Rút gọn: \(A=\frac{3a^2-2ab-b^2}{2a^2+ab-b^2}:\frac{3a^2-4ab+b^2}{3a^2+2ab-b^2}\)
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R = \(\frac{3a^2-2ab-b^2}{2a^2+ab-b^2}:\frac{3a^2-4ab+b^2}{3a^2+2ab-b^2}\)
\(R=\frac{3a^2-2ab-b^2}{2a^2+ab-b^2}:\frac{3a^2-4ab+b^2}{3a^2+2ab-b^2}\)
\(R=\frac{3a^2-2ab-b^2}{2a^2+ab-b^2}.\frac{3a^2+2ab-b^2}{3a^2-4ab+b^2}\)
\(R=\frac{\left(3a+b\right)\left(a-b\right)}{\left(a+b\right)\left(2a-b\right)}.\frac{\left(a+b\right)\left(3a-b\right)}{\left(a-b\right)\left(3a-b\right)}\)
\(R=\frac{3a+b}{2a-b}\)
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Bài 1:
Tìm x, y, z biết\(\hept{\begin{cases}xy+x+y=1\\yz+y+z=3\\zx+z+x=7\end{cases}}\)
Bài 2:
Rút gọn A = \(\frac{3a^2-2ab-b^2}{2a+ab-b^2}\): \(\frac{3a^2-4ab+b^2}{3a^2+2ab-b^2}\)
Ta có: \(\hept{\begin{cases}xy+x+y=1\\yz+y+z=3\\xz+x+z=7\end{cases}}\Rightarrow\hept{\begin{cases}xy+x+y+1=2\\yz+y+z+1=4\\xz+x+z+1=8\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+1\right)\left(y+1\right)=2\\\left(y+1\right)\left(z+1\right)=4\\\left(x+z\right)\left(z+1\right)=8\end{cases}}\)
Nhân theo vế:
\(\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=64\Rightarrow\orbr{\begin{cases}\left(x+1\right)\left(y+1\right)\left(z+1\right)=8\\\left(x+1\right)\left(y+1\right)\left(z+1\right)=-8\end{cases}}\)
Thay vào từng trường hợp tìm x;y;z
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cho A=2a^2-3ab+4b^2
'B=3a^+4ab-b^2
C=a^2+2ab+b^2
tính A-B+C
Rút gọn các phân thức sau
a) N = \(\frac{a^4-5a^2+a}{a^4-a^2+4a-4}\)
b) M =\(\frac{a^3-3a+2}{2a^3-7a^2+8a-3}\)
c) P = \(\frac{a^2-2ab+b^2-c^2}{a^2+b^2+c^2-2ab-2bc+2ac}\)
c)\(P=\)\(\frac{\left(a-b\right)^2-c^2}{\left(a-b+c\right)^2}=\frac{\left(a-b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)^2}=\frac{a-b-c}{a-b+c}\)
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b)\(M\)\(=\frac{\left(a+2\right)\left(a-1\right)^2}{\left(2a-3\right)\left(a-1\right)^2}=\frac{a+2}{2a-3}\)
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1)Rút gọn các phân thức sau
a)N = \(\frac{a^4-5a^2+4}{a^4-a^2+4a-4}\)
b)M = \(\frac{a^3-3a+2}{2a^3-7a^2+8a-3}\)
c)P= \(\frac{a^2-2ab+b^2-c^2}{a^2+b^2+c^2-2ab-2bc+2ac}\)
a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)
\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)
\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)
\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)
\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)
\(=\left(a^2-a+2\right)\left(a+2\right)\)
\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)
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Bài 1: Cho a,b thỏa mãn \(a^2\) +\(ab^2-2b^4=0\) ; a,b≠ 0; \(b^2≠ 3a ; b≠ 0 ; b≠-2a\)
Tính A= \(\frac{a+2b^2}{3a-b^2}+\frac{ab-3b^2}{2ab+b^2}\)
Cho 2a+b=7m và 3a-b=3m Tính: P=\(\frac{a^2-2ab}{a^2+b^2}\left(a^2+b^2\ne0\right)\)
\(3a-b=3m\Rightarrow b=3a-3m\)
\(\Rightarrow2a+\left(3a-3m\right)=7m\Rightarrow5a=10m\Rightarrow a=2m\)
\(\Rightarrow b=3a-3m=6m-3m=3m\)
\(\Rightarrow P=\frac{\left(2m\right)^2-2.2m.3m}{\left(2m\right)^2+\left(3m\right)^2}=\frac{4m^2-12m^2}{4m^2+9m^2}=\frac{-8m^2}{13m^2}=\frac{-8}{13}\)
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Rút gọn các phân thức sau
a) N = \(\frac{a^4-5a^2+a}{a^4-a^2+4a-4}\)
b) M =\(\frac{a^3-3a+2}{2a^3-7a^2+8a-3}\)
c) P = \(\frac{a^2-2ab+b^2-c^2}{a^2+b^2+c^2-2ab-2bc+2ac}\)
CMR:
a)a(a-b)-b(a+b)-a^2+b^2=-2ab
b)(a-1)(a+1)-(a+2)(a-2)=3
c)(a+b)^2-(a-b)^2=4ab
d)(2a-5)(3a-4)-(5a+3)(a-1)=a^2-21a+23