a,

\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)+\sqrt{2}\cdot\dfrac{\sqrt{2^5}}{1-\sqrt{9}}\)

\(=2^2-\left(\sqrt{3}\right)^2+\dfrac{\sqrt{2}\cdot\sqrt{2^5}}{1-3}=4-3+\dfrac{\sqrt{2^6}}{-2}=1+\dfrac{8}{-2}=1+\left(-4\right)=-3\)

b,

\(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\right)\cdot\dfrac{49}{50}\)

\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\cdot\dfrac{49}{50}\)

\(=\left(1-\dfrac{1}{50}\right)\cdot\dfrac{49}{50}=\dfrac{49}{50}\cdot\dfrac{49}{50}=\dfrac{49^2}{50^2}=\dfrac{2401}{2500}\)

.

\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

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

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

đoạn cuối thiếu dấu"+"

\(A=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5}-\sqrt{6}}{5-6}+....+\dfrac{\sqrt{34}-\sqrt{35}}{34-35}+\dfrac{\sqrt{35}-\sqrt{36}}{335-36}\)

\(A=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+....+\sqrt{35}-\sqrt{36}}{-1}=\dfrac{\sqrt{4}-\sqrt{36}}{-1}\)

\(A=\sqrt{36}-\sqrt{4}=6-2=4\)

\(\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+...+\dfrac{1}{\sqrt{34}+\sqrt{35}}+\dfrac{1}{\sqrt{35}+\sqrt{36}}\)

\(=-\sqrt{4}+\sqrt{5}-\sqrt{5}+\sqrt{6}-...-\sqrt{35}+\sqrt{36}\)

\(=6-2=4\)

\(D=\left(4\sqrt{2}-\dfrac{4}{3}\sqrt{10}+\dfrac{9}{7}\sqrt{2}\right)\cdot\dfrac{\sqrt{2}}{2}\)

\(=\dfrac{37}{7}-\dfrac{4}{3}\sqrt{5}\)

\(a,\)

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

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

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

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

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

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

Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)

\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :

\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)

\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)

\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)

\(\Leftrightarrow-3\sqrt{x}+11=0\)

\(\Leftrightarrow-3\sqrt{x}=-11\)

\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)

\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)

\(\Leftrightarrow x=\dfrac{121}{9}\)

Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)

Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

`=>(\sqrt{3x^2})^2=(x+2)^2`

`=>3x^2=x^2+4x+4`

`=>3x^2-x^2-4x-4=0`

`=>2x^2-4x-4=0`

`=>x^2-2x-2=0`

`=>(x^2-2x+1)-3=0`

`=>(x-1)^2=3`

`=>(x-1)^2=(\pm \sqrt{3})^2`

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

Bài 1 

a, `3x-7\sqrt{x}+4=0`            ĐKXĐ : `x>=0`

`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`

`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`

`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`

TH1 :

`3\sqrt{x}-4=0`

`<=>\sqrt{x}=4/3`

`<=>x=16/9` ( tm )

TH2

`\sqrt{x}-1=0`

`<=>\sqrt{x}=1` (tm)

Vậy `S={16/9;1}`

b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17`     ĐKXĐ : `x>=1`

`<=>(1/2-9/2+3)\sqrt{x-1}=-17`

`<=>-\sqrt{x-1}=-17`

`<=>\sqrt{x-1}=17`

`<=>x-1=289`

`<=>x=290` ( tm )

Vậy `S={290}`

Bài 1: 

a) Ta có: \(3x-7\sqrt{x}+4=0\)

\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

b) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\cdot\left(-1\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

ĐK: \(a\ge0;a\ne4\)

a) ⇔ \(P=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

⇔ \(P=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)

b) \(P< 1\Leftrightarrow\dfrac{\sqrt{a}-4}{\sqrt{a}-2}< 1\)

\(\Leftrightarrow\dfrac{\sqrt{a}-4}{\sqrt{a}-2}-1< 0\Leftrightarrow\dfrac{-2}{\sqrt{a}-2}< 0\)

Do \(-2< 0\) ⇔ \(\sqrt{a}-2< 0\Leftrightarrow a< 4\)

Kết hợp điều kiện ban đầu, ta có: \(0< a< 4\)

Vậy khi \(0< a< 4\) thì \(P< 1\)

a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)

\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)

\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)

b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)

c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

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

a, \(\sqrt{25}-3\sqrt{\dfrac{4}{9}}=5-3.\dfrac{2}{3}=3\)

b, \(\left(2-\dfrac{5}{3}\right):\left(\dfrac{2}{7}+\dfrac{5}{21}-1\right)\)

\(=\dfrac{1}{3}:\dfrac{6+5-21}{21}\)

\(=-\dfrac{1}{3}.\dfrac{21}{10}\)

\(=-\dfrac{7}{10}\)