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1)Nhân vào ta sẽ đc VT=\(x^4-y^4+x^2y^2-x^2y^2+xy^3-x^3y-xy^3+x^3y=x^4-y^4\)
2) \(x\left(x+2\right)\left(x^2+2x+2\right)+1=\left(x^2+2x\right)\left(x^2+2x+2\right)\)
Đặt y=\(x^2+2x\).Ta sẽ đc : \(y\left(y+2\right)+1=y^2+2y+1=\left(y+1\right)^2=\left(x^2+2x+1\right)^2=\left(x+1\right)^4\)
3/Theo đề ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)
Vậy Ta có \(a^4+b^4+c^4=3a^4=3\Rightarrow a=b=c=1\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
Bài 1:
a) \(\dfrac{15xy}{10x^2y}\)
= \(\dfrac{3.5xy}{2.5xyx}\)
= \(\dfrac{3}{2x}\)
d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)
= \(\dfrac{3\left(x+5\right)^2}{x}\)
a) \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}\)
\(=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)
b) \(\dfrac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x^2-4\right)\left(x+1\right)}\)
\(=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x-2\right)}{x+2}\)
c) \(\dfrac{x^3-x^2y+xy^2}{x^3+y^3}=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\dfrac{x}{x+y}\)
d) \(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}\)
\(=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)
\(=\dfrac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a-b+c\right)\left(a+b+c\right)}\)
\(=\dfrac{a+b-c}{a-b+c}\)
e) \(\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\dfrac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
\(=\dfrac{2x^2-x-15}{3x^2-10x+3}\)
\(=\dfrac{\left(x-3\right)\left(2x+5\right)}{\left(x-3\right)\left(3x-1\right)}\)
\(=\dfrac{2x+5}{3x-1}\)
Bài 1 rút gọn bc tự làm :
\(B=\dfrac{3y^3-7y^2+5y-1}{2y^3-y^2-4y+3}\)
\(B=\dfrac{3x^3-3y^2-4y^2+4y+y-1}{2y^3-2y^2+y^2-y+3y-3}\)
\(B=\dfrac{3y^2\left(y-1\right)-4y\left(y-1\right)+\left(y-1\right)}{2y^2\left(y-1\right)+y\left(y-1\right)-3\left(y-1\right)}\)
\(B=\dfrac{\left(3y^2-4y+1\right)\left(y-1\right)}{\left(2y^2+y-3\right)\left(y-1\right)}\)
\(B=\dfrac{3y^2-3y-y+1}{2y^2-2y+3y-3}=\dfrac{3y\left(y-1\right)-\left(y-1\right)}{2y\left(y-1\right)+3\left(y-1\right)}\)
\(B=\dfrac{\left(3y-1\right)\left(y-1\right)}{\left(3y+2\right)\left(y-1\right)}=\dfrac{3y-1}{3y+2}\)
Bài 2 )
a ) \(x+\dfrac{1}{x}=3\)
\(\Leftrightarrow x^2+2x\dfrac{1}{x}+\dfrac{1}{x^2}=9\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}=1\)
b ) \(\left(x+\dfrac{1}{x}\right)^3=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+\dfrac{3}{x}+3x=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3\left(\dfrac{1}{x}+x\right)=27\)
\(\Leftrightarrow x^3+\dfrac{1}{x^3}=18\)
\(a,\dfrac{x^3-3x^2-x+3}{x^2-3x}=\dfrac{x^2\left(x-3\right)-\left(x-3\right)}{x\left(x-3\right)}=\dfrac{\left(x-3\right)\left(x^2-1\right)}{x\left(x-3\right)}=\dfrac{x^2-1}{x}\)
\(b,\dfrac{x^3y+xy^3+xy}{x^3+y^3+x^2y+xy^2+x+y}\)
\(=\dfrac{xy\left(x^2+y^2+1\right)}{\left(x^3+xy^2+x\right)+\left(y^3+x^2y+y\right)}\)
\(=\dfrac{xy\left(x^2+y^2+1\right)}{x\left(x^2+y^2+1\right)+y\left(x^2+y^2+1\right)}\)
\(=\dfrac{xy\left(x^2+y^2+1\right)}{\left(x^2+y^2+1\right)\left(x+y\right)}\)
\(=\dfrac{xy}{x+y}\)
\(c,\dfrac{\left(3x+2\right)^2-\left(x+2\right)^2}{x^3-x^2}\)
\(=\dfrac{\left(3x+2-x-2\right)\left(3x+2+x+2\right)}{x\left(x^2-1\right)}\)
\(=\dfrac{2x.\left(4x+4\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{8\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{8}{x-1}\)
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a/ \(\dfrac{x^3-x^2y+xy^2-y^3}{x^2y+xy^2-x^3-y^3}\)
\(=\dfrac{x^2\left(x-y\right)+y^2\left(x-y\right)}{x^2\left(y-x\right)+y^2\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(y-x\right)\left(y+x\right)}\)
\(=\dfrac{x^2+y^2}{\left(y-x\right)\left(y+x\right)}\)