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\(DK:x\ge0\)
\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+1}}{x-x-1}+\frac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}+\frac{\sqrt{x+2}-\sqrt{x+3}}{x+2-x-3}=1\)
\(\Leftrightarrow-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}=1\)
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x+3}=1+\sqrt{x}\)
\(\Leftrightarrow x+3=x+2\sqrt{x}+1\)
\(\Leftrightarrow x=1\)
Vay nghiem cua PT la \(x=1\)
a: ĐKXĐ: x>=0
b: \(\Leftrightarrow\dfrac{2\sqrt{2}-2\sqrt{2-\sqrt{x}}+\sqrt{2x}-\sqrt{x\left(2-\sqrt{x}\right)}+2\sqrt{2}+2\sqrt{2+\sqrt{x}}-\sqrt{2x}-\sqrt{x\left(2+\sqrt{x}\right)}}{2-2+\sqrt{x}}=\sqrt{2}\)
\(\Leftrightarrow4\sqrt{2}-2\sqrt{x\left(\sqrt{x}+2\right)}=\sqrt{2x}\)
\(\Leftrightarrow\sqrt{4x\left(\sqrt{x}+2\right)}=4\sqrt{2}-\sqrt{2x}\)
\(\Leftrightarrow4x\left(\sqrt{x}+2\right)=32-16\sqrt{x}+2x\)
\(\Leftrightarrow4x\sqrt{x}+8x-32+16\sqrt{x}-2x=0\)
=>\(x\in\left\{0;1.2996\right\}\)
ĐK: \(0< x\le4\)
Đặt \(\sqrt{2+\sqrt{x}}=a\left(a>0\right)\) ; \(\sqrt{2-\sqrt{x}}=b\left(b\ge0\right)\)
=> \(a^2+b^2=2+\sqrt{x}+2-\sqrt{x}=4\) (1)
Ta có: \(\dfrac{a^2}{\sqrt{2}+a}+\dfrac{b^2}{\sqrt{2}-b}=\sqrt{2}\)
<=> \(\dfrac{a^2.\sqrt{2}-a^2b+b^2.\sqrt{2}+ab^2}{2+\sqrt{2}\left(a-b\right)-ab}=\sqrt{2}\)
<=> \(\left(a^2+b^2\right)\sqrt{2}+ab\left(b-a\right)=2\sqrt{2}+2\left(a-b\right)-ab.\sqrt{2}\)
<=> \(4\sqrt{2}+ab\left(b-a\right)=2\sqrt{2}+2\left(a-b\right)-ab.\sqrt{2}\) ( Theo 1)
<=> \(\left(a-b\right)\left(2+ab\right)=2\sqrt{2}+ab.\sqrt{2}\)
<=> \(\left(a-b-\sqrt{2}\right)\left(ab+2\right)=0\)
<=> \(\left[{}\begin{matrix}ab+2=0\\a-b-\sqrt{2}=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}ab=-2\\a-b=\sqrt{2}\end{matrix}\right.\) mà a2 + b2 = 4
Xét \(\left\{{}\begin{matrix}ab=-2\\a^2+b^2=4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left(a-b\right)^2=8\\\left(a+b\right)^2=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}a-b=\pm\sqrt{8}\\a+b=0\end{matrix}\right.\) ( Loại vì \(a>0;b\ge0\) )
Xét \(\left\{{}\begin{matrix}a-b=\sqrt{2}\\a^2+b^2=4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left(b+\sqrt{2}\right)^2+b^2=4\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\2b^2+2b.\sqrt{2}-2=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\b^2+b.\sqrt{2}-1=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left[{}\begin{matrix}b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\\b=\dfrac{-\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)
#Lề: Bn lấy cái đề ở đâu hay v?
\(\sqrt{x-\sqrt{x-2}}+\sqrt{x+\sqrt{x-2}}=3\)
\(\Leftrightarrow2x+2\sqrt{\left(x-\sqrt{2-x}\right)\left(x+\sqrt{x-2}\right)}=9\)
\(\Leftrightarrow2\sqrt{\left(x-\sqrt{x-2}\right)\left(x+\sqrt{x+2}\right)}=9-2x\)
\(\Leftrightarrow4\left(x-\sqrt{x-2}\right)\left(x+\sqrt{x-2}\right)=\left(9-2x\right)^2\)
\(\Leftrightarrow4x^2-4x+8=81-36x+4x^2\)
\(\Leftrightarrow-4x+8=81-36x\)
\(\Leftrightarrow-4x=81-36x-8\)
\(\Leftrightarrow-4x=-36x+73\)
\(\Leftrightarrow-4x+36x=73\)
\(\Leftrightarrow32x=73\)
\(\Leftrightarrow x=\frac{73}{32}\)
Vậy: nghiệm phương trình là: \(\left\{\frac{73}{32}\right\}\)
Lỗi sai ngu người nhất của Chihiro.Quên viết ĐKXĐ ak em
\(\sqrt{x-\sqrt{x-2}}+\sqrt{x+\sqrt{x-2}}=3\)
\(ĐKXĐ:x\ge2\)
Bình phương 2 vế của pt ta được
\(2x+2\sqrt{\left(x-\sqrt{x-2}\right)\left(x+\sqrt{x-2}\right)}=9\)
\(\Leftrightarrow2\sqrt{x^2-x+2}=9-2x\)
\(\Leftrightarrow\hept{\begin{cases}9-2x\ge0\Leftrightarrow\frac{9}{2}\ge x\\4\left(x^2-x+2\right)=81-36x+4x^2\left(2\right)\end{cases}}\)
\(\left(2\right)\Leftrightarrow32x-73=0\Leftrightarrow x=\frac{73}{32}\left(tmDK\right)\)
Vậy \(S=\left\{\frac{73}{32}\right\}\)
p/s:học hỏi đi con.
a, \(\sqrt{2}x-\sqrt{6}=0\Leftrightarrow\sqrt{2}x=\sqrt{6}\Leftrightarrow x=\sqrt{3}\)
b, \(\frac{x^2}{\sqrt{3}}-\sqrt{12}=0\Leftrightarrow\frac{x^2}{\sqrt{3}}=\sqrt{12}\Leftrightarrow x^2=\sqrt{12}.\sqrt{3}\Leftrightarrow x^2=\sqrt{36}\Leftrightarrow x=36\)
c, \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}x=\sqrt{12}+\sqrt{27}-\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}x=2\sqrt{3}+3\sqrt{3}-\sqrt{3}\Leftrightarrow\sqrt{3}x=4\sqrt{3}\Leftrightarrow x=4\)
ĐKXĐ: \(x\ge\frac{1}{2}\)
Đề \(\Rightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}+8-2x^2-\left(\sqrt{2x-1}-\sqrt{3}\right)=0\)
Nhân liên hợp ta được:
\(\frac{\left(\sqrt{\frac{x+7}{x+1}}-\sqrt{3}\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{\left(\sqrt{2x-1}-\sqrt{3}\right)\left(\sqrt{2x+1}+\sqrt{3}\right)}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{2x-1-3}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(2-x\right)\left(2+x\right)-\frac{2x-4}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\left(x-2\right)\left[\frac{-2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}\right]=0\)
mà \(-\frac{2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}< 0\)
=> x - 2 = 0 => x = 2
Vậy x = 2
Xin lỗi bạn nhiều nhiều lắm mình không biết làm bài này vì mình chưa học
không cần đâu bạn à bài này mình giải được rồi