a) A xác định khi:

x - 3 ≥ 0 và 4 - x > 0

⇔ x ≥ 3 và x < 4

⇔ 3 ≤ x < 4

b) B xác định khi x - 1 > 0 và x - 2 ≠ 0

⇔ x > 1 và x ≠ 2

a) \(A=\sqrt[]{x-3}-\sqrt[]{\dfrac{1}{4-x}}\left(1\right)\)

\(\left(1\right)xđ\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\4-x>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x< 4\end{matrix}\right.\)

\(\Leftrightarrow3\le x< 4\)

b) \(B=\dfrac{1}{\sqrt[]{x-1}}+\dfrac{2}{\sqrt[]{x^2-4x+4}}\left(1\right)\)

\(\left(1\right)xđ\Leftrightarrow\left\{{}\begin{matrix}x-1>0\\x^2-4x+4>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>1\\\left(x-2\right)^2>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>1\\x\ne2\end{matrix}\right.\)

Ta có: a+b+c=0\(\Leftrightarrow\)b+c=-a

Bình phương hai vế có: (b+c)²=a²

⇔ b²+2bc+c²=a²\(\Leftrightarrow\) b²+c²-a²=-2bc

Tương tự, ta có: c²+a²-b²=-2ca

                           a²+b²-c²=-2ab

→ A=\(-\dfrac{1}{2bc}-\dfrac{1}{2ca}-\dfrac{1}{2ab}=\dfrac{-\left(a+b+c\right)}{2abc}=0\)(vì a+b+c=0)

Vậy A=0

\(a)\sqrt{2x-1}=3\\ \Leftrightarrow2x-1=9\\ \Leftrightarrow2x=10\\ \Leftrightarrow x=5\)

\(b)\sqrt{4x+8}+3\sqrt{9x+18}-2\sqrt{16x+32}+5=7\\ \Leftrightarrow2\sqrt{x+2}+9\sqrt{x+2}-8\sqrt{x+2}+5=7\\ \Leftrightarrow\sqrt{x+2}\left(2+9-8\right)+5=7\\ \Leftrightarrow3\sqrt{x+2}+5=7\\ \Leftrightarrow3\sqrt{x+2}=2\\ \Leftrightarrow9x+18=4\\ \Leftrightarrow9x=-14\\ \Leftrightarrow x=-\dfrac{14}{9}\)

a) \(\sqrt[]{2x-1}=3\)

\(\Leftrightarrow2x-1=9\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\)

b) \(\sqrt[]{4x+8}+3\sqrt[]{9x+18}-2\sqrt[]{16x+32}+5=7\)

\(\Leftrightarrow\sqrt[]{4\left(x+2\right)}+3\sqrt[]{9\left(x+2\right)}-2\sqrt[]{16\left(x+2\right)}=2\)

\(\Leftrightarrow2\sqrt[]{x+2}+9\sqrt[]{x+2}-8\sqrt[]{x+2}=2\)

\(\Leftrightarrow3\sqrt[]{x+2}=2\)

\(\Leftrightarrow\sqrt[]{x+2}=\dfrac{2}{3}\)

\(\Leftrightarrow x+2=\dfrac{4}{9}\)

\(\Leftrightarrow x=\dfrac{4}{9}-2=-\dfrac{14}{4}=-\dfrac{7}{2}\)

ĐKXĐ: x + 1 ≥ 0 và 1 - x ≥ 0

⇔ x ≥ -1 và x ≤ 1

⇔ -1 ≤ x ≤ 1

Để \(\sqrt{\dfrac{1}{2-x}}\) xác định khi:

\(2-x>0\)

\(\Leftrightarrow-x>-2\)

\(\Leftrightarrow x< 2\)

\(c)ĐK:x^2+x+1\ge0\\ \sqrt[]{x^2+x+1}=1\\ \Leftrightarrow x^2+x+1=1\\ \Leftrightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ Vậy.S=\left\{-1;0\right\}\\ d)\sqrt{x^2+1}=-3\left(\varnothing\forall x\right)\\ Vậy.S=\varnothing\)

