Bài 7: Biến đối đơn giản biểu thức chứa căn bậc hai (Tiếp theo)

Bài 1 :

a) \(A=\dfrac{3\sqrt[]{x}-1}{\sqrt[]{x}+2}\left(x\ge0\right)\)

\(\Leftrightarrow A=\dfrac{3\sqrt[]{x}+6-7}{\sqrt[]{x}+2}\)

\(\Leftrightarrow A=\dfrac{3\left(\sqrt[]{x}+2\right)-7}{\sqrt[]{x}+2}\)

\(\Leftrightarrow A=3-\dfrac{7}{\sqrt[]{x}+2}\left(1\right)\)

Ta lại có :

\(\sqrt[]{x}\ge0\)

\(\Leftrightarrow\sqrt[]{x}+2\ge2\)

\(\Leftrightarrow\dfrac{1}{\sqrt[]{x}+2}\ge\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{7}{\sqrt[]{x}+2}\le-\dfrac{7}{2}\)

\(\left(1\right)\Leftrightarrow\Leftrightarrow A=3-\dfrac{7}{\sqrt[]{x}+2}\le3-\dfrac{7}{2}=-\dfrac{1}{2}\)

Dấu "=" xảy ra khi và chỉ khi \(\sqrt[]{x}=0\Leftrightarrow x=0\)

Vậy \(GTLN\left(A\right)=-\dfrac{1}{2}\left(khi.x=0\right)\)

\(\left(5-\dfrac{7-\sqrt{7}}{1-\sqrt{7}}\right)\left(\dfrac{\sqrt{14}+\sqrt{7}}{1+\sqrt{2}}-5\right)\\ =\left(5-\dfrac{\sqrt{7}\left(\sqrt{7}-1\right)}{1-\sqrt{7}}\right)\left(\dfrac{\sqrt{7}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-5\right)\\ =\left(5+\dfrac{\sqrt{7}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}\right)\left(\sqrt{7}-5\right)\\ =\left(5+\sqrt{7}\right)\left(\sqrt{7}-5\right)\\ =\left(\sqrt{7}+5\right)\left(\sqrt{7}-5\right)\\ =\left(\sqrt{7}\right)^2-5^2\\ =7-25\\ =-18\)

\(\left(5-\dfrac{7-\sqrt{7}}{1-\sqrt{7}}\right)\left(\dfrac{\sqrt{14}+\sqrt{7}}{1+\sqrt{2}}-5\right)\)

\(=\left(5+\dfrac{\sqrt{7}-7}{1-\sqrt{7}}\right)\left(\dfrac{\sqrt{14}+\sqrt{7}}{1+\sqrt{2}}-5\right)\)

\(=\left[5+\dfrac{\sqrt{7}\left(1-\sqrt{7}\right)}{1-\sqrt{7}}\right]\left[\dfrac{\sqrt{7}\left(1+\sqrt{2}\right)}{1+\sqrt{2}}-5\right]\)

\(=\left(\sqrt{7}+5\right)\left(\sqrt{7}-5\right)\)

\(=\left(\sqrt{7}\right)^2-5^2\)

\(=7-25\)

\(=-18\)

a: \(x=\dfrac{8}{3-\sqrt{5}}=\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\left(3+\sqrt{5}\right)=\left(\sqrt{5}+1\right)^2\)

Khi \(x=\left(\sqrt{5}+1\right)^2\) thì \(A=\dfrac{3+\sqrt{5}+3}{3+\sqrt{5}-3}=\dfrac{6+\sqrt{5}}{\sqrt{5}}\)

b: \(B=\dfrac{\sqrt{x}-7-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}-7-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

c: B>-1

=>B+1>0

=>\(\dfrac{\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}>0\)

=>\(\dfrac{2\sqrt{x}-3}{\sqrt{x}-3}>0\)

=>căn x-3>0 hoặc 2căn x-3<0

=>căn x>3 hoặc căn x<3/2

=>0<=x<9/4 hoặc x>9

d: B=-4/5

=>\(\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{-4}{5}\)

=>\(5\sqrt{x}=-4\sqrt{x}+12\)

=>\(9\sqrt{x}=12\)

=>\(\sqrt{x}=\dfrac{4}{3}\)

=>x=16/9

a) \(6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9x-9}+\dfrac{7}{2}\sqrt{4x-4}=24\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9\left(x-1\right)}+\dfrac{7}{2}\sqrt{4\left(x-1\right)}=24\)

\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot3\sqrt{x-1}+\dfrac{7}{2}\cdot2\sqrt{x-1}=24\)

\(\Leftrightarrow6\sqrt{x-1}-\sqrt{x-1}+7\sqrt{x-1}=24\)

\(\Leftrightarrow12\sqrt{x-1}=24\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{24}{12}\)

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

\(\Leftrightarrow x-1=4\)

