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a: \(VT=\dfrac{a^2\left(a-4\right)-\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}\)
\(=\dfrac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-1\right)\left(a^2-5a+4\right)}\)
\(=\dfrac{\left(a-4\right)\left(a+1\right)}{\left(a-4\right)\left(a-1\right)}=\dfrac{a+1}{a-1}=VP\)
b: \(VT=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2}{x^2+1}=VP\)
P=\(\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{\left(a^3-4a^2\right)-\left(a-4\right)}{\left(a^3-8\right)-\left(7a^2-14a\right)}\)
\(=\frac{a^2\left(a-4\right)-\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}\)
\(=\frac{\left(a-4\right)\left(a^2-1\right)}{\left(a-2\right)\left(a^2-5a+4\right)}\)
\(=\frac{\left(a-4\right)\left(a^2-1\right)}{\left(a-2\right)\left(\left(a^2-4a\right)-\left(a-4\right)\right)}\)
\(=\frac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-2\right)\left(a\left(a-4\right)-\left(a-4\right)\right)}\)
\(=\frac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-2\right)\left(a-4\right)\left(a-1\right)}\)
\(=\frac{a+1}{a-2}\)
Chúc bạn học giỏi, k cho mình nhé!!!
Có \(\text{VT }=\) \(\dfrac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}\)
\(\Rightarrow VT=\dfrac{a^2\left(a-4\right)-\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}\)
\(\Rightarrow VT=\dfrac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-2\right)\left(a^2-5a+4\right)}\)
\(\Rightarrow VT=\dfrac{\left(a+1\right)\left(a^2-5a+4\right)}{\left(a-2\right)\left(a^2-5a+4\right)}\)
\(\Rightarrow\dfrac{a+1}{a-2}\)
\(\Rightarrow VT=VP\)
\(\Rightarrowđpcm\)
\(\frac{a^3-4a^2-a+a}{a^3-7a^2+14a-8}=\frac{a^3-4a^2}{a^3-4a^2-3a^2+12a+2a-8}\)
\(=\frac{a^2\left(a-4\right)}{a^2\left(a-4\right)-3a\left(a-4\right)+2\left(a-4\right)}=\frac{a^2\left(a-4\right)}{\left(a-4\right)\left(a^2-3a+2\right)}\)
\(=\frac{a^2}{a^2-3a+2}=\frac{a^2}{a\left(a-2\right)-\left(a-2\right)}=\frac{a^2}{\left(a-2\right)\left(a-1\right)}\)
Ủng hộ mik nhé!!!!
Lời giải:
ĐK....................
a)
\(\frac{a^3-4a^2-a+4}{a^3-7a^3+14a-8}=\frac{(a^3-4a^2)-(a-4)}{(a^3-4a^2)-(3a^2-12a)+(2a-8)}=\frac{a^2(a-4)-(a-4)}{a^2(a-4)-3a(a-4)+2(a-4)}\)
\(=\frac{(a-4)(a^2-1)}{(a-4)(a^2-3a+2)}=\frac{a^2-1}{a^2-3a+2}=\frac{(a-1)(a+1)}{(a-1)(a-2)}=\frac{a+1}{a-2}\) (đpcm)
b)
\(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{(x^4+x^3)+(x+1)}{(x^4+x^2)-(x^3+x)+x^2+1}=\frac{x^3(x+1)+(x+1)}{x^2(x^2+1)-x(x^2+1)+(x^2+1)}=\frac{(x+1)(x^3+1)}{(x^2+1)(x^2-x+1)}\)
\(=\frac{(x+1)(x+1)(x^2-x+1)}{(x^2+1)(x^2-x+1)}=\frac{(x+1)^2}{x^2+1}\) (đpcm)
Lời giải:
ĐK....................
a)
\(\frac{a^3-4a^2-a+4}{a^3-7a^3+14a-8}=\frac{(a^3-4a^2)-(a-4)}{(a^3-4a^2)-(3a^2-12a)+(2a-8)}=\frac{a^2(a-4)-(a-4)}{a^2(a-4)-3a(a-4)+2(a-4)}\)
\(=\frac{(a-4)(a^2-1)}{(a-4)(a^2-3a+2)}=\frac{a^2-1}{a^2-3a+2}=\frac{(a-1)(a+1)}{(a-1)(a-2)}=\frac{a+1}{a-2}\) (đpcm)
b)
\(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{(x^4+x^3)+(x+1)}{(x^4+x^2)-(x^3+x)+x^2+1}=\frac{x^3(x+1)+(x+1)}{x^2(x^2+1)-x(x^2+1)+(x^2+1)}=\frac{(x+1)(x^3+1)}{(x^2+1)(x^2-x+1)}\)
\(=\frac{(x+1)(x+1)(x^2-x+1)}{(x^2+1)(x^2-x+1)}=\frac{(x+1)^2}{x^2+1}\) (đpcm)
ĐKXĐ: \(a\ne\pm1;2;4\)
\(P=\frac{a^3-5a^2+4a+a^2-5a+4}{a^3-5a^2+4a-2a^2+10a-8}=\frac{a\left(a^2-5a+4\right)+\left(a^2-5a+4\right)}{a\left(a^2-5a+4\right)-2\left(a^2-5a+4\right)}\)
\(P=\frac{\left(a+1\right)\left(a^2-5a+4\right)}{\left(a-2\right)\left(a^2-5a+4\right)}=\frac{a+1}{a-2}\)
b/ \(P=\frac{a+1}{a-2}=1+\frac{3}{a-2}\)
\(P\) nguyên khi \(a-2=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(a-2=-3\Rightarrow a=-1\left(l\right)\)
\(a-2=-1\Rightarrow a=1\left(l\right)\)
\(a-2=1\Rightarrow a=3\)
\(a-2=3\Rightarrow a=5\)
Vậy \(\left[{}\begin{matrix}a=3\\a=5\end{matrix}\right.\) thì P nguyên
\(P=\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{\left(a-4\right)\left(a+1\right)\left(a-1\right)}{\left(a-1\right)\left(a-2\right)\left(a-4\right)}=\frac{a+1}{a-2}\)
b \(P=\frac{a-2+3}{a-2}=1+\frac{3}{a-2}\)
Để P nhận giá trị nguyên \(\left(a-2\right)\inƯ\left(3\right)=\left\{1;-1;-3;3\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}a-2=1\\a-2=-1\\a-2=3\\a-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3\\a=1\\a=5\\a=-1\end{matrix}\right.\)
Phân tích phương trình:
\(\frac{x^3+x^2-4\cdot x-4}{x^3+8\cdot x^2+17\cdot x+10}=\frac{x^2\cdot\left(x+1\right)-4\cdot\left(x+1\right)}{x^2\cdot\left(x+1\right)+7\cdot x\cdot\left(x+1\right)+10\cdot\left(x+1\right)}\)
\(=\frac{\left(x+1\right)\cdot\left(x^2-4\right)}{\left(x+1\right)\cdot\left(x^2+7\cdot x+10\right)}\)
\(=\frac{\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x-2\right)}{\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x+5\right)}=\frac{x-2}{x+5}\)
Vậy \(a=-2;b=5\)
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