Bài 8: Rút gọn biểu thức chứa căn bậc hai

rút gọn hở bạn?

đkxđ: x>0 ; x≠1

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

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

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

\(\dfrac{2\sqrt{x}}{\sqrt{x}}+\dfrac{2\cdot2\sqrt{x}}{\sqrt{x}}=\dfrac{6\sqrt{x}}{\sqrt{x}}=6\)

1)

DKCĐ: a>0,\(a\ne1\)

\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{a}-\dfrac{1}{a}\right)\)\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\right)\)\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{1+a+1-a+2\sqrt{\left(1+a\right)\left(1-a\right)}}{\left(1+a\right)-\left(1-a\right)}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\)\(=\dfrac{2\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)}{2a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\sqrt{\left(1+a\right)\left(1-a\right)}+1}{a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)\left(\sqrt{\left(1+a\right)\left(1-a\right)}-1\right)}{a^2}\\ =\dfrac{\left(1+a\right)\left(1-a\right)-1}{a^2}\\ =\dfrac{1-a^2-1}{a^2}\\ =\dfrac{-a^2}{a^2}\\ =-1\)

điều kiện xác định : \(x>0;x\ne1\)

ta có : \(B=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)

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

\(\Leftrightarrow B=\left(\dfrac{x-1}{2\sqrt{x}}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

M = \(\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\times\sqrt{2}+\sqrt{20}\)

= \(2-\sqrt{6-2\sqrt{5}}+\sqrt{20}\)

= \(2-\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{20}\)

= 2 \(-\sqrt{5}+1+\sqrt{20}\)

= 3 + \(\sqrt{5}\)

\(M=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\sqrt{2}+\sqrt{20}=2-\sqrt{5-2\sqrt{5}+1}+\sqrt{20}=2-\sqrt{5}-1+2\sqrt{5}=1+\sqrt{5}\) \(N=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(5-2\right)=-3\)

Bạn @Phùng Khánh Linh làm hơi tắt rồi để mình giúp bạn làm cho dễ hiểu nhá !

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

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

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

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

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

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

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

\(A=x-2\sqrt{x}+7=x-2\sqrt{x}+1+6=\left(\sqrt{x}-1\right)^2+6\ge6\left(x\ge0\right)\)

\(\Rightarrow A_{Min}=6."="\Leftrightarrow x=1\left(TM\right)\)

gtnn là 6 tại x=1

\(3\sqrt{2a}-4\sqrt{8a}-\sqrt{32a}=3\sqrt{2a}-8\sqrt{2a}-4\sqrt{2a}=-9\sqrt{2a}\left(a\ge0\right)\)

\(3\sqrt{2a}-4\sqrt{8a}-\sqrt{32a}=3\sqrt{2a}-8\sqrt{2a}-4\sqrt{2a}=-9\sqrt{2a}\)

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

Ta có:

+/\(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)

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

\(\Rightarrow y+\sqrt{y^2+3}=\sqrt{x^2+3}-x\)

\(\Leftrightarrow x+y=\sqrt{x^2+3}-\sqrt{y^2+3}\) (1)

+/\(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)

Mà \(\left(\sqrt{y^2+3}-y\right)\left(y+\sqrt{y^2+3}\right)=y^2+3-y^2=3\)

\(\Rightarrow x+\sqrt{x^2+3}=\sqrt{y^2+3}-y\)

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

Từ (1), (2)

\(\Rightarrow2\left(x+y\right)=0\)

\(\Leftrightarrow x+y=0\)

Vậy \(x+y=0\)

đk: \(2008\le x\le2010\)

ta có: \(\left(\sqrt{2010-x}+\sqrt{x-2008}\right)^2=2+2\sqrt{\left(2010-x\right)\left(x-2008\right)}\)

\(\le2+2010-x+x-2008=4\) (bđt Cauchy)

=> \(VT^2\le4\Rightarrow VT\le2\)

Mà \(x^2-4018x+4036083=\left(x-2009\right)^2+2\ge2\)

Do đó pt có nghiệm khi VT=VP=2 => x=2009 (tm)