phương châm lịch sự

Đặt: \(\sqrt{2x-1}=a;\sqrt{x-2}=b\Rightarrow\sqrt{x+1}=\sqrt{\left(2x-1\right)-\left(x-2\right)}=\sqrt{a^2-b^2}\)

\(pt\Leftrightarrow a+b=\sqrt{a^2-b^2}\)

\(\Leftrightarrow a^2+2ab+b^2=a^2-b^2\)

\(\Leftrightarrow2b^2+2ab=0\Leftrightarrow2b\left(a+b\right)=0\)

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\(Q=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\dfrac{x-y}{\sqrt{x}-\sqrt[]{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right)\)

\(=\dfrac{x+y-2\sqrt{xy}+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)

\(=\dfrac{x+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+y+\sqrt{xy}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)

\(=\dfrac{x+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-x-y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{x+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\dfrac{x+y+2\sqrt{xy}-x-y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}=\dfrac{x+y-\sqrt{xy}}{\sqrt{xy}}\)

a ơi sao e đóng góp nội dung bài hc r mà chx đc duyệt v ạ

a ơi sao em chx thấy câu hỏi đâu ạ

Đặt \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

\(\Rightarrow A^2=3+\sqrt{5}+3-\sqrt{5}+2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

\(=6+2\sqrt{9-5}=6+2.2=10\)

\(B=2+\sqrt{5}\Rightarrow B^2=\left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

\(>9+1=10=A^2\)

\(\Rightarrow B^2>A^2\Rightarrow B>A\)

Vậy, B>A

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\(\Leftrightarrow\left(5x+1\right)^2=16\Leftrightarrow\left[{}\begin{matrix}5x+1=16\\5x+1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{17}{5}\end{matrix}\right.\)

KL:.....

g) \(x^2-4x+4=16\)

\(\Leftrightarrow x^2-2.2x+2^2=16\)

\(\Leftrightarrow\left(x-2\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

KL:....

h) \(\left(x+1\right)^2=x+1\)

Xét 2 TH:

TH1: \(x+1=0\Leftrightarrow x=-1\)

Thử lại ta thấy x=-1 t/m 

TH2: \(x+1\ne0\Leftrightarrow x\ne-1\)

Khi đó: \(\left(x+1\right)^2=x+1\)

\(\Leftrightarrow x+1=1\Leftrightarrow x=0\)

KL:....

i) \(4x\left(x-3\right)-x\left(3-x\right)=0\)

\(\Leftrightarrow4x\left(x-3\right)+x\left(x-3\right)=0\)

\(\Leftrightarrow5x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

KL:....

j) \(25x^2+10x+1=16\)

\(\Leftrightarrow\left(5x\right)^2+2.5x.1+1^2=16\)

\(\Leftrightarrow\left(5x+1\right)^2=16\Leftrightarrow\left[{}\begin{matrix}5x+1=16\\5x+1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=15\\5x=-17\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=3\\x=-\dfrac{17}{5}\end{matrix}\right.\)