Lời giải:

Gọi biểu thức trên là $A$

Đặt \(\sqrt[3]{15\sqrt{3}-26}=a; \sqrt[3]{15\sqrt{3}+26}=b\). Ta có:

\(a^3-b^3=-52\)

\(ab=-1\)

\(A^3=(a-b)^3=a^3-3ab(a-b)-b^3=-52+3A\)

\(\Leftrightarrow A^3-3A+52=0\)

\(\Leftrightarrow A^2(A+4)-4A(A+4)+13(A+4)=0\)

\(\Leftrightarrow (A+4)(A^2-4A+13)=0\)

Dễ thấy $A^2-4A+13>0$ nên $A+4=0$

$\Leftrightarrow A=-4$

Lời giải:

\(\frac{2x-2\sqrt{x}+2}{x-\sqrt{x}}=\frac{2(x-\sqrt{x})+2}{x-\sqrt{x}}=\frac{2(x-\sqrt{x})+2}{x-\sqrt{x}}=2+\frac{2}{x-\sqrt{x}}\)

\(\dfrac{2x-2\sqrt{x}+2}{x\sqrt{x}+1}=\dfrac{2}{\sqrt{x}+1}\)

a: 3/4=12/16=6/8=9/12

b: 7/9=14/18=21/27=35/45

21/15=21:3/15:3=7/5

11/121=11:11/121:11=1/11

27/72=27:9/72:9=3/8

\(\dfrac{21}{15}=\dfrac{7}{5};\dfrac{11}{121}=\dfrac{1}{11};\dfrac{27}{72}=\dfrac{3}{8}\)

\(\dfrac{21}{15}=\dfrac{21:3}{15:3}=\dfrac{7}{5}\)

\(\dfrac{11}{121}=\dfrac{11:11}{121:11}=\dfrac{1}{11}\)

\(\dfrac{27}{72}=\dfrac{27:9}{72:9}=\dfrac{3}{8}\)

\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(=\dfrac{1}{\sqrt{x}+1}\)

\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge0;x\ne25\right)\\ A=\dfrac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\\ A=\dfrac{5+\sqrt{x}}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)

\(C=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(đk:x\ge0,x\ne25\right)\)

\(=\dfrac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)

\(ĐK:x\ge0;x\ne25\)

\(C=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\\ C=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\)

DKXD: \(x\ne5;x>0\)

\(C=\left(\dfrac{15-\sqrt[]{x}}{x-25}+\dfrac{2}{\sqrt[]{x}+5}\right):\dfrac{\sqrt[]{x+1}}{\sqrt[]{x}-5}\)
\(C=\left(\dfrac{15-\sqrt[]{x}}{\left(\sqrt[]{x}—5\right)\left(\sqrt{x}+5\right)}+\dfrac{2\left(\sqrt[]{x}-5\right)}{\left(\sqrt[]{x}-5\right)\left(\sqrt{x+5}\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)
\(C=\left(\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(C=\dfrac{5+\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)
\(C=\dfrac{1}{\sqrt{x}+1}\)

gọn các phân số sau:

a) 7.5⁹.9⁵ / 15¹⁰

b) 3.7.8.13/ 14.15.26

/ là gạch ngang trong phân số 

Giải chi tiết giúp mình

a)=35

b)=2/5

k nha

a)=35

b)=2/5

k nha