Question

gpt : a) \(\frac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\frac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)

b) \(\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}=0\)

c) \(\sqrt[4]{1-x^2}+\sqrt[4]{1+x}+\sqrt[4]{1-x}=3\)

Answer 1

b) Nhẩm thấy \(x=-2\) là nghiệm, ta xét trường hợp:

* Với \(x>-2\) thì

\(\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}>-1+0+1=0=VP\)

* Với \(x< -2\) thì

\(\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}< -1+0+1=0=VP\)

Do đó pt có nghiệm duy nhất \(x=-2\)

Answer 2

c) Đặt \(\sqrt[4]{1-x}=a;\sqrt[4]{1+x}=b\)

\(\Rightarrow a^4+b^4=2\)

Theo đề bài \(a+b+ab=3\Rightarrow a+b=3-ab\)

Cần giải cái hệ (đợi một xíu em ăn xong em làm tiếp hoặc là nếu bận thì thứ 6 tuần này em làm):v \(\left\{{}\begin{matrix}a^4+b^4=3\\a+b=3-ab\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a^2+b^2\right)^2=3+2a^2b^2\\ab=3-a-b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(a+b\right)^2-2ab\right]^2=3+2\left(3-a-b\right)^2\\ab=3-a-b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(a+b\right)^2-2\left(3-a-b\right)\right]^2=3+2\left(3-a-b\right)^2\\ab=3-a-b\end{matrix}\right.\)

Answer 3

tth, Hoàng Tử Hà, Bonking, Akai Haruma, @Nguyễn Việt Lâm

Quoc Tran Anh Le

giúp mk vs!

mk cảm ơn nhiều!

Answer 4

a/ ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{2+\sqrt{x}}=a>0\\\sqrt{2-\sqrt{x}}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=4\)

\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=4\\\frac{a^2}{\sqrt{2}+a}+\frac{b^2}{\sqrt{2}-b}=\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{2}a^2-a^2b+\sqrt{2}b^2+ab^2=\sqrt{2}\left(2+\sqrt{2}a-\sqrt{2}b-ab\right)\)

\(\Leftrightarrow\sqrt{2}\left(a^2+b^2\right)-ab\left(a-b\right)=2\sqrt{2}+2\left(a-b\right)-\sqrt{2}ab\)

\(\Leftrightarrow2\left(a-b\right)-\sqrt{2}ab+ab\left(a-b\right)-2\sqrt{2}=0\)

\(\Leftrightarrow\left(a-b\right)\left(ab+2\right)-\sqrt{2}\left(ab+2\right)=0\)

\(\Leftrightarrow\left(a-b-\sqrt{2}\right)\left(ab+2\right)=0\)

\(\Leftrightarrow a=b+\sqrt{2}\)

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

\(\Leftrightarrow2+\sqrt{x}=2-\sqrt{x}+2+2\sqrt{4-2\sqrt{x}}\)

\(\Leftrightarrow\sqrt{x}-1=\sqrt{4-2\sqrt{x}}\) (\(x\ge1\))

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

\(\Leftrightarrow x=3\)

Answer 5

c/ ĐKXĐ: \(-1\le x\le1\)

Đặt \(\left\{{}\begin{matrix}\sqrt[4]{1-x}=a\ge0\\\sqrt[4]{1+x}=b\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^4+b^4=2\\a+b+ab=3\end{matrix}\right.\)

Ta có:

\(2=a^4+b^4\ge\frac{1}{2}\left(a^2+b^2\right)^2\ge\frac{1}{2}\left[\frac{\left(a+b\right)^2}{2}\right]^2=\frac{1}{8}\left(a+b\right)^4\)

\(\Rightarrow\left(a+b\right)^4\le16\Rightarrow a+b\le2\) (1)

Mặt khác \(3=a+b+ab\le a+b+\frac{\left(a+b\right)^2}{4}\)

\(\Rightarrow\left(a+b\right)^2+4ab\ge12\Leftrightarrow\left(a+b\right)^2+4ab-12\ge0\)

\(\Leftrightarrow\left(a+b-2\right)\left(a+b+6\right)\ge0\)

\(\Leftrightarrow a+b\ge2\) (2)

Từ (1) và (2) \(\Rightarrow a+b=2\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt[4]{1-x}=1\\\sqrt[4]{1+x}=1\end{matrix}\right.\) \(\Rightarrow x=0\)