giai p.t :\(\sqrt{\frac{x^3+1}{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)
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Đặt \(\sqrt[3]{1-x}=a;\text{ }\sqrt[3]{1+x}=b\Rightarrow a^3+b^3=2\)
Pt đã cho trở thành \(a+b=1\Leftrightarrow b=1-a\)
Suy ra: \(a^3+\left(1-a\right)^3=2\Leftrightarrow3a^2-3a-1=0\Leftrightarrow a=\frac{3\pm\sqrt{21}}{6}\)
\(\Leftrightarrow\sqrt[3]{1-x}=\frac{3\pm\sqrt{21}}{6}\Leftrightarrow x=1-\left(\frac{3\pm\sqrt{21}}{6}\right)^3\)
ĐKXĐ: \(x>2;y>1\)
Khi đó Pt \(\Leftrightarrow\)\(\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}+\frac{4}{\sqrt{y-1}}+\sqrt{y-1}=28\)
theo BĐT Cô si ta có \(\frac{36}{\sqrt{x-2}}+4\sqrt{x-2}\ge2.\sqrt{\frac{36}{\sqrt{x-2}}.4\sqrt{x-2}=24}\)
và \(\frac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge2\sqrt{\frac{4}{\sqrt{y-1}}.\sqrt{y-1}}=4\)
Pt đã cho có VT>= 28 Dấu "=" xảy ra \(\Leftrightarrow\)
\(\frac{36}{\sqrt{x-2}}=4\sqrt{x-2}\Leftrightarrow x=11\)
và \(\frac{4}{\sqrt{y-1}}=\sqrt{y-1}\Leftrightarrow y=5\)
Đối chiếu với ĐK thì x=11; y=5 là nghiệm của PT
\(DK:x\ge0\)
\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+1}}{x-x-1}+\frac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}+\frac{\sqrt{x+2}-\sqrt{x+3}}{x+2-x-3}=1\)
\(\Leftrightarrow-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}=1\)
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x+3}=1+\sqrt{x}\)
\(\Leftrightarrow x+3=x+2\sqrt{x}+1\)
\(\Leftrightarrow x=1\)
Vay nghiem cua PT la \(x=1\)
1.
đặt \(a=\sqrt{2+\sqrt{x}}\),\(b=\sqrt{2-\sqrt{x}}\)\(\left(a,b>0\right)\)
có \(a^2+b^2=4\)
pt thành \(\frac{a^2}{\sqrt{2}+a}+\frac{b^2}{\sqrt{2}-b}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}\left(a^2+b^2\right)-ab\left(a-b\right)=\sqrt{2}\left(\sqrt{2}+a\right)\left(\sqrt{2}-b\right)\)
\(\Leftrightarrow2\sqrt{2}+\sqrt{2}ab-ab\left(a-b\right)-2\left(a-b\right)=0\)
\(\Leftrightarrow\left(ab+2\right)\left(\sqrt{2}-a+b\right)=0\)
vì a,b>o nên \(a-b=\sqrt{2}\)
\(\Rightarrow\sqrt{2+\sqrt{x}}-\sqrt{2-\sqrt{x}}=\sqrt{2}\)
Bình phương 2 vế:
\(4-2\sqrt{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=2\)
\(\Leftrightarrow\sqrt{4-x}=1\)
\(\Rightarrow x=3\)
ĐK: \(x^4-4x^3+14x-11\ge0\) (*)
\(PT\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3+14x-11=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3-x^2+16x-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)(tm)
e/ ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow x+3-\sqrt{x-1}=4\)
\(\Leftrightarrow\sqrt{x-1}=x-1\)
\(\Leftrightarrow x-1=x^2-2x+1\)
\(\Leftrightarrow x^2-3x+2=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
f/ \(\Leftrightarrow\left\{{}\begin{matrix}x+5\ge0\\x^3+x^2+6x+28=\left(x+5\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x^3-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\\left(x-1\right)\left(x^2+x-3\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1\pm\sqrt{13}}{2}\\\end{matrix}\right.\)
ĐK: \(x\ne0;\pm\sqrt{2}\)
Đặt \(x=a;\text{ }\sqrt{2-x^2}=b\Rightarrow a^2+b^2=2\text{ (1)}\)
pt đã cho: \(\frac{1}{a}+\frac{1}{b}=2\Leftrightarrow a+b=2ab\)
\(\left(1\right)\Leftrightarrow\left(a+b\right)^2-2ab=2\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-2=0\)
\(\Leftrightarrow a+b=-1\text{ hoặc }a+b=2\)
\(+TH1:\text{ }a+b=-1\Rightarrow x+\sqrt{2-x^2}=-1\Leftrightarrow\sqrt{2-x^2}=-x-1\)
\(\Rightarrow2-x^2=\left(-x-1\right)^2\Leftrightarrow2x^2+2x-1=0\)
\(\Leftrightarrow x=\frac{-1\pm\sqrt{3}}{2}\)
\(TH2:\text{ }a+b=2\) tương tự
Do dùng khá nhiều phép suy ra nên phải thử lại các nghiệm trước khi kết luận.
ĐK: \(x\ge-1\)
\(\frac{pt\Leftrightarrow\sqrt{x+1}\sqrt{x^2-x+1}}{\sqrt{x+3}}+\sqrt{x+1}=\sqrt{x^2-x+1}+\sqrt{x+3}\)
\(\Leftrightarrow\frac{\sqrt{x+1}}{\sqrt{x+3}}\left(\sqrt{x^2-x+1}+\sqrt{x+3}\right)=\sqrt{x^2-x+1}+\sqrt{x+3}\)
\(\Leftrightarrow\frac{\sqrt{x+1}}{\sqrt{x+3}}=1\text{ (do }\sqrt{x^2-x+1}>0\text{)}\)
\(\Leftrightarrow...\)