\(a,ĐKXĐ:x\ge0\\ Ta.có:\sqrt{x}-1=3\\ \Rightarrow\sqrt{x}=3+1=4\\ \Leftrightarrow x=4^2=16\\ ---\\ b,ĐKXĐ:x\ge0\\ Ta.có:\sqrt{x}=\sqrt{3}\\ \Rightarrow\left|x\right|=3\\ Vậy:x=3\)

a) \(\sqrt[]{48}+\sqrt[]{5\dfrac{1}{3}}+2\sqrt[]{75}-5\sqrt[]{1\dfrac{1}{3}}\)

\(=4\sqrt[]{3}+\sqrt[]{\dfrac{16}{3}}+2.5\sqrt[]{3}-5\sqrt[]{\dfrac{4}{3}}\)

\(=4\sqrt[]{3}+10\sqrt[]{3}+\dfrac{4}{\sqrt[]{3}}+\dfrac{10}{\sqrt[]{3}}\)

\(=14\sqrt[]{3}+\dfrac{14}{\sqrt[]{3}}\)

\(=14\left(\sqrt[]{3}+\dfrac{1}{\sqrt[]{3}}\right)\)

b) \(\sqrt[]{\left(2-\sqrt[]{3}\right)^2}+\sqrt[]{4-2\sqrt[]{3}}\)

\(=\left|2-\sqrt[]{3}\right|+\sqrt[]{3-2\sqrt[]{3}+1}\)

\(=\left|2-\sqrt[]{3}\right|+\sqrt[]{\left(\sqrt[]{3}-1\right)^2}\)

\(=\left|2-\sqrt[]{3}\right|+\left|\sqrt[]{3}-1\right|\)

\(=2-\sqrt[]{3}+\sqrt[]{3}-1\left(2>\sqrt[]{3};\sqrt[]{3}>1\right)\)

\(=1\)

c) \(\left(\sqrt[]{10}-\sqrt[]{2}\right)\sqrt[]{3+\sqrt[]{5}}\)

\(=\sqrt[]{2}\left(\sqrt[]{5}-1\right)\sqrt[]{3+\sqrt[]{5}}\)

\(=\left(\sqrt[]{5}-1\right)\sqrt[]{6+2\sqrt[]{5}}\)

\(=\left(\sqrt[]{5}-1\right)\sqrt[]{5+2\sqrt[]{5}+1}\)

\(=\left(\sqrt[]{5}-1\right)\sqrt[]{\left(\sqrt[]{5}+1\right)^2}\)

\(=\left(\sqrt[]{5}-1\right)\left|\sqrt[]{5}+1\right|\)

\(=\left(\sqrt[]{5}-1\right)\left(\sqrt[]{5}+1\right)=5-1=4\)

e) \(3\sqrt[]{\left(a-2\right)^2}-\sqrt[]{9a^2}+\sqrt[]{\dfrac{4a^2}{25}}\left(a< 0\right)\)

\(=3\left|a-2\right|-\sqrt[]{\left(3a\right)^2}+\sqrt[]{\left(\dfrac{2a}{5}\right)^2}\)

\(=3\left|a-2\right|-\left|3a\right|+\left|\dfrac{2a}{5}\right|\)

\(=3\left(2-a\right)+3a-\dfrac{2a}{5}\left(a< 0\right)\)

\(=6-3a+3a-\dfrac{2a}{5}\)

\(=6-\dfrac{2a}{5}=2\left(3-\dfrac{a}{5}\right)\)

a: ĐKXĐ: x>0; x<>1; x<>4

\(P=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{1}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)

\(P-1=\dfrac{\sqrt{x}-2-3\sqrt{x}}{3\sqrt{x}}=\dfrac{-2\sqrt{x}-2}{3\sqrt{x}}< 0\)

=>P<1

b: Khi x=1/4 thì \(P=\dfrac{\dfrac{1}{2}-2}{3\cdot\dfrac{1}{2}}=-\dfrac{3}{2}:\dfrac{3}{2}=-1\)

a)\(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{x^2+y^2-2xy}{xy}=\dfrac{\left(x-y\right)^2}{xy}\)\(\ge0\)

Vậy \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)

b) ta có: A=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

A\(\ge\)\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+2\)

=\(\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\)