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Lời giải:
1.
\(\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{a^2(a-4)-(a-4)}{(a^3-8)-(7a^2-14a)}=\frac{(a-4)(a^2-1)}{(a-2)(a^2+2a+4)-7a(a-2)}\)
\(=\frac{(a-4)(a-1)(a+1)}{(a-2)(a^2-5a+4)}=\frac{(a-4)(a-1)(a+1)}{(a-2)(a-1)(a-4)}=\frac{a+1}{a-2}\)
2.
\(\frac{x^2y^2+1+(x^2-y)(1-y)}{x^2y^2+1+(x^2+y)(1+y)}=\frac{x^2y^2+1+x^2-x^2y-y+y^2}{x^2y^2+1+x^2+x^2y+y+y^2}\)
\(=\frac{(x^2y^2-x^2y+x^2)+(y^2-y+1)}{(x^2y^2+x^2y+x^2)+(y^2+y+1)}\)
\(=\frac{x^2(y^2-y+1)+(y^2-y+1)}{x^2(y^2+y+1)+(y^2+y+1)}=\frac{(x^2+1)(y^2-y+1)}{(x^2+1)(y^2+y+1)}=\frac{y^2-y+1}{y^2+y+1}\)
\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)
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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)
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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)
Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!
ĐKXĐ: \(x\ne\pm y\)
\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)}-\dfrac{2xy^2}{x^4-2x^2y^2+y^4}+\dfrac{7^2}{\left(x^2-y^2\right)\left(x+y\right)}\)
\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)}-\dfrac{2xy^2}{\left(\left(x+y\right).\left(x-y\right)\right)^2}+\dfrac{49}{\left(x+y\right)^2.\left(x-y\right)}\)
\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)^{ }}-\dfrac{2xy^2}{\left(x-y\right)^2.\left(x+y\right)^2}+\dfrac{49}{\left(x+y\right)^2.\left(x-y\right)}\)
\(A=\dfrac{x^2.\left(x+y\right)-2xy^2+49.\left(x-y\right)}{\left(x-y\right)^2.\left(x+y\right)^2}\)
\(A=\dfrac{x^3+x^2y-2xy^2+49x-49y}{\left(x-y\right)^2.\left(x+y\right)^2}\)
ĐKXĐ: \(x\ne\pm1\)
\(B=\dfrac{x+3}{x+1}-\dfrac{2x-1}{x-1}-\dfrac{x-3}{x-1}\)
\(B=\dfrac{\left(x+3\right).\left(x-1\right)-\left(2x-1\right).\left(x+1\right)-\left(x-3\right)\left(x+1\right)}{\left(x+1\right).\left(x-1\right)}\)
\(B=\dfrac{x^2-x+3x-3-2x^2-2x+x+1-x^2-x+3x+3}{\left(x+1\right).\left(x-1\right)}\)
\(B=\dfrac{-4x^2+4x+1}{\left(x+1\right).\left(x-1\right)}=\dfrac{1+4x-4x^2}{\left(x+1\right).\left(x-1\right)}=\dfrac{\left(1-2x\right)^2}{\left(x+1\right).\left(x-1\right)}\)
\(\left(\dfrac{1}{2}x^2-\dfrac{1}{3}y\right)\left(\dfrac{1}{4}x^4+\dfrac{1}{6}x^2y+\dfrac{1}{9}y^2\right)\\ =\left(\dfrac{1}{2}x^2\right)^3-\left(\dfrac{1}{3}y\right)^3\\ =\dfrac{1}{8}x^6-\dfrac{1}{27}y^3.\)
d)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)
=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)
ĐKXĐ: y<>0
\(y^2\left[\dfrac{1}{y\left(y-1\right)+1}-\dfrac{1}{y\left(y+1\right)+1}\right]=\dfrac{3}{y\left(y^4+y^2+1\right)}+\dfrac{2y-2}{y^2-y+1}\)
=>\(y^2\cdot\dfrac{y\left(y+1\right)+1-y\left(y-1\right)-1}{\left(y^2-y+1\right)\left(y^2+y+1\right)}=\dfrac{3}{y\left(y^2-y+1\right)\left(y^2+y+1\right)}+\dfrac{2y-2}{y^2-y+1}\)
=>\(y^2\cdot\dfrac{y\left(y+1-y+1\right)}{\left(y^2-y+1\right)\left(y^2+y+1\right)}=\dfrac{3+\left(2y-2\right)\cdot y\left(y^2+y+1\right)}{y\left(y^2-y+1\right)\left(y^2+y+1\right)}\)
=>\(y^2\cdot\dfrac{y\cdot2\cdot y}{\left(y^2-y+1\right)\cdot\left(y^2+y+1\right)\cdot y}=\dfrac{3+2y\left(y-1\right)\left(y^2+y+1\right)}{y\left(y^2-y+1\right)\left(y^2+y+1\right)}\)
=>\(2y^2\cdot y^2=3+2y\left(y^3-1\right)\)
=>\(2y^4=3+2y^4-2y\)
=>3-2y=0
=>2y=3
=>\(y=\dfrac{3}{2}\left(nhận\right)\)
Em cảm ơn ạ.