Bài 8: Rút gọn biểu thức chứa căn bậc hai

\(A=\sqrt{\dfrac{18-3\sqrt{3}}{11}}-\sqrt{2+\sqrt{3}}\)

\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{6}+\sqrt{2}}{2}\)

\(=\dfrac{2\sqrt{11\left(18-3\sqrt{3}\right)}-11\sqrt{6}-11\sqrt{2}}{22}\)

b: \(=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-4x-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}\)

Để A là số nguyên thì 5 căn x chia hết cho 2 căn x+1

\(\Leftrightarrow10\sqrt{x}+5-5⋮2\sqrt{x}+1\)

\(\Leftrightarrow2\sqrt{x}+1\in\left\{1;5\right\}\)

hay \(x\in\left\{0;4\right\}\)

\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}-\dfrac{5-\sqrt{5}}{5+\sqrt{5}}=\dfrac{\left(5+\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}-\dfrac{\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}=\dfrac{\left(5+\sqrt{5}\right)^2-\left(5-\sqrt{5}\right)^2}{25-5}=\dfrac{\left(5+\sqrt{5}-5+\sqrt{5}\right)\left(5+\sqrt{5}+5-\sqrt{5}\right)}{20}=\dfrac{\sqrt{5^2}.10}{20}=\dfrac{50}{20}=\dfrac{5}{2}\)

Mọi ngươi giúp em với ạ chứ em làm câu a Bài 1 và 2 ra kết quả dài quá :(

Bài 1: 

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

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

b: Để P<1 thì P-1<0

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

=>căn a-2>0

=>a>4

Lời giải:

Đặt \(\sqrt{x}=a(a\geq 0)\)

Khi đó:\(M=\frac{a^3-4a^2-a+4}{2a^3-14a^2+28a-16}\)

a) Điều kiện để M có nghĩa:

\(2a^3-14a^2+28a-16\neq 0\Leftrightarrow 2a^2(a-1)-12a(a-1)+16(a-1)\neq 0\)

\(\Leftrightarrow (a-1)(2a^2-12a+16)\neq 0\)

\(\Leftrightarrow (a-1)[2a(a-4)-4(a-4)]\neq 0\)

\(\Leftrightarrow 2(a-1)(a-2)(a-4)\neq 0\Leftrightarrow a\neq 1; a\neq 2; a\neq 4\)

Suy ra điều kiện để M có nghĩa là $x\geq 0; x\neq 1; x\neq 4; x\neq 16$

b)

\(M=\frac{a^3-4a^2-a+4}{2a^3-14a^2+28a-16}=\frac{a^2(a-4)-(a-4)}{2(a-1)(a-2)(a-4)}\)

\(=\frac{(a^2-1)(a-4)}{2(a-1)(a-2)(a-4)}=\frac{(a-1)(a+1)(a-4)}{2(a-1)(a-2)(a-4)}=\frac{a+1}{2(a-2)}=\frac{\sqrt{x}+1}{2(\sqrt{x}-2)}\)

c)

Để M nhận giá trị nguyên thì \(\sqrt{x}+1\vdots 2(\sqrt{x}-2)\)

\(\Rightarrow \sqrt{x}+1\vdots \sqrt{x}-2\)

\(\Rightarrow \sqrt{x}-2+3\vdots \sqrt{x}-2\)

\(\Rightarrow 3\vdots \sqrt{x}-2\Rightarrow \sqrt{x}-2\in\left\{\pm 1;\pm 3\right\}\)

\(\Rightarrow \sqrt{x}\in\left\{1; 3; 5\right\}\Rightarrow x\in\left\{1;9;25\right\}\)

Thử lại thấy đều thỏa mãn.

đkxđ: x≠9

\(P=\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+2}+\dfrac{4x\sqrt{x}+3x+9}{x-\sqrt{x}-6}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}+3}{x+5\sqrt{x}+6}\right)\)

\(=\left[\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\dfrac{4x\sqrt{x}+3x+9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\right]:\left[\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\right]\)

\(=\dfrac{x-9+4x\sqrt{x}+3x+9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}:\dfrac{x+2\sqrt{x}+2\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)

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

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

b/ \(P=48\Leftrightarrow\dfrac{4x}{\sqrt{x}-3}=48\)

\(\Leftrightarrow4x=48\sqrt{x}-144\)

\(\Leftrightarrow4x-48\sqrt{x}+144=0\)

\(\Leftrightarrow\left(2\sqrt{x}-12\right)^2=0\)

\(\Leftrightarrow2\sqrt{x}-12=0\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)(TM)

Vậy................

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

\(\Leftrightarrow\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)

\(\Leftrightarrow\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(x-1\right)-\left(x-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{1}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

.

\(\dfrac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\dfrac{\sqrt{2}.\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}{\sqrt{2}.\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}=\dfrac{2+\sqrt{4+2\sqrt{3}}}{2-\sqrt{4-2\sqrt{3}}}=\dfrac{2+\sqrt{3+2\sqrt{3}+1}}{2-\sqrt{3-2\sqrt{3}+1}}=\dfrac{2+\sqrt{\left(\sqrt{3}+1\right)^2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{2+\left|\sqrt{3}+1\right|}{2-\left|\sqrt{3}-1\right|}=\dfrac{2+\sqrt{3}+1}{2-\sqrt{3}+1}=\dfrac{3+\sqrt{3}}{3-\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}=\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\dfrac{4+2\sqrt{3}}{2}=2+\sqrt{3}\)

\(a.\left(\dfrac{\sqrt{6}-\sqrt{3}}{5\sqrt{2}-5}+\dfrac{\sqrt{5}}{5}\right):\dfrac{2}{\sqrt{5}-\sqrt{3}}=\left[\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{5\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}}{5}\right]:\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}=\dfrac{\sqrt{3}+\sqrt{5}}{5}.\dfrac{1}{\sqrt{5}+\sqrt{3}}=\dfrac{1}{5}\)

\(b.\dfrac{\sqrt{6}-3}{\sqrt{3}-\sqrt{2}}-\dfrac{4}{\sqrt{3}+1}+3\sqrt{3}=\dfrac{-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{4}{\sqrt{3}+1}+3\sqrt{3}=2\sqrt{3}-\dfrac{4}{\sqrt{3}+1}=\dfrac{2\sqrt{3}\left(\sqrt{3}+1\right)-4}{\sqrt{3}+1}=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=\dfrac{2\left(1+\sqrt{3}\right)}{1+\sqrt{3}}=2\)