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\(\sqrt{51-7\sqrt{8}}=\sqrt{7^2-7.2\sqrt{2}+\left(\sqrt{2}\right)^2}=\sqrt{\left(7-\sqrt{2}\right)^2}=7-\sqrt{2}\)
(vì\(7=\sqrt{49}>\sqrt{2}\Rightarrow7-\sqrt{2}>0\))
\(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left(\dfrac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{a-4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
\(\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\)
\(\Leftrightarrow\dfrac{1105-1100}{x+5}=2\)
\(\Leftrightarrow\dfrac{5}{x-5}=2\)
\(\Leftrightarrow5=2\left(x-5\right)\)
\(\Leftrightarrow5=2x-10\)
\(\Leftrightarrow2x=15\)
\(\Leftrightarrow x=\dfrac{15}{2}=7,5\)
\(\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\left(ĐK:x\ne0;x\ne-5\right)\\ \Leftrightarrow\dfrac{1100\left(x+5\right)-1100x}{x\left(x+5\right)}=\dfrac{2x\left(x+5\right)}{x\left(x+5\right)}\\ \Leftrightarrow2x^2+10x-5500=0\\ \Leftrightarrow2x^2-100x+110x-5500=0\\ \Leftrightarrow2x.\left(x-50\right)+110.\left(x-50\right)=0\\ \Leftrightarrow\left(2x+110\right).\left(x-50\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+110=0\\x-50=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-55\left(TM\right)\\x=50\left(TM\right)\end{matrix}\right.\)
Vậy: S={-55;50}
\(\left\{{}\begin{matrix}2\left(x+y\right)+\sqrt{x+1}=4\\\left(x+y\right)-3\sqrt{x+1}=-5\end{matrix}\right.\left(x\ge-1\right)\)
Đặt \(\left\{{}\begin{matrix}a=x+y\\b=\sqrt{x+1}\end{matrix}\right.\left(b\ge0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}2a+b=4\\a-3b=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a+b=4\left(1\right)\\2a-6b=-10\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\Rightarrow7b=14\Rightarrow b=2\Rightarrow2a=4-2=2\Rightarrow a=1\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=1\\\sqrt{x+1}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Câu 1:
c.
PT $(1)\Leftrightarrow x=1+2my$. Thay vô PT $(2)$:
$m(1+2my)+y=2$
$\Leftrightarrow y(2m^2+1)=2-m$
$\Leftrightarrow y=\frac{2-m}{2m^2+1}$
$x=1+2my=1+\frac{4m-2m^2}{2m^2+1}=\frac{4m+1}{2m^2+1}$
Vậy hpt có nghiệm duy nhất $(x,y)=(\frac{4m+1}{2m^2+1}, \frac{2-m}{2m^2+1})$
Để $x,y$ nguyên thì:
$4m+1\vdots 2m^2+1$ và $2-m\vdots 2m^2+1$
$\Rightarrow 4m+1+4(2-m)\vdots 2m^2+1$
$\Leftrightarrow 9\vdots 2m^2+1$
$\Rightarrow 2m^2+1\in\left\{1;3;9\right\}$
$\Rightarrow m\in\left\{0; 1; -1;2;-2\right\}$
Thử lại thì thấy $m=0; -1;2$ thỏa mãn.
Đề 3:
Câu 1:
a) Ta có: \(2x^2-3x-3=0\)
\(\Delta=\left(-3\right)^2-4\cdot2\cdot\left(-3\right)=9+24=33\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3+\sqrt{33}}{2}\\x_2=\dfrac{3-\sqrt{33}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3+\sqrt{33}}{2};\dfrac{3-\sqrt{33}}{2}\right\}\)