Bài 6: Biến đối đơn giản biểu thức chứa căn bậc hai

thay \(x=3-2\sqrt{2}\) vào P ta có:

\(\dfrac{x+8}{\sqrt{x}+1}=\dfrac{3-2\sqrt{2}+8}{\sqrt{3-2\sqrt{2}}+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}-1+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}\)

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

Thay vào P, ta được:

\(P=\dfrac{3-2\sqrt{2}+8}{\sqrt{\left(\sqrt{2}-1\right)^2}+1}=\dfrac{11-2\sqrt{2}}{\sqrt{2}}=\dfrac{11\sqrt{2}-4}{2}\)

b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:

\(P=\dfrac{3+2\sqrt{2}+8}{2+\sqrt{2}}=\dfrac{18-7\sqrt{2}}{2}\)

Vd7:

a: Ta có: \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

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

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

b: Thay \(x=11-6\sqrt{2}\) vào M, ta được:

\(M=\dfrac{3+\sqrt{2}+1}{3+\sqrt{2}-3}=\dfrac{4+\sqrt{2}}{\sqrt{2}}=2\sqrt{2}+1\)

c: Để M=2 thì \(\sqrt{x}+1=2\sqrt{x}-6\)

\(\Leftrightarrow-\sqrt{x}=-7\)

hay x=49

Vd7: 

d: Để M<1 thì M-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\sqrt{x}-3< 0\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)

e: Để M nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Leftrightarrow4⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;-1;1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;4;5;7\right\}\)

hay \(x\in\left\{1;16;25;49\right\}\)

a: Ta có: \(5\sqrt{a}+6\sqrt{\dfrac{a}{4}}-a\sqrt{\dfrac{4}{a}}+\sqrt{5}\)

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

\(=6\sqrt{a}+\sqrt{5}\)

b: Ta có: \(5\sqrt{a}-4b\sqrt{25a^3}+5a\sqrt{16ab^2}-\sqrt{9a}\)

\(=5\sqrt{a}-20ab\sqrt{a}+20ab\sqrt{a}-3\sqrt{a}\)

\(=2\sqrt{a}\)

a: Ta có: \(\sqrt{x^2-9}+\sqrt{x+3}=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

b: Ta có: \(\sqrt{9-x^2}+\sqrt{x^2+3x}=0\)

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

\(\Leftrightarrow x+3=0\)

hay x=-3

2: Ta có: \(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)

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

\(\Leftrightarrow2x-1=25\)

hay x=13

1) đk: \(x\ge5\)\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\Leftrightarrow2\sqrt[]{x-5}+\sqrt{x-5}-\dfrac{1}{3}.3\sqrt{x-5}=4\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\)3) đk: \(x\ge2\)\(\sqrt{x^2-4}=\sqrt{x-2}\Leftrightarrow\sqrt{\left(x-2\right)\left(x+2\right)}=\sqrt{x-2}\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x+2=1\Leftrightarrow x=-1\)

1: Ta có: \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

\(\Leftrightarrow x-5=4\)

hay x=9

Ta có: \(\sqrt{\dfrac{3\sqrt{5}-1}{2\sqrt{5}+3}}-\sqrt{\dfrac{\sqrt{5}+11}{7-2\sqrt{5}}}\)

\(=\sqrt{\dfrac{\left(3\sqrt{5}-1\right)\left(2\sqrt{5}-3\right)}{11}}-\sqrt{\dfrac{\left(11+\sqrt{5}\right)\left(7+2\sqrt{5}\right)}{29}}\)

\(=\sqrt{\dfrac{11\cdot\left(30-9\sqrt{5}-2\sqrt{5}+3\right)}{121}}-\sqrt{29\cdot\left(\dfrac{77+22\sqrt{5}+7\sqrt{5}+20}{841}\right)}\)

\(=\dfrac{\sqrt{363-121\sqrt{5}}}{11}+\dfrac{\sqrt{2813+841\sqrt{5}}}{29}\)

\(=\dfrac{\sqrt{726-242\sqrt{5}}}{11\sqrt{2}}+\dfrac{\sqrt{5626+1682\sqrt{5}}}{29\sqrt{2}}\)

a) \(2\sqrt{23}=\sqrt{4}.\sqrt{23}=\sqrt{4.23}=\sqrt{92}\)

\(3\sqrt{10}=\sqrt{9}.\sqrt{10}=\sqrt{90}\)

Vì \(92>90\Rightarrow\sqrt{92}>\sqrt{90}\Rightarrow2\sqrt{23}>3\sqrt{10}\)

b) \(2\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1}{5}.4}=\sqrt{\dfrac{4}{5}}=\sqrt{\dfrac{20}{25}}\)

\(\dfrac{1}{5}\sqrt{21}=\sqrt{\dfrac{1}{25}.21}=\sqrt{\dfrac{21}{25}}\)

Vì \(\dfrac{20}{25}< \dfrac{21}{25}\Rightarrow\sqrt{\dfrac{20}{25}}< \sqrt{\dfrac{21}{25}}\Rightarrow2\sqrt{\dfrac{1}{5}}< \dfrac{1}{5}\sqrt{21}\)

Bài 16:

a: Ta có: \(\sqrt{4u-20}+3\sqrt{\dfrac{u-5}{9}}-\dfrac{1}{3}\sqrt{9u-45}=4\)

\(\Leftrightarrow2\sqrt{u-5}+3\cdot\dfrac{\sqrt{u-5}}{3}-\dfrac{1}{3}\cdot3\sqrt{u-5}=4\)

\(\Leftrightarrow u-5=4\)

hay u=9

b: Ta có: \(\dfrac{2}{3}\sqrt{9u-9}-\dfrac{1}{4}\sqrt{16u-16}+27\sqrt{\dfrac{u-1}{81}}=4\)

\(\Leftrightarrow2\sqrt{u-3}-4\sqrt{u-3}+3\sqrt{u-1}=4\)

\(\Leftrightarrow u-3=16\)

hay u=19

a) Ta có: \(\left(\sqrt{12}+2\sqrt{27}\right)\cdot\dfrac{\sqrt{3}}{2}-\sqrt{150}\)

\(=8\sqrt{3}\cdot\dfrac{\sqrt{3}}{2}-5\sqrt{6}\)

\(=12-5\sqrt{6}\)

b) Ta có: \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)+2\sqrt{21}\)

\(=\sqrt{7}-2\sqrt{3}+2\sqrt{21}\)