-1/2)2-9/4
=(x-1/2)2-(3/2)2
=(x-1/2-3/2).(x-1/2+3/2)
=(x-2)(x+1)
a/ ĐK: $x\ne -5$
$\dfrac{6x^2+30x}{4}=\dfrac{6x(x+5)}{4}=\dfrac{3x(x+5)}{2}$
Đề này sai
b/ ĐK: $x\ne \pm 1$
$\dfrac{(x+2)(x+1)}{x^2-1}\\=\dfrac{(x+2)(x+1)}{(x-1)(x+1)}\\=\dfrac{x+2}{x-1}$
$\to$ ĐPCM
a) \(\dfrac{\left(x+2\right)P}{x-2}=\dfrac{\left(x-1\right)Q}{x^2-4}\)
\(\Leftrightarrow\left(x^2-4\right)\left(x+2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2\left(x-2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2P=\left(x-1\right)Q\)
\(\Leftrightarrow P=x-1\)
\(Q=\left(x+2\right)^2=x^2+4x+4\)
b)\(\dfrac{\left(x+2\right)P}{x^2-1}=\dfrac{\left(x-2\right)Q}{x^2-2x+1}\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)P=\left(x+1\right)\left(x-1\right)\left(x-2\right)Q\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)P=\left(x+1\right)\left(x-2\right)Q\)
\(\Leftrightarrow P=\left(x+1\right)\left(x-2\right)=x^2-x-2\)
\(Q=\left(x-1\right)\left(x+2\right)=x^2+x-2\)
\(\Leftrightarrow\dfrac{x+1}{\left(x-3\right)\left(x+2\right)\cdot B}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)
\(\Leftrightarrow B=\dfrac{x-1}{\left(x-3\right)\left(x+2\right)}\)
ý mình là vì sao được kết quả đó , giải thích ra giúp mình nha
a)\(\dfrac{x+5}{3x-2}=\dfrac{x\left(x+5\right)}{x\left(3x-2\right)}\) b)\(\dfrac{2x-1}{4}=\dfrac{\left(2x-1\right)\left(2x+1\right)}{8x+4}\) c)\(\dfrac{2x\left(x-2\right)}{x^2-4x+4}=\dfrac{2x}{x-2}\) d) \(\dfrac{5x^2+10x}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x}{x-2}\)
a) \(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{9}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\)
b) \(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}-2=\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\)
a, \(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{9}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{2}^2\)
\(=\left(x-\dfrac{1}{2}-\dfrac{3}{2}\right)\left(x-\dfrac{1}{2}+\dfrac{3}{2}\right)\)
\(=\left(x-2\right)\left(x+1\right)\)
b, \(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}-2\)
\(=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{9}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{2}^2\)
\(=\left(x-\dfrac{1}{2}-\dfrac{3}{2}\right)\left(x-\dfrac{1}{2}+\dfrac{3}{2}\right)\)
\(=\left(x-2\right)\left(x+1\right)\)
Chúc bạn học tốt!!!