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a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne\pm2\\x\ne0\end{matrix}\right.\)
Ta có : \(\frac{x-4}{x\left(x+2\right)}-\frac{1}{x\left(x-2\right)}=-\frac{2}{\left(x+2\right)\left(x-2\right)}\)
=> \(\frac{\left(x-4\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}-\frac{x+2}{x\left(x-2\right)\left(x+2\right)}=-\frac{2x}{x\left(x+2\right)\left(x-2\right)}\)
=> \(\left(x-4\right)\left(x-2\right)-x-2=-2x\)
=> \(x^2-4x-2x+8-x-2=-2x\)
=> \(x^2-5x+6=0\)
=> \(\left(x-2\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}x=2\\x=3\left(TM\right)\end{matrix}\right.\)
=> x = 3 .
Vậy phương trình trên có tập nghiệm là \(S=\left\{3\right\}\)
b, ĐKXĐ : \(x\ne0,-3,-6,-9,-12\)
Ta có : \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+12\right)}=\frac{1}{16}\)
=> \(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+12}=\frac{1}{16}\)
=> \(\frac{1}{x}-\frac{1}{x+12}=\frac{1}{16}\)
=> \(\frac{x+12}{x\left(x+12\right)}-\frac{x}{x\left(x+12\right)}=\frac{1}{16}\)
=> \(x\left(x+12\right)=192\)
=> \(x^2+12x-192=0\)
=> \(x^2+2x.6+36-228=0\)
=> \(\left(x+6\right)^2=288\)
=> \(\left[{}\begin{matrix}x=\sqrt{288}-6\\x=-\sqrt{288}-6\end{matrix}\right.\) ( TM )
Vậy phương trình có tập nghiệm là \(S=\left\{\pm\sqrt{288}-6\right\}\)
Ta có: \(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{8099}{8100}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{89.91}{90.90}\)
\(=\frac{1.2.3.4...89}{2.3.4...90}.\frac{3.4.5.91}{2.3.4...90}\)
\(=\frac{1}{90}.\frac{91}{2}\)
\(=\frac{91}{180}>\frac{90}{180}=\frac{1}{2}\)
=>\(A>\frac{1}{2}\)
a) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-3\right)\left(x^2+x\cdot3+3^2\right)\)
\(=x^3-3^3=x^3-27\)
b) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x-2\right)\left(x^2+x\cdot2+2^2\right)\)
\(=x^3-2^3=x^3-8\)
c) Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)\)
\(=\left(x+4\right)\left(x^2-x\cdot4+4^2\right)\)
\(=x^3+4^3=x^3+64\)
d) Ta có: \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=\left(x-3y\right)\left[x^2+x\cdot3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3=x^3-27y^3\)
e) Ta có: \(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)\)
\(=\left(x^2-\frac{1}{3}\right)\left[\left(x^2\right)^2+x^2\cdot\frac{1}{3}+\left(\frac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\frac{1}{3}\right)^3\)
\(=x^6-\frac{1}{27}\)
f) Ta có: \(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)\)
\(=\left(\frac{1}{3}x+2y\right)\left[\left(\frac{1}{3}x\right)^2-\frac{1}{3}x\cdot2y+\left(2y\right)^2\right]\)
\(=\left(\frac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\frac{1}{27}x^3+8y^3\)
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Phải là (x+9)(x+10) mới đúng chứ
