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\(M=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\right)\left(x^4-\frac{\left(x^2-1\right)\left(x^2+1\right)}{x^2+1}\right)\)
\(=\left(\frac{x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\right)\left(x^4-x^2+1\right)\)
\(=\frac{x^2-2}{x^2+1}\)
b/ \(M=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
Do \(x^2+1\ge1\Rightarrow\frac{3}{x^2+1}\le3\Rightarrow1-\frac{3}{x^2+1}\ge1-3=-2\)
\(\Rightarrow M_{min}=-2\) khi \(x=0\)
\(\left(\frac{4\sqrt{a}}{\sqrt{a}+2}+\frac{8a}{4-a}\right):\left(\frac{\sqrt{a}-1}{a-2\sqrt{a}}-\frac{2}{\sqrt{a}}\right)\) (ĐKXĐ : \(a>0;a\ne4;a\ne9\))
\(=\left[\frac{4\sqrt{a}\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\frac{8a}{a-4}\right]:\left[\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-2\right)}-\frac{2\left(\sqrt{a}-2\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}\right]\)
\(=\frac{4a-8\sqrt{a}-8a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}:\frac{\sqrt{a}-1-2\sqrt{a}+4}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{-4\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}:\frac{-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{-4\sqrt{a}}{\sqrt{a}-2}.\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{3-\sqrt{a}}=-\frac{4a}{3-\sqrt{a}}\)
\(A=\left(\dfrac{x^2-2x+1}{x^2+x+1}-\dfrac{-2x^2+4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right):\dfrac{2x}{x^3+x}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}=\dfrac{x^2+1}{2}\)
a) sửa đề: \(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
=\(\frac{-x^2\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-y^2\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-z^2\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(-\frac{\left\lbrace x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
xét tử số:
Tử=\(x^2y-x^2z+y^2z-y^2x+z^2x-z^2y\)
=\(x^2\left(y-z\right)-x\left(y^2-z^2\right)+yz\left(y-z\right)\)
=\(x^2\left(y-z\right)-x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\)
=\(\left(y-z\right)\left\lbrace x^2-x\left(y+z\right)+yz\right\rbrace\)
=\(\left(y-z\right)\left\lbrace x\left(x-y\right)-z\left(x-y\right)\right\rbrace\)
=\(\left(y-z\right)\left(x-y\right)\left(x-z\right)\)
=\(-\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
thay lại vào biểu thức cũ:
\(\Rightarrow-\frac{\left\lbrace-\left(x-y\right)\left(y-z\right)\left(z-x\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(1\)
b) \(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\)
=\(\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(A=\left(\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}+\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}\right):\dfrac{x-1}{x^3}\)
\(=\dfrac{x^2+3}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)^2}\)
A = \(\frac{t^2+6t-2t-12}{t^2+t+2t+2-12}\)
<=> A = \(\frac{t\left(t+6\right)-2\left(t+6\right)}{t^2+3t-10}\)
<=> A = \(\frac{\left(t+6\right)\left(t-2\right)}{t^2-2t+5t-10}\)
<=> A = \(\frac{\left(t+6\right)\left(t-2\right)}{t\left(t-2\right)+5\left(t-2\right)}\)
<=> A = \(\frac{\left(t+6\right)\left(t-2\right)}{\left(t-2\right)\left(t+5\right)}\)
<=> A = \(\frac{t+6}{t+5}\)
Chúc bạn làm bài tốt :)