Căn bậc hai. Căn bậc ba - Hỏi đáp

a) \(\sqrt{\left(x-3\right)^2}=9\)

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

\(\Leftrightarrow\left(x-3\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)

b) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{4x^2+4x+1}^2=6^2\)

\(\Leftrightarrow4x^2+4x+1-36=0\)

\(\Leftrightarrow4x^2+4x-35=0\)

\(\Leftrightarrow4\cdot\left(x-\dfrac{5}{2}\right)\cdot\left(x+\dfrac{7}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

a, \(\sqrt{\left(x-3\right)^2}=9\)

\(\Leftrightarrow\)I x-3 I=3

<=>\(\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)

b, \(\sqrt{4x^2+4x+1}=6\)

<=>\(\sqrt{\left(2x+1\right)^2}=6\)

<=> I 2x+1 I =6

<=>\(\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)

\(a.P=\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right):\left(\dfrac{2}{2-\sqrt{x}}-\dfrac{\sqrt{x}+3}{2\sqrt{x}-x}\right)=\dfrac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{4-x}:\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{8\sqrt{x}+4x}{4-x}.\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right).\sqrt{x}}{\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=\dfrac{4x}{\sqrt{x}-3}\) ( x # 4 ; x # 9 ; x > 0 )

\(b.\) \(P=\dfrac{4x}{\sqrt{x}-3}=\dfrac{4\left(x-9\right)+36}{\sqrt{x}-3}=\dfrac{4\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}-3}+\dfrac{36}{\sqrt{x}-3}=4\sqrt{x}+12+\dfrac{36}{\sqrt{x}-3}=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)

Áp dụng BĐT Cauchy cho các số dương , ta có :

\(4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}\text{≥}2\sqrt{4\left(\sqrt{x}-3\right).\dfrac{36}{\sqrt{x}-3}}=2.2.6=24\) ⇔ \(4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\) ≥ \(24+24=48\)

⇒ \(P_{MIN}=48."="\text{⇔}x=36\)

\(a.\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{9}}+\dfrac{2}{4+\sqrt{4-2\sqrt{3}}}=\dfrac{\sqrt{3}-1}{3}+\dfrac{2}{4+\sqrt{3-2\sqrt{3}+1}}=\dfrac{\sqrt{3}-1}{3}+\dfrac{2}{3+\sqrt{3}}=\dfrac{3-1+2\sqrt{3}}{3\left(\sqrt{3}+1\right)}=\dfrac{2\left(\sqrt{3}+1\right)}{3\left(\sqrt{3}+1\right)}=\dfrac{2}{3}\)

\(b.\dfrac{\sqrt{\sqrt{5}+\sqrt{2}}}{\sqrt{3\sqrt{5}-3\sqrt{2}}}=\sqrt{\dfrac{\sqrt{5}+\sqrt{2}}{3\left(\sqrt{5}-\sqrt{2}\right)}}=\sqrt{\dfrac{\left(\sqrt{5}+\sqrt{2}\right)^2}{3.3}}=\dfrac{\sqrt{5}+\sqrt{2}}{3}\)

Akai Haruma giúp e với

Akai Haruma giúp e với

e) \(E=A-\sqrt{2}\)

\(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

\(A^2=8-2\sqrt{16-7}=8-6=2\)

\(A>0=>A=\sqrt{2}\)

\(E=A-\sqrt{2}=0\)

e) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\right)}{\sqrt{2}}\)

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

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

\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1-2}{\sqrt{2}}\)

\(=\dfrac{0}{\sqrt{2}}=0\)

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

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

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

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

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

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

b) thay \(x=7-4\sqrt{3}\) vào A, ta được:

\(A=\dfrac{1}{\sqrt{7-4\sqrt{3}}-\left(7-4\sqrt{3}\right)}=\dfrac{1}{2-\sqrt{3}-7+4\sqrt{3}}=\dfrac{1}{3\sqrt{3}-5}\)

c) Đk: x > 0, x khác 1

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=-1\left(vn\right)\end{matrix}\right.\) \(\Leftrightarrow x=4\) (N)

Kl:........

\(a.P=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}=\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\) ( x ≥ 0 ; x # 1 )

\(b.P=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\text{≥ 0 }\)

⇒ \(P_{MIN}=0."="\) ⇔ \(x=0\) .

Đk: x khác 4

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

Ta có: \(\sqrt{x}\ge0\Leftrightarrow-\sqrt{x}\le0\Leftrightarrow2-\sqrt{x}\le2\Leftrightarrow\dfrac{1}{2-\sqrt{x}}\le\dfrac{1}{2}\Leftrightarrow-\dfrac{1}{2-\sqrt{x}}\ge-\dfrac{1}{2}\Leftrightarrow1-\dfrac{1}{2-\sqrt{x}}\ge1-\dfrac{1}{2}=\dfrac{1}{2}\)

Vậy P đạt GTNN bằng 1/2 <=> x=0