\(\sqrt[3]{x+4}+\sqrt[3]{x-3}=1\). GPT
GPT sau: \(\sqrt[3]{x+4}=\sqrt{x-1}+2x-3\)
gpt:
\(\sqrt{x}+\sqrt[4]{x\left(1-x\right)}+\sqrt[4]{\left(1-x\right)^3}=\sqrt{1-x}+\sqrt[4]{x^3}+\sqrt[4]{x^2\left(1-x\right)}\)
GPT \(4\sqrt{x+1}-4\sqrt{1-x}+3-x=3\sqrt{1-x^2}\)
tui giải khác không biết phải không =]]
=>4 \(\left(\sqrt{x+1}\right)^2\)- 4 \(\left(\sqrt{1-x}\right)^2\)+(3 - x) = 3\(\left(\sqrt{1-x}\right)^2\)
= >4(x+1) -4(1-x) + (3-x) = 3(1-x)
=>4x +4 -4 +4x +3 -x = 3 - 3x
=>10x = 0
=> x=0 => pt VN
Gpt: \(\sqrt{x-3}+\sqrt[3]{x+4}=3\)
ĐKXĐ: \(x\ge3\)
Đặt \(\left\{{}\begin{matrix}\sqrt[]{x-3}=a\ge0\\\sqrt[3]{x+4}=b\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}a+b=3\\b^3-a^2=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=3-a\\b^3-a^2=7\end{matrix}\right.\)
\(\Rightarrow\left(3-a\right)^3-a^2=7\)
\(\Leftrightarrow\left(a-1\right)\left(a^2-7a+20\right)=0\)
\(\Leftrightarrow...\)
`sqrt{x-3}+root{3}{x+4}=3(x>=3)`
`<=>sqrt{x-3}-1+root{3}{x+4}-2=0`
`<=>(x-3-1)/(sqrt{x-3}+1)+(x+4-8)/(root{3}{(x+4)^2}+2root{3}{x+4}+4)=0`
`<=>(x-4)/(sqrt{x-3}+1)+(x-4)/(root{3}{(x+4)^2}+2root{3}{x+4}+4)=0`
`<=>(x-4)(1/(sqrt{x-3}+1)+1/(root{3}{(x+4)^2}+2root{3}{x+4}+4))=0`
Mà `1/(sqrt{x-3}+1)+1/(root{3}{(x+4)^2}+2root{3}{x+4}+4)>0AAx>=3`
`<=>x-4=0<=>x=4(tmdk)`
`->S={4}`
gpt\(\sqrt[4]{1-x^2}+\sqrt[4]{1+x}-\sqrt[4]{1-x}=3\)
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) \(\sqrt[4]{1-x^2}+\sqrt[4]{1+x}+\sqrt[4]{1-x}=3\)
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 \(x=-2\)
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 \(\left\{{}\begin{matrix}a^4+b^4=3\\a+b=3-ab\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a^2+b^2\right)^2=3+2a^2b^2\\ab=3-a-b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(a+b\right)^2-2ab\right]^2=3+2\left(3-a-b\right)^2\\ab=3-a-b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(a+b\right)^2-2\left(3-a-b\right)\right]^2=3+2\left(3-a-b\right)^2\\ab=3-a-b\end{matrix}\right.\)
GPT: \(3.\sqrt{1-x^2}=4\sqrt{1+x}-4\sqrt{1-x}+\left(3-x\right)\)
GPT \(\sqrt{3-x}\sqrt{4-x}+\sqrt{4-x}\sqrt{5-x}+\sqrt{3-x}\sqrt{ 5-x}=x\)
Gpt: \(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+4\sqrt{x^4-1}\)
Gpt:
\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=\sqrt{x^4-1}+1\)