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\(a,\dfrac{x+1}{5}-\dfrac{2}{x}\)
\(=\dfrac{x\left(x+1\right)-2.5}{5x}=\dfrac{x^2+x-10}{5x}\)
\(b,\dfrac{x+y}{9x}:\dfrac{x+y}{3x}\)
\(=\dfrac{x+y}{9x}.\dfrac{3x}{x+y}=\dfrac{1}{3}\)
a. \(\dfrac{x+1}{5}\)-\(\dfrac{2}{x}\)=\(\dfrac{x\left(x+1\right)-2.5}{5x}\)=\(\dfrac{x^2+x-10}{5x}\)
b. \(\dfrac{x+y}{9x}:\dfrac{x+y}{3x}\)=\(\dfrac{x+y}{9x}.\dfrac{3x}{x+y}=\dfrac{1}{3}\)
ĐKXĐ: \(\left\{{}\begin{matrix}3x\ne-y\\3x\ne y\end{matrix}\right.\)
a. \(\dfrac{x}{3x+y}+\dfrac{x}{3x-y}-\dfrac{2xy}{y^2-9x^2}\)
\(=\dfrac{x.\left(3x-y\right)}{\left(3x+y\right).\left(3x-y\right)}+\dfrac{x.\left(3x+y\right)}{\left(3x+y\right).\left(3x-y\right)}+\dfrac{2xy}{9x^2-y^2}\)
\(=\dfrac{x.\left(3x+y+3x-y\right)+2xy}{\left(3x-y\right).\left(3x+y\right)}\)
\(=\dfrac{6x^2+2xy}{\left(3x-y\right).\left(3x+y\right)}\)
\(=\dfrac{2x\left(3x+y\right)}{\left(3x+y\right).\left(3x-y\right)}\)
\(=\dfrac{2x}{3x-y}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\)
b. \(\dfrac{4x+5}{x^2+5x}-\dfrac{3}{x+5}\)
\(=\dfrac{4x+5}{x.\left(x+5\right)}-\dfrac{3x}{x.\left(x+5\right)}\)
\(=\dfrac{x+5}{x.\left(x+5\right)}\)
\(=\dfrac{1}{x}\)
Bài 2:
1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)
\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)
\(=x^3+2^3-2\left(x^2-1\right)\)
\(=x^3+8-2x^2+2=x^3-2x^2+10\)
\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)
\(=\left(-2y\right)^2+4\left(y+2\right)\)
\(=4y^2+4y+8\)
2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)
3: \(B=4y^2+4y+8\)
\(=4y^2+4y+1+7\)
\(=\left(2y+1\right)^2+7>=7>0\forall y\)
=>B luôn dương với mọi y
Bài 1:
5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)
\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)
\(=2x^3-x+x^2-y\)
6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)
\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)
\(=6x^2+23x-55-6x^2-84x-294\)
=-61x-349
\(\frac{4}{x-3}+\frac{5}{x+3}-\frac{13-9x^2}{x^2-9}\)
ĐKXĐ : \(x\ne\pm3\)
\(=\frac{4}{x-3}+\frac{5}{x+3}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{5\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}+\frac{5x-15}{\left(x+3\right)\left(x-3\right)}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4x+12+5x-15-13+9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{9x^2+9x-16}{\left(x+3\right)\left(x-3\right)}=\frac{9x^2+9x-16}{x^2-9}\)
\(\dfrac{2\left(x-y\right)}{4-9x^2}\cdot\dfrac{3x+2}{2\left(y-x\right)}\)
\(=\dfrac{2\left(x-y\right)}{2\left(x-y\right)}\cdot\dfrac{3x+2}{9x^2-4}=\dfrac{3x+2}{9x^2-4}\)
\(=\dfrac{3x+2}{\left(3x+2\right)\left(3x-2\right)}=\dfrac{1}{3x-2}\)