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

a) \(2\sqrt{3}-4\sqrt{3x}+27-3\sqrt{3x}\)

= \(\left(2\sqrt{3}+27\right)-\left(4\sqrt{3x}+3\sqrt{3x}\right)\)

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

=\(5\sqrt{3}-\sqrt{3x}\)

=\(\sqrt{3}\left(5-\sqrt{x}\right)\)

b)\(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+28\)

=\(3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28\)

=\(\sqrt{2x}\left(3-10+21\right)+28\)

=\(14\sqrt{2x}+28\)

=\(14\sqrt{2}\left(\sqrt{x}+\sqrt{2}\right)\)

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

\(\Leftrightarrow\left(\sqrt{3x+1}+\sqrt{2-x}\right)^2=9\)

\(\Leftrightarrow3x+1+2.\sqrt{\left(3x+1\right)\left(2-x\right)}+2-x=0\)

\(\Leftrightarrow2x+3+2\sqrt{\left(3x+1\right)\left(2-x\right)}=9\)

\(\Leftrightarrow2x+2\sqrt{\left(3x+1\right)\left(2-x\right)}=6\)

\(\Leftrightarrow2\left(x+\sqrt{6x-3x^2+2-x}\right)=6\)

\(\Leftrightarrow x+\sqrt{5x-3x^2+2}\) = 3

\(\Leftrightarrow\sqrt{5x-3x^2+2}=3-x\)

\(\Leftrightarrow5x-3x^2+2=\left(3-x\right)^2\)

\(\Leftrightarrow5x-3x^2+2=9-6x+x^2\)

\(\Leftrightarrow5x-3x^2+2-9+6x-x^2=0\)

\(\Leftrightarrow11x-4x^2-7=0\)

\(\Leftrightarrow11x-11x^2+7x^2-7=0\)

\(\Leftrightarrow11x\left(1-x\right)-7\left(1-x^2\right)=0\)

\(\Leftrightarrow11x\left(1-x\right)-7\left(1+x\right)\left(1-x\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(11x-7-7x\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(4x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}1-x=0\\4x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{4}\end{matrix}\right.\)

Sai thì thông cảm nha!!!hate

a) ĐK: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+2\ne0\\3-\sqrt{x}\ne0\\x-\sqrt{x}-6\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne9\end{matrix}\right.\)

\(C=\dfrac{3-\sqrt{x}}{-x+\sqrt{x}+6}+\dfrac{5}{-x+\sqrt{x}+6}-\dfrac{x-4}{-x+\sqrt{x}+6}=\dfrac{-x-\sqrt{x}+12}{-x+\sqrt{x}+6}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=1+\dfrac{2}{\sqrt{x}+2}\)

b) Đặt \(T=\dfrac{2}{\sqrt{x}+2}\)

Để C đạt gtln thì T đạt gtln hay \(\sqrt{x}+2\) đạt gtnn

Ta có: \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\)

=> gtnn của \(\sqrt{x}+2\) là 2 khi x=0 (N)

\(\Rightarrow Max_T=\dfrac{2}{2}=1\)

Thay vào C, ta được: \(Max_C=1+1=2\)

Kl: Gtln của C là 2 tại x=0

Bài 1: 

\(A=\dfrac{2}{\sqrt{2017}+\sqrt{2015}}\)

\(B=\dfrac{2}{\sqrt{2019}+\sqrt{2017}}\)

mà \(\sqrt{2015}< \sqrt{2019}\)

nên A>B

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

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

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

\(ab\sqrt{\dfrac{a}{b}}=\sqrt{ab.\dfrac{a}{b}}=\sqrt{a^2}=a\)

Chúc bạn học tốt!!!

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\Sigma\dfrac{1}{2x+3y+3z}\le\Sigma\dfrac{1}{16}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)

\(\Rightarrow P\le\dfrac{4}{16}\Sigma\left(\dfrac{1}{x+y}\right)=\dfrac{2017}{4}\)

Dấu " = " xảy ra khi \(x=y=z=\dfrac{3}{4034}\)

câu b bạn phân tích x² +5x + 6 =(x+2 )(x+3) và 3x -x² = x(3-x ) rồi đặtnhân tử chung tương tự câu a ,

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

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

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

\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{2}-\sqrt{5}\)

=\(\sqrt{8+\sqrt{2.4}+\sqrt{5.4}+\sqrt{10.4}}-\sqrt{2}-\sqrt{5}\)

=\(\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{2}-\sqrt{5}\)

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

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

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

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

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

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

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

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

\(\left\{{}\begin{matrix}x\ge0\\A=\left(1-\sqrt{x}\right)\left(\left(\sqrt{x}\right)^2+\sqrt{x}+1\right)=1-\sqrt{x^3}\end{matrix}\right.\)