= 12/11

Thanks bạn!

1) Vì x=25 thỏa mãn ĐKXĐ nên Thay x=25 vào biểu thức \(A=\dfrac{\sqrt{x}-2}{x+1}\), ta được:

\(A=\dfrac{\sqrt{25}-2}{25+1}=\dfrac{5-2}{25+1}=\dfrac{3}{26}\)

Vậy: Khi x=25 thì \(A=\dfrac{3}{26}\)

2) Ta có: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}+\dfrac{2x+8\sqrt{x}-6}{x-\sqrt{x}-2}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-5\sqrt{x}+6+2x+8\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3x+3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)

câu 3 chứ

bài 7
\(x^{n+2}-x^n\)
\(=x^{n+2-n}=x^2\)
\(\left(b\right)x^{x+3}-x^{x+1}=x^{x+3-x-1}=X^2\)
c)
\(x^{2m}+x^m=x^{2m+m}=x^{2m}\)
d)
\(x^{2n+1}-x^{4n}=x^{2n+1-4n}=x^{1-2n}\)

f: \(3ab-6a+b-2\)

\(=3a\left(b-2\right)+\left(b-2\right)\)

\(=\left(b-2\right)\left(3a+1\right)\)

đù mik cx có 1 bài như này

a: \(-145\cdot\left(13-57\right)+57\left(10-145\right)\)

\(=-145\cdot13+145\cdot57+57\cdot10-57\cdot145\)

\(=-145\cdot13+57\cdot10\)

=-1885+570

=-1315

b: \(17\left(15-16\right)+16\left(17-20\right)\)

\(=17\cdot15-17\cdot16+16\cdot17-16\cdot20\)

\(=17\cdot15-16\cdot20=255-320=-65\)

c: \(-38\left(25-4\right)+25\left(-4+38\right)\)

\(=-38\cdot25+38\cdot4-25\cdot4+25\cdot38\)

\(=38\cdot4-25\cdot4=4\cdot13=52\)

Giúp mình nhé

Nhanh lên à