ai gúp m vs

\(\sqrt{2a}-\sqrt{18^3}+4\sqrt{\dfrac{a}{2}}=\sqrt{2}.\sqrt{a}-54\sqrt{2}+2\sqrt{2}.\sqrt{a}=3\sqrt{2}.\sqrt{a}-54\sqrt{2}\)

\(\sqrt{\dfrac{a}{1+2b+b^2}}.\sqrt{\dfrac{4a+8ab+4ab^2}{225}}=\sqrt{\dfrac{a}{\left(b+1\right)^2}}.\sqrt{\dfrac{4a\left(1+2b+b^2\right)}{225}}=\dfrac{\sqrt{a}}{\left|b+1\right|}.\dfrac{\sqrt{4a\left(b+1\right)^2}}{15}=\dfrac{\sqrt{a}}{\left|b+1\right|}.\dfrac{2\sqrt{a}\left|b+1\right|}{15}=\dfrac{2a}{15}\)

Xét \(x^3=2a+3x.\sqrt[3]{a^2-\left(\dfrac{a+1}{3}\right)^2.\dfrac{8a-1}{3}}\)

\(\Leftrightarrow x^3=2a+3x.\sqrt[3]{\dfrac{\left(1-2a\right)^3}{27}}\)

\(\Leftrightarrow x^3=2a+x.\left(1-2a\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+2a\right)=0\)

Dễ thấy \(x^2+x+2a=\left(x+\dfrac{1}{2}\right)^2+\dfrac{8a-1}{4}>0\) (vì \(a>\dfrac{1}{8}\))

Nên x=1 hay x là số nguyên.

a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=1+\sqrt{2}\)

b: \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{6a^2}{24}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{a}{2}\)

c: \(\sqrt{5a\cdot45a}-3a=-15a-3a=-18a\)

\(a,\left(\sqrt{8}-3.\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)

\(=\sqrt{8}.\sqrt{2}-3\sqrt{2}.\sqrt{2}+\sqrt{10}.\sqrt{2}-\sqrt{5}\)

\(=\sqrt{16}-3.2+\sqrt{20}-\sqrt{5}\)

\(=\sqrt{4^2}-6+\sqrt{2^2.5}-\sqrt{5}\)

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

\(=-2+\sqrt{5}\)

\(b,\)

\(0,2\sqrt{\left(-10^2\right).3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)

\(=0,2.\left|-10\right|.\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|\)

\(=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)

\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)

\(=2\sqrt{5}\)

a) $xy-y\sqrt{x}+\sqrt{x}-1$

$=y\cdot \sqrt{x}\cdot \sqrt{x}-y\sqrt{x}+\sqrt{x}-1$

$=y\sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)$

$=(\sqrt{x}-1)(y\sqrt{x}+1)$.

b) $\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}$

$=(\sqrt{ax}+\sqrt{bx})-(\sqrt{ay}+\sqrt{by})$

$=(\sqrt{a}\cdot \sqrt{x}+\sqrt{b}\cdot \sqrt{x})-(\sqrt{a}\cdot \sqrt{y}+\sqrt{b}\cdot \sqrt{y})$

$=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})$

$=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})$.

c) $\sqrt{a+b}+\sqrt{a^2-b^2}$

$=\sqrt{a+b}+\sqrt{(a+b)(a-b)}$

$=\sqrt{a+b}+\sqrt{a+b}\cdot \sqrt{a-b}$

$=\sqrt{a+b}(1+\sqrt{a-b})$.

d) $12-\sqrt{x}-x$

$=12-4\sqrt{x}+3\sqrt{x}-x$

$=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})$

$=(3-\sqrt{x})(4+\sqrt{x})$.

a) x y − y √ x + √ x − 1 = y ⋅ √ x ⋅ √ x − y √ x + √ x − 1 = y √ x ( √ x − 1 ) + ( √ x − 1 ) = ( √ x − 1 ) ( y √ x + 1 ) . b) √ a x − √ b y + √ b x − √ a y = ( √ a x + √ b x ) − ( √ a y + √ b y ) = ( √ a ⋅ √ x + √ b ⋅ √ x ) − ( √ a ⋅ √ y + √ b ⋅ √ y ) = √ x ( √ a + √ b ) − √ y ( √ a + √ b ) = ( √ a + √ b ) ( √ x − √ y ) . c) √ a + b + √ a 2 − b 2 = √ a + b + √ ( a + b ) ( a − b ) = √ a + b + √ a + b ⋅ √ a − b = √ a + b ( 1 + √ a − b ) . d) 12 − √ x − x = 12 − 4 √ x + 3 √ x − x = 4 ( 3 − √ x ) + √ x ( 3 − √ x ) = ( 3 − √ x ) ( 4 + √ x

\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)

\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)

\(C=\dfrac{1}{2}\left(11+2\sqrt{30}\right)-\dfrac{\sqrt{30}}{2}-\dfrac{\sqrt{30}}{2}\\ =\dfrac{11}{2}+\sqrt{30}-\sqrt{30}\\ =\dfrac{11}{2}\)

\(=\sqrt{\dfrac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}\cdot\sqrt{2}\left(3+\sqrt{7}\right)\\ =\sqrt{\dfrac{2\left(3+\sqrt{7}\right)^2}{8+3\sqrt{7}}}=\sqrt{\dfrac{32+12\sqrt{7}}{8+3\sqrt{7}}}\\ =\sqrt{\dfrac{4\left(8+3\sqrt{7}\right)}{8+3\sqrt{7}}}=\sqrt{4}=2\)

a: \(=3xy\cdot\dfrac{\sqrt{2}}{\sqrt{xy}}=3\sqrt{2}\sqrt{xy}\)

b: \(=x\cdot\dfrac{\sqrt{6}}{\sqrt{x}}+\dfrac{\sqrt{6}}{3}\sqrt{x}\)

\(=\sqrt{6}\sqrt{x}+\dfrac{\sqrt{6}}{3}\sqrt{x}=\dfrac{4\sqrt{6}}{3}\cdot\sqrt{x}\)

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

\(=\sqrt{xy}+\sqrt{xy}-\sqrt{xy}=\sqrt{xy}\)

\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)

\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)

\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)

\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)

\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)

a: \(=\dfrac{6}{4+\sqrt{3}-1}=\dfrac{6}{3+\sqrt{3}}=3-\sqrt{3}\)

b: \(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}=\sqrt{6}\cdot\dfrac{1}{\sqrt{6}}\left(\dfrac{1}{2}-2\right)=-\dfrac{3}{2}\)