\(14\cdot\sqrt{x}-5\cdot\sqrt{x}< \frac{15}{2}\)

\(\Leftrightarrow9\cdot\sqrt{x}< \frac{15}{2}\Leftrightarrow\sqrt{x}< \frac{5}{6}\Leftrightarrow x< \left(\frac{5}{6}\right)^2=\frac{25}{36}\)

Ta có  14 \(\sqrt{x}\)-  5  \(\sqrt{x}\)<  \(\frac{15}{2}\)

          => \(\sqrt{x}\)(14-5)    < \(\frac{15}{2}\)

          =>\(\sqrt{x}\)9   <    \(\frac{15}{2}\)

          => \(\sqrt{x}\)<   \(\frac{15}{2}\):9

          => x  <  \(\left(\frac{5}{6}\right)^2\)

         => x < \(\frac{25}{36}\)

Vậy x <  \(\frac{25}{36}\)

đk: x\(x\ge2,y\ge-1999,z\ge2000\)

pt <-> 2VT=x+y+z

<-> (x-2-\(2\sqrt{x-2}\)+1)+(y+1999-\(2\sqrt{y+1999}\)+1)+(z-2000-\(2\sqrt{z-2000}\)+1)=0

<-> \(\left(\sqrt{x-2}-1\right)^2\)+\(\left(\sqrt{y+1999}-1\right)^2\)+\(\left(\sqrt{z-2000}-1\right)^2\)=0

<-> \(\hept{\begin{cases}\sqrt{x-2}-1=0\\\sqrt{y+1999}-1=0\\\sqrt{z-2000}-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=-1998\\z=2001\end{cases}}}\)(tm)

1.\(DK:x\le\frac{1}{3}\)

2.\(DK:x\ge-1\)

3.\(DK:-1\le x< 1\)

\(A=\frac{-7x^2}{\sqrt{x-3}-2}\)

\(đkxđ\Leftrightarrow\hept{\begin{cases}\sqrt{x-3}-2\ne0\\x-3>0\end{cases}}\)

\(\sqrt{x-3}-2\ne0\Rightarrow\sqrt{x-3}\ne2\)

\(\Rightarrow x-3\ne4\Leftrightarrow x\ne7\)

\(x-3>0\Leftrightarrow x>3\)

Vậy điều kiện xác định của A là \(\hept{\begin{cases}x>3\\x\ne7\end{cases}}\)

ĐKXĐ:

\(\sqrt{x-3}\ge0\Rightarrow\sqrt{x-3}-2\ge-2\)

\(\Rightarrow x\ge3\) 

Mà \(\sqrt{x-3}-2\ne0\) \(\Rightarrow x\ne7\)

Vậy \(x\ge3\) và \(x\ne7\)

\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)

=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

b/ Để R<-1   => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)

<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)

<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)

Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)   là sao vậy ạ?

Thì \(\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x}-3}=\frac{\sqrt{x}-3}{\sqrt{x}-3}=1\)

\(\frac{11.3^{29}-\left(3.3\right)^{15}}{2^2.3^{28}}\)=\(\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}.\left(11-3\right)}{3^{28}.2^2}=\frac{3.8}{2^2}=3.2\)=6

thank you bn nha!!!!!!!!!!!!

Thuy Duong Nguyen đánh đề cẩn thận hơn bạn nhé

Lời giải :

a) ĐKXĐ : \(x\ne1\)

 \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+3\right)\left(2-3\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\frac{15\sqrt{x}-11-3x+6-7\sqrt{x}-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

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

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

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

\(\Leftrightarrow\sqrt{x}=\sqrt{2}-1\)

Khi đó \(A=\frac{2-5\left(\sqrt{2}-1\right)}{\sqrt{2}-1+3}\)

\(A=\frac{2-5\sqrt{2}+5}{\sqrt{2}+2}=\frac{7-5\sqrt{2}}{\sqrt{2}+2}\)

c) \(A=\frac{1}{2}\)

\(\Leftrightarrow\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2}\)

\(\Leftrightarrow2\left(2-5\sqrt{x}\right)=\sqrt{x}+3\)

\(\Leftrightarrow4-10\sqrt{x}-\sqrt{x}-3=0\)

\(\Leftrightarrow1-11\sqrt{x}=0\)

\(\Leftrightarrow11\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x}=\frac{1}{11}\)

\(\Leftrightarrow x=\frac{1}{121}\)( thỏa )

d) A nguyên \(\Leftrightarrow2-5\sqrt{x}⋮\sqrt{x}+3\)

\(\Leftrightarrow-5\left(\sqrt{x}+3\right)+17⋮\sqrt{x}+3\)

Vì \(-5\left(\sqrt{x}+3\right)⋮\sqrt{x}+3\)

\(\Rightarrow17⋮\sqrt{x}+3\)

\(\Rightarrow\sqrt{x}+3\inƯ\left(17\right)=\left\{17\right\}\)( vì \(\sqrt{x}+3\ge3\))

\(\Leftrightarrow\sqrt{x}=14\)

\(\Leftrightarrow x=196\)( thỏa )

Vậy....

\(a,ĐKXĐ:\orbr{\begin{cases}x+2\sqrt{x}+3\ne0\\\sqrt{x}+3\ne0\end{cases}}\)

\(\Leftrightarrow\orbr{ }\sqrt{x}\ne-3\)

Rút gọn: p/s: sau phân số thứ 2 ở mẫu ko có x à? Bạn chép đề sai?

Hình như đề phần rút gọn sai nhé! 

\(x+2\sqrt{x}+3\) không thể tách được 

Và đa số mình làm mẫu sẽ không như này :\(1-\sqrt{2}\) ,phải có x nữa .

Bạn xem lại đề rồi mình sẽ làm tiếp