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a.Hệ thứ nhất kì quặc thật:
\(\Leftrightarrow\sqrt{y^2+xy}+\sqrt{x+y}=\sqrt{x^2+y^2}+2\)
\(\Leftrightarrow\sqrt{x^2+y^2}-\sqrt{y^2+xy}=\sqrt{x+y}-2\)
\(\Leftrightarrow\dfrac{x\left(x-y\right)}{\sqrt{x^2+y^2}+\sqrt{y^2+xy}}=\dfrac{x+y-4}{\sqrt{x+y}+2}\)
\(\Rightarrow\left(x-y\right)\left(x+y-4\right)=\left(\dfrac{\sqrt{x^2+y^2}+\sqrt{y^2+xy}}{x\sqrt{x+y}+2x}\right)\left(x+y-4\right)^2\ge0\) (1)
\(2.\dfrac{x}{2}\sqrt{y-1}+2.\dfrac{y}{2}\sqrt{x-1}\le\dfrac{x^2}{4}+y-1+\dfrac{y^2}{4}+x-1\)
\(\Rightarrow\dfrac{x^2+4y-4}{2}\le\dfrac{x^2+y^2+4x+4y-8}{4}\)
\(\Leftrightarrow x^2-y^2+4y-4x\le0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y-4\right)\le0\) (2)
(1);(2) \(\Rightarrow\left(x-y\right)\left(x+y-4\right)=0\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=2\)
b.
\(x^3-x^2y+2y^2-2xy=0\)
\(\Leftrightarrow x^2\left(x-y\right)-2y\left(x-y\right)=0\)
\(\Leftrightarrow\left(x^2-2y\right)\left(x-y\right)=0\)
\(\Leftrightarrow y=x\) (loại \(x^2-2y=0\) do ĐKXĐ \(x^2-2y-1\ge0\))
Thế vào pt dưới
\(2\sqrt{x^2-2x-1}+\sqrt[3]{x^3-14}=x-2\)
\(\Leftrightarrow2\sqrt{x^2-2x-1}+\dfrac{x^3-14-\left(x-2\right)^3}{\sqrt[3]{\left(x^3-14\right)^2}+\left(x-2\right)\sqrt[3]{x^3-14}+\left(x-2\right)^2}=0\)
\(\Leftrightarrow\sqrt[]{x^2-2x-1}\left(2+\dfrac{6\sqrt[]{x^2-2x-1}}{\sqrt[3]{\left(x^3-14\right)^2}+\left(x-2\right)\sqrt[3]{x^3-14}+\left(x-2\right)^2}\right)=0\)
\(\Leftrightarrow\sqrt{x^2-2x-1}=0\)
Sửa lại câu c) đặt \(\sqrt{x}+1=\)t \(\Rightarrow\left[2\left(t+\dfrac{1}{2}\right)\right]\left(t-3\right)\)=7⇒\(\left\{{}\begin{matrix}t=3\\t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{9}{4}\end{matrix}\right.\)
a) \(\left(\sqrt{4-3x}\right)^2=8^2\)\(\Leftrightarrow4-3x=64\Rightarrow x=-20\)
b) \(\sqrt{4x-8}+1=12\sqrt{\dfrac{x-2}{9}}\Leftrightarrow2\sqrt{x-2}+1\)\(=\left(12\sqrt{\left(x-2\right).\dfrac{1}{9}}\right)\)
\(\Leftrightarrow2t+1=12.\dfrac{1}{3}t\) (Đặt t = \(\sqrt{x-2}\))
\(\Rightarrow t=\dfrac{1}{2}\) \(\Rightarrow\sqrt{x-2}=\dfrac{1}{2}\)\(\Rightarrow x=\dfrac{9}{4}\)
c) pt\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+1=7\\\sqrt{x}-2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)
Dùng liên hợp.
pt <=> \(\left(x-\sqrt{2}\right)\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(1+\sqrt{3}\right)\)
\(-3\left(x-1\right)\left(x-\sqrt{3}\right)\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}\right)\)
\(+2\left(x-1\right)\left(x-\sqrt{2}\right)\left(\sqrt{3}+1\right)\left(\sqrt{3}+\sqrt{2}\right)=3x-1\)
<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left[\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)-\left(x-1\right)\left(\sqrt{2}+\sqrt{3}\right)\right]\)
\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left[\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)-\left(x-\sqrt{2}\right)\left(1+\sqrt{3}\right)\right]\)
\(=3x-1\)
<=> \(\left(x-\sqrt{3}\right)\left(1+\sqrt{2}\right)\left(x+\sqrt{3}\right)\left(1-\sqrt{2}\right)\)
\(-2\left(x-1\right)\left(\sqrt{3}+\sqrt{2}\right)\left(x+1\right)\left(\sqrt{2}-\sqrt{3}\right)=3x-1\)
<=> \(3-x^2-2\left(1-x^2\right)=3x-1\)
<=> \(x^2-3x+2=0\) phương trình bậc 2.
Em làm tiếp nhé!
a/ \(\text{ĐK: }....\Leftrightarrow x\le-3\text{ hoặc }x\ge0\)
+TH1: \(x\ge0\)
\(pt\Leftrightarrow\sqrt{x}\left(\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow x=0\text{ hoặc }\sqrt{x+1}+\sqrt{x+2}=\sqrt{x+3}\text{ (1)}\)
\(\left(1\right)\Leftrightarrow x+1+x+2+2\sqrt{\left(x+1\right)\left(x+2\right)}=x+3\)
\(\Leftrightarrow x+2\sqrt{\left(x+1\right)\left(x+2\right)}=0\text{ (vô nghiệm do }x\ge0\text{ nên }x+\sqrt{\left(x+1\right)\left(x+2\right)}>0\text{)}\)
\(+TH2:\text{ }x\le-3\)
\(pt\Leftrightarrow\sqrt{-x}\left(\sqrt{-x-1}+\sqrt{-x-2}-\sqrt{-x-3}\right)=0\)
\(\Leftrightarrow\sqrt{-x-1}+\sqrt{-x-2}=\sqrt{-x-3}\text{ }\left(do\text{ }x\le-3\Rightarrow\sqrt{-x}>\sqrt{3}\right)\)
\(\Leftrightarrow-x-1-x-2+2\sqrt{\left(-x-1\right)\left(-x-2\right)}=-x-3\)
\(\Leftrightarrow2\sqrt{\left(-x-1\right)\left(-x-2\right)}-x=0\text{ (vô nghiệm do }-x\ge3\text{)}\)
Vậy \(x=0\)
b/
\(\text{ĐK: }x\ge1\)
\(\text{Đặt }\sqrt{x-1}=t;\text{ }t\ge0\)
\(pt\text{ thành: }\left(t+1\right)^3+2t+t^2-1=0\)
\(\Leftrightarrow t^3+4t^2+5t=0\Leftrightarrow t\left(t^2+4t+5\right)=0\)
\(\Leftrightarrow t=0\vee t^2+4t+5=0\text{ (Vô nghiệm)}\)
\(pt\text{ đã cho }\Leftrightarrow\sqrt{x-1}=0\Leftrightarrow x=1\)