Căn bậc hai. Căn bậc ba - Hỏi đáp

a) \(\sqrt{9\left(a-3\right)^2}=\sqrt{\left[3\left(a-3\right)\right]^2}=\left|3\left(a-3\right)\right|=3a-3\) (vì \(a\ge3\))

b) \(\sqrt{36\left(a-3\right)^2}=\sqrt{\left[6\left(a-3\right)\right]^2}=\left|6a-18\right|=18-6a\) (vì \(a\le3\))

c) \(\sqrt{a^2\left(a+2\right)^2}=\sqrt{\left[a\left(a+2\right)\right]^2}=\left|a^2+2a\right|=a^2+2a\) (vì a > 0)

\(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\) ( Sửa đề )

\(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\left(x\ge1;y\ge2;z\ge3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(TM\right)\\y=6\left(TM\right)\\z=12\left(TM\right)\end{matrix}\right.\)

KL..........

bn có ghi sai đề k v

\(\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right).\left(2-\sqrt{2+\sqrt{3}}\right)}\\ \sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}\\ \sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\\ \sqrt{\left(2+\sqrt{3}\right).\left(2-\sqrt{3}\right)}\\ \sqrt{4-3}\\ \sqrt{1}\\ 1\)

Mysterious Person giup mk nha

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

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

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

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

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

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

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

mk chép sai đề bài ở trên nha

https://loga.vn/hoi-dap/tinh-can-1-1-2-2-1-3-2-can-1-1-2-2-1-3-2-tinh-sqrt-1-dfrac-1-2-2-dfrac-1-3-2-sqrt-1-dfrac-1-2-2-19838

câu 1 tham khảo bn nhé

https://hoc24.vn/hoi-dap/question/841612.html

câu 1 sai đầu bài à bạn

\(\sqrt{\left(2+\sqrt{7}\right)^2}-\sqrt{\left(2-\sqrt{7}\right)^2}=2+\sqrt{7}-\sqrt{7}+2\left(vì:\sqrt{7}>\sqrt{4}=2\right)=4\)

ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x\le-1\end{matrix}\right.\)

\(\Leftrightarrow\frac{2x+2}{x-1}=x-1\)

\(\Leftrightarrow2x+2=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-4x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{5}\\x=2-\sqrt{5}\left(l\right)\end{matrix}\right.\)

\(a+b+c\ge3\sqrt[3]{abc}+\left(\sqrt{a}-\sqrt{b}\right)^2\)

\(\Leftrightarrow c+2\sqrt{ab}\ge3\sqrt[3]{abc}\)

Mà ta có:

\(c+2\sqrt{ab}=c+\sqrt{ab}+\sqrt{ab}\ge3\sqrt[3]{abc}\left(ĐPCM\right)\)

ĐKXĐ: ....

\(\Leftrightarrow x^2-8x+16+3x+4-8\sqrt{3x+4}+16=0\)

\(\Leftrightarrow\left(x-4\right)^2+\left(\sqrt{3x+4}-4\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\\sqrt{3x+4}-4=0\end{matrix}\right.\) \(\Rightarrow x=4\)

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