gpt:
\(\sqrt[3]{x-4}+\sqrt{x+4}=4\)
Những câu hỏi liên quan
GPT \(\sqrt{3-x}\sqrt{4-x}+\sqrt{4-x}\sqrt{5-x}+\sqrt{3-x}\sqrt{ 5-x}=x\)
GPT\(\sqrt[4]{3}x^4-4\sqrt[4]{2}x^3+18.\sqrt[4]{3}=0\)
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\(\sqrt[4]{1-x^2}+\sqrt[4]{1+x}-\sqrt[4]{1-x}=3\)
21 tháng 9 2021 lúc 21:18
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:
\(\left\{{}\begin{matrix}a+b=3\\b^3-a^2=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=3-a\\b^3-a^2=7\end{matrix}\right.\)
\(\Rightarrow\left(3-a\right)^3-a^2=7\)
\(\Leftrightarrow\left(a-1\right)\left(a^2-7a+20\right)=0\)
\(\Leftrightarrow...\)
Đúng 3
Bình luận (0)
`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}`
Đúng 1
Bình luận (0)
GPT sau: \(\sqrt[3]{x+4}=\sqrt{x-1}+2x-3\)
Gpt: \(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+4\sqrt{x^4-1}\)
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
Đúng 0
Bình luận (0)
gpt \(\sqrt{x^4-x^2+4}+\sqrt{x^4+20x^2+4}=7x\)
