\(\frac{x^3+125}{x^2-3x-40}=\frac{x^3+5^3}{\left(x^2+5x\right)-\left(8x+40\right)}=\frac{\left(x+5\right)\left(x^2-5x+25\right)}{x\left(x+5\right)-8\left(x+5\right)}\)

\(=\frac{\left(x+5\right)\left(x^2-5x+25\right)}{\left(x+5\right)\left(x-8\right)}=\frac{x^2-5x+25}{x-8}\)

a) Số:

\(\dfrac{12}{18}\) = \(\dfrac{6}{9}\)  = \(\dfrac{2}{3}\) 

b) Rút gọn các phân số:
\(\dfrac{12}{48}\) = \(\dfrac{1}{4}\)

\(\dfrac{80}{100}\)= \(\dfrac{4}{5}\) 

\(\dfrac{75}{125}\) = \(\dfrac{3}{5}\) 

b: Ta có: \(\left(x+5\right)^3-x^3-125\)

\(=x^3+15x^2+75x+125-x^3-125\)

\(=15x^2+75x\)

30,001x3​=3(0,1x)3​=0,1x;

\sqrt[3]{-125 a^{12}}=\sqrt[3]{\left(-5 a^{4}\right)^{3}}=-5 a^{4};3−125a12​=3(−5a4)3​=−5a4;

\sqrt[3]{27 x^{6}}=\sqrt[3]{\left(3 x^{2}\right)^{3}}=3 x^{2};327x6​=3(3x2)3​=3x2;

\sqrt[3]{-0,343 a^{3}}=\sqrt[3]{(-0,7 a)^{3}}=-0,7 a;3−0,343a3​=3(−0,7a)3​=−0,7a;

Ta rút gọn các biểu thức như sau:

\(\sqrt[3]{0,001x^3}=\sqrt[3]{\left(0,1x\right)^3}=0,1x.\)

\(\sqrt[3]{-125a^{12}}=\sqrt[3]{\left(-5a^4\right)^3}=-5a^4\)

\(\sqrt[3]{27x^6}=\sqrt[3]{\left(3x^2\right)^3}=3x^2\)

\(\sqrt[3]{-0,343a^3}=\sqrt[3]{\left(-0,7a\right)^3}=-0,7a\)

\(\sqrt[3]{0,001x^3}=0,1x\)   ,   \(\sqrt[3]{-125a^{12}}=-5a^4\) ,  \(\sqrt[3]{27x^6}=3x^2\) , \(\sqrt[3]{-0,343a^3}=-0,7a\)     

a) (x+5)³-x³-125=x³+15x²+75x+125-x³-125=15x²+75x=15x(x+5)

Bài 1 :

a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)

Mà \(x^2+1\ge1>0\)

\(\Rightarrow2x+1\ge0\)

\(\Rightarrow x\ge-\dfrac{1}{2}\)

Vậy ...

b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)

\(=-3+4-\left(-4\right)=-3+4+4=5\)

Bài 2 :

\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)

\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)

\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)

\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)

\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)

\(=3\)

\(b,=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\left(x\ne-y\right)\\ c,=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\left(x\ne-1;x\ne\pm2;x\ne0\right)\)

b: \(\dfrac{x^3-x^2y+xy^2}{x^3+y^3}=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\)

c: \(\dfrac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\)

ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)

Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)