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GIAO LUU

\(Lim_{x\rightarrow vc}=\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\ \)

\(\Leftrightarrow Lim_{x\rightarrow vc}=\frac{\sqrt{\frac{x+\sqrt{x}}{x}}}{\sqrt{\frac{x+\sqrt{x+\sqrt{x}}}{x}}+1}=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{x+\sqrt{x}}{x^2}}}+1}\\ \)

\(\Leftrightarrow\frac{Lim}{x\rightarrow+vc}=\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x^3}}}}+1}=\frac{\sqrt{1+\frac{1}{+vc}}}{\sqrt{1+\sqrt{\frac{1}{+vc}+\frac{1}{+vc}}}+1}=\frac{\sqrt{1+0}}{\sqrt{1+\sqrt{0+0}}+1}=\frac{1}{2}\)

1. $x+\sqrt{x}=\sqrt{x}(\sqrt{x}+1)$

2. $x-\sqrt{x}=\sqrt{x}(\sqrt{x}-1)$

3. $a+3\sqrt{a}-10=(a-2\sqrt{a})+(5\sqrt{a}-10)$

$=\sqrt{a}(\sqrt{a}-2)+5(\sqrt{a}-2)=(\sqrt{a}+5)(\sqrt{a}-2)$

4. $x\sqrt{x}+\sqrt{x}-x-1=(x\sqrt{x}+\sqrt{x})-(x+1)=\sqrt{x}(x+1)-(x+1)$

$=(x+1)(\sqrt{x}-1)$

5. $x+\sqrt{x}-2=(x-\sqrt{x})+(2\sqrt{x}-2)$

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

6. $x-5\sqrt{x}+6=(x-2\sqrt{x})-(3\sqrt{x}-6)=\sqrt{x}(\sqrt{x}-2)-3(\sqrt{x}-2)=(\sqrt{x}-2)(\sqrt{x}-3)$

7. $x\sqrt{x}-1=(\sqrt{x})^3-1^3=(\sqrt{x}-1)(x+\sqrt{x}+1)$

8. $x\sqrt{x}-x+\sqrt{x}-1=x(\sqrt{x}-1)+(\sqrt{x}-1)=(\sqrt{x}-1)(x+1)$

9. $x+2\sqrt{x}-15=(x-3\sqrt{x})+(5\sqrt{x}-15)=\sqrt{x}(\sqrt{x}-3)+5(\sqrt{x}-3)=(\sqrt{x}-3)(\sqrt{x}+5)$

10. $x-2\sqrt{x}-3=(x+\sqrt{x})-(3\sqrt{x}+3)=\sqrt{x}(\sqrt{x}+1)-3(\sqrt{x}+1)=(\sqrt{x}+1)(\sqrt{x}-3)$

\(x+\sqrt{x}=\sqrt{x}\left(\sqrt{x}+1\right)\\ x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)\\ a+3\sqrt{a}-10=a+5\sqrt{a}-2\sqrt{a}-10=\sqrt{a}\left(\sqrt{a}+5\right)-2\left(\sqrt{a}+5\right)=\left(\sqrt{a}-2\right)\left(\sqrt{a}+5\right)\)

\(x\sqrt{x}+\sqrt{x}-x-1=\left(x\sqrt{x}-x\right)+\left(\sqrt{x}-1\right)=x\left(\sqrt{x}-1\right)+\sqrt{x}-1=\left(\sqrt{x}-1\right)\left(x+1\right)\\ x+\sqrt{x}-2=x+2\sqrt{x}-\sqrt{x}-2=\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\\ x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}-6=\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)

Mấy bạn còn lại tương tự những bài trên nhé. Nếu còn thắc mắc ở chỗ nào bạn có thể liên hệ mình nhé. Nhớ lần sau bạn tách ra nha, chứ nhiều câu quá.

hỏi j v

\(\left(\dfrac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\) (ĐK: \(x>0\))

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

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

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

\(=\left(\sqrt{x}+1\right)^2\)

c: 

b; 

Sửa đề: \(\dfrac{x\sqrt{x}+26\sqrt{x}-19}{x+2\sqrt{x}-3}-\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)

a: \(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}\cdot\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{4xy}{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

b: \(=\sqrt{x}+\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)-2\sqrt{y}\)

\(=\sqrt{x}-\sqrt{y}-\sqrt{x}+\sqrt{y}=0\)

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

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

#)Hỏi j đi bn, bn ph hỏi cái j chứ làm lun rùi còn để cộng đồng ngắm ak ???

Bó cả tay lẫn chân !!! Bất lực như gặp cực hình !

Chắc là bạn ấy hỏi bạn ấy làm có đúng ko ha gì đó ?

\(A=\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=x-\sqrt{xy}+y\)

\(B=\dfrac{\sqrt{x}-\sqrt{y}}{x\sqrt{x}-y\sqrt{y}}=\dfrac{1}{x+\sqrt{xy}+y}\)

\(C=\dfrac{3\sqrt{3}+x\sqrt{x}}{3-\sqrt{3x}+x}=\sqrt{x}+\sqrt{3}\)

\(D=\dfrac{x+\sqrt{5x}+5}{x\sqrt{x}-5\sqrt{5}}=\dfrac{1}{\sqrt{x}-\sqrt{5}}\)

\(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

\(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)

\(=\frac{\left(\sqrt{x}\right)^3-1}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)

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

\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)

\(=\frac{\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)+\left(x+1\right)}{\sqrt{x}}\)

\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)

\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

Để xem nào ...

Ta có HĐT : \(\hept{\begin{cases}a\sqrt{a}+b\sqrt{b}=\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\\a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\end{cases}\left(a,b\ge0\right)}\)

\(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)

ĐKXĐ : x > 0 ; x khác 1

\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)

\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)

\(=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)

\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)