\(\Leftrightarrow x=4+1\)

\(\Leftrightarrow x=5\left(tm\right)\)

b) \(\dfrac{1}{2}\sqrt{4x+8}-2\sqrt{x+2}-\dfrac{3}{7}\sqrt{49x+98}=-8\) (ĐK: \(x\ge-2\))

\(\Leftrightarrow\dfrac{1}{2}\cdot2\sqrt{x+2}-2\sqrt{x+2}-\dfrac{3}{7}\cdot7\sqrt{x+2}=-8\)

\(\Leftrightarrow\sqrt{x+2}-2\sqrt{x+2}-3\sqrt{x+2}=-8\)

\(\Leftrightarrow-4\sqrt{x+2}=-8\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{-8}{-4}\)

\(\Leftrightarrow\sqrt{x+2}=2\)

\(\Leftrightarrow x+2=4\)

\(\Leftrightarrow x=4-2\)

\(\Leftrightarrow x=2\left(tm\right)\)

a) \(15\sqrt{\dfrac{4}{3}}-5\sqrt{48}+2\sqrt{12}-6\sqrt{\dfrac{1}{3}}\)

\(=\sqrt{15^2\cdot\dfrac{4}{3}}-5\cdot4\sqrt{3}+2\cdot2\sqrt{3}-\sqrt{6^2\cdot\dfrac{1}{3}}\)

\(=\sqrt{\dfrac{225\cdot4}{3}}-20\sqrt{3}+4\sqrt{3}-\sqrt{\dfrac{36}{3}}\)

\(=\sqrt{75\cdot4}-16\sqrt{3}-\sqrt{12}\)

\(=10\sqrt{3}-16\sqrt{3}-2\sqrt{3}\)

\(=-8\sqrt{3}\)

b) \(\dfrac{15}{\sqrt{6}+1}-\dfrac{3}{\sqrt{7}-\sqrt{2}}-15\sqrt{6}+3\sqrt{7}\)

\(=\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\dfrac{3\left(\sqrt{7}+\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}-15\sqrt{6}+3\sqrt{7}\)

\(=\dfrac{15\left(\sqrt{6}-1\right)}{6-1}-\dfrac{3\sqrt{7}+3\sqrt{2}}{7-2}-15\sqrt{6}+3\sqrt{7}\)

\(=3\left(\sqrt{6}-1\right)-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)

\(=3\sqrt{6}-3-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)

\(=-12\sqrt{6}-3+3\sqrt{7}-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}\)

\(=\dfrac{-60\sqrt{6}-15+15\sqrt{7}-3\sqrt{7}-3\sqrt{2}}{5}\)

\(=\dfrac{-60\sqrt{6}-15+12\sqrt{7}-3\sqrt{2}}{5}\)

\(A=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x+9\sqrt{x}}{x-9}\left(dkxd:x\ge0,x\ne9\right)\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)-\left(x+9\sqrt{x}\right)}{x-9}\\ =\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{x-9}\\ =\dfrac{x-3\sqrt{x}}{x-9}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(B=\dfrac{x+5\sqrt{x}}{x-25}\left(dkxd:x\ne25,x\ge0\right)\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)} \\ =\dfrac{\sqrt{x}}{\sqrt{x}-5}\)

Sửa đề: \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)

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

\(=\dfrac{x+3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\)

\(=\dfrac{9\sqrt{x}-9}{x-9}\)

a: \(=\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{3\left(\sqrt{3}+\sqrt{7}\right)}=\dfrac{\sqrt{2}}{3}\)

b: \(=\dfrac{\sqrt{7}\left(3-\sqrt{2}\right)}{3\left(3-\sqrt{2}\right)}=\dfrac{\sqrt{7}}{3}\)

a) \(\dfrac{\sqrt{6}+\sqrt{14}}{3\sqrt{3}+\sqrt{63}}\)

\(=\dfrac{\sqrt{2}\cdot\sqrt{3}+\sqrt{2}\cdot\sqrt{7}}{3\sqrt{3}+3\sqrt{7}}\)

\(=\dfrac{\sqrt{2}\cdot\left(\sqrt{3}+\sqrt{7}\right)}{3\cdot\left(\sqrt{3}+\sqrt{7}\right)}\)

\(=\dfrac{\sqrt{2}}{3}\)

b) \(\dfrac{3\sqrt{7}-\sqrt{14}}{9-\sqrt{18}}\)

\(=\dfrac{3\sqrt{7}-\sqrt{2}\cdot\sqrt{7}}{9-3\sqrt{2}}\)

\(=\dfrac{\sqrt{7}\cdot\left(3-\sqrt{2}\right)}{3\cdot\left(3-\sqrt{2}\right)}\)

\(=\dfrac{\sqrt{7}}{3}\)

Yêu cầu đề là gì vậy bạn?