Giải pt: \(\sqrt{1+\sqrt{1-x^2}}=\left[\left(\sqrt{1+x}\right)^3-\left(\sqrt{1-x}\right)^3\right]=2+\sqrt{1-x^2}\)
Những câu hỏi liên quan
giải pt
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-2\right)}\)=3x-1
Áp dụng nội suy niu tơn để giải pt sau
\(\frac{2\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{2}\right)}=3x-1\)
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
giải pt :a,\(\left(2x+6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
18 tháng 6 2021 lúc 21:19
Giải pt:
\(x^3+\left(x+1\right)\sqrt{x+1}+2\sqrt{2}=\left(x+\sqrt{x+1}+\sqrt{2}\right)^3\)
Đk:\(x\ge-1\)
Đặt \(\left(a,b,c\right)=\left(x;\sqrt{x+1};\sqrt{2}\right)\)
Pt tt: \(a^3+b^3+c^3=\left(a+b+c\right)^3\)
\(\Leftrightarrow a^3+b^3+c^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(\Leftrightarrow0=3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)
\(\Leftrightarrow3\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\)
\(\Leftrightarrow3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\b+c=0\\a+c=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x+1}=0\\\sqrt{x+1}+\sqrt{2}=0\left(vn\right)\\x+\sqrt{2}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}=-x\\x=-\sqrt{2}\left(ktm\right)\end{matrix}\right.\)\(\Rightarrow\)\(\sqrt{x+1}=-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le0\\x+1=x^2\end{matrix}\right.\)\(\Rightarrow x=\dfrac{1-\sqrt{5}}{2}\) (tm)
Vậy...
Đúng 3
Bình luận (0)
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
Giải phương trình:
\(\frac{2\left(x-\sqrt{3}\right)\left(x-\sqrt{2}\right)}{\left(1-\sqrt{2}\right)\left(1-\sqrt{3}\right)}+\frac{3\left(x-1\right)\left(x-\sqrt{3}\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{4\left(x-1\right)\left(x-\sqrt{2}\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}-\sqrt{2}\right)}=3x-1\)
giải pt :
a,\(\left(\sqrt{5x-1}+\sqrt{x-1}\right)\left(3x-1-\sqrt{5x^2-6x+1}\right)=4x\)
b,\(2\left(\sqrt{x}-\sqrt{x-1}\right)\left(1+\sqrt{x^2-1}\right)=x\sqrt{x}\)
a, ĐK: \(x\ge1\)
Đặt \(\sqrt{5x-1}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)
\(pt\Leftrightarrow\left(a+b\right)\left(\dfrac{a^2+b^2}{2}-ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2=2\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)\left(a-b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=b+2\end{matrix}\right.\)
TH1: \(a=b\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}\Leftrightarrow x=0\left(l\right)\)
TH2: \(a=b+2\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}+2\)
\(\Leftrightarrow5x-1=x-1+4+4\sqrt{x-1}\)
\(\Leftrightarrow4x-4-4\sqrt{x-1}=0\)
\(\Leftrightarrow4x-4-4\sqrt{x-1}+1=1\)
\(\Leftrightarrow\left(2\sqrt{x-1}-1\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x-1}-1=1\\2\sqrt{x-1}-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-1}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Đúng 2
Bình luận (0)
giải pt :
a, \(\sqrt[3]{2-x}=1-\sqrt{x-1}\)
b, \(2\sqrt[3]{3x-2}+3\sqrt{6-5x}-8=0\)
c, \(\left(x+3\right)\sqrt{-x^2-8x+48}=x-24\)
d, \(\sqrt[3]{\left(2-x\right)^2}+\sqrt[3]{\left(7+x\right)\left(2-x\right)}=3\)
e, \(\dfrac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)