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a: Sửa đề: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
Để A là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3+4⋮\sqrt{x}-3\)
=>\(4⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7\right\}\)
=>\(x\in\left\{16;4;25;1;49\right\}\)
b: 
a: ĐKXĐ: x>0
Để A là số nguyên thì \(7⋮\sqrt{x}\)
=>\(\sqrt{x}\in\left\{1;7\right\}\)
=>\(x\in\left\{1;49\right\}\)
b: ĐKXĐ: x>1
Để B là số nguyên thì \(3⋮\sqrt{x-1}\)
=>\(\sqrt{x-1}\in\left\{1;3\right\}\)
=>\(x-1\in\left\{1;9\right\}\)
=>\(x\in\left\{2;10\right\}\)
c: ĐKXĐ: x>3
Để C là số nguyên thì \(2⋮\sqrt{x-3}\)
=>\(\sqrt{x-3}\in\left\{1;2\right\}\)
=>\(x-3\in\left\{1;4\right\}\)
=>\(x\in\left\{4;7\right\}\)
a)\(\sqrt{x}=4\Leftrightarrow x=4^2\Leftrightarrow x=16\)
b)\(\sqrt{x-2}=3\Leftrightarrow x-2=3^2\Leftrightarrow x=9-2=7\)
c)\(\sqrt{\dfrac{x}{3}-\dfrac{7}{6}}=\dfrac{1}{6}\Leftrightarrow\dfrac{x}{3}-\dfrac{7}{6}=\dfrac{1}{36}\Leftrightarrow\dfrac{x}{3}=-\dfrac{41}{36}\Leftrightarrow x=-\dfrac{41}{12}\)
d)\(x^2=7vớix< 0\)
\(\Leftrightarrow\left(-x\right)^2=7\Leftrightarrow-x=\sqrt{7}\Leftrightarrow x=-\sqrt{7}\)
e)\(x^2-4=0với>0\)
\(\Leftrightarrow x^2=4\Leftrightarrow x=\sqrt{4}=2\)
f)\(\left(2x+7\sqrt{7}\right)^2=7\)
\(\Leftrightarrow4x^2+\sqrt{5488}+343=7\)
\(\Leftrightarrow4x^2+\sqrt{5488}=-336\)
\(\Leftrightarrow4x^2=28\left(12-\sqrt{7}\right)\Leftrightarrow x^2=\dfrac{28\left(12-\sqrt{7}\right)}{4}=7\left(12-\sqrt{7}\right)\)
\(\Leftrightarrow x=\sqrt{7\left(12-\sqrt{7}\right)}=\sqrt{84-7\sqrt{7}}\)
Lời giải:
a. $\frac{2-x}{4}=\frac{3x-1}{3}$
$\Rightarrow 3(2-x)=4(3x-1)$
$\Rightarrow 6-3x=12x-4$
$\Rightarrow 6+4=12x+3x$
$\Rightarrow 10=15x$
$\Rightarrow x=\frac{10}{15}=\frac{2}{3}$
b.
$\frac{x}{7}=\frac{x+16}{35}$
$\Rightarrow \frac{5x}{35}=\frac{x+16}{35}$
$\Rightarrow 5x=x+16$
$\Rightarrow 4x=16$
$\Rightarrow x=4$
c.
$\sqrt{x^2+1}=3$
$\Rightarrow x^2+1=9$
$\Rightarrow x^2=8\Rightarrow x=\pm \sqrt{8}=\pm 2\sqrt{2}$
a: Đặt \(\sqrt{x^2+x+3}=a\)
Ta sẽ có \(\dfrac{a^2}{a}+\dfrac{1}{a}=a+\dfrac{1}{a}\ge2\cdot\sqrt{a\cdot\dfrac{1}{a}}=2\left(đpcm\right)\)
b: Đặt \(\sqrt{x^2+x+3}=b\)
Ta sẽ có \(\dfrac{b^2+4}{b}=b+\dfrac{4}{b}\ge2\cdot\sqrt{b\cdot\dfrac{4}{b}}=4\)
a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)
=>\(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}x+\dfrac{1}{2}\)
=>\(-\dfrac{3}{2}x-\dfrac{1}{2}x=\dfrac{1}{2}-\dfrac{1}{4}\)
=>\(-2x=\dfrac{1}{4}\)
=>\(2x=-\dfrac{1}{4}\)
=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)
b: ĐKXĐ: x>=0
\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)
=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
\(\sqrt{x}=1\Leftrightarrow x=1\\ \sqrt{x}=3\Leftrightarrow x=9\\ \sqrt{x}=5\Leftrightarrow x=25\\ \sqrt{x}=7\Leftrightarrow x=49\\ \sqrt{x}=9\Leftrightarrow x=81\\ \sqrt{x+1}=11\\ \Leftrightarrow x+1=121\\ \Leftrightarrow x=120\)
\(a,x-\dfrac{5}{7}=\dfrac{19}{21}\\ x=\dfrac{34}{21}\\ b,\dfrac{5}{3}-\left|x-\dfrac{1}{5}\right|=\dfrac{1}{3}\\ \left|x-\dfrac{1}{5}\right|=\dfrac{4}{3}\\ TH1:x-\dfrac{1}{5}=\dfrac{4}{3}\\ x=\dfrac{23}{15}\\ TH2:x-\dfrac{1}{5}=-\dfrac{4}{3}\\ x=-\dfrac{17}{15}\\ c,x-\dfrac{2}{5}=\dfrac{1}{4}\\ x=\dfrac{13}{20}\\ d,5\sqrt{x}-30=15\\ 5\sqrt{x}=45\\ \sqrt{x}=9\\ x=9^2=81\)