a, = \(3\sqrt{2}+3\sqrt{3}+9\sqrt{2}-5\sqrt{3}\)

= \(12\sqrt{2}-2\sqrt{3}\)

b, = 5 - 7 - 8 + 2

= - 8

a. ĐKXĐ: x < 2

Có

\(\frac{1}{2}\sqrt{72}+\frac{3}{4}\sqrt{48}+\sqrt{162}-\sqrt{75}=3\sqrt{2}+3\sqrt{3}+9\sqrt{2}-5\sqrt{3}=12\sqrt{2}-2\sqrt{3}\)

\(\sqrt[3]{125}+\sqrt[3]{-343}-2\sqrt[3]{64}+\frac{1}{3}\sqrt[3]{126}=5-7-8+\frac{1}{3}\sqrt[3]{126}=\frac{1}{3}\sqrt[3]{126}-10\)

Bài 1 :

a, ĐKXĐ : \(\dfrac{1}{2-x}\ge0\)

Mà 1 > 0

\(\Rightarrow2-x>0\)

\(\Rightarrow x< 2\)

Vậy ...

b, Ta có : \(\sqrt[3]{125}.\sqrt[3]{216}-\sqrt[3]{512}.\sqrt[3]{\dfrac{1}{8}}\)

\(=5.6-\dfrac{8.1}{2}=26\)

1a) Để căn thức bậc 2 có nghĩa thì \(\dfrac{1}{2-x}\ge0\Rightarrow2-x>0\Rightarrow x< 2\)

b) \(\sqrt[3]{125}.\sqrt[3]{-216}-\sqrt[3]{512}.\sqrt[3]{\dfrac{1}{8}}=\sqrt[3]{5^3}.\sqrt[3]{\left(-6\right)^3}-\sqrt[3]{8^3}.\sqrt[3]{\left(\dfrac{1}{2}\right)^3}\)

\(=5.\left(-6\right)-8.\dfrac{1}{2}=-34\)

\(\dfrac{\sqrt{ab}-b}{b}-\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{b}\right)^2}-\dfrac{\sqrt{a}}{\sqrt{b}}=\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{b}}-\dfrac{\sqrt{a}}{\sqrt{b}}\)

\(=-\dfrac{\sqrt{b}}{\sqrt{b}}=-1< 0\)

Bài 1 :

a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)

Mà \(x^2+1\ge1>0\)

\(\Rightarrow2x+1\ge0\)

\(\Rightarrow x\ge-\dfrac{1}{2}\)

Vậy ...

b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)

\(=-3+4-\left(-4\right)=-3+4+4=5\)

Bài 2 :

\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)

\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)

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

\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)

\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)

\(=3\)

a: \(=\dfrac{5}{4}\cdot3=\dfrac{15}{4}\)

b: \(=\sqrt[5]{\dfrac{98}{64}\cdot343}=\sqrt[5]{\left(\dfrac{7}{2}\right)^5}=\dfrac{7}{2}\)

\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)

\(A=6\sqrt{3^2.3}-2\sqrt{5^2.3}-\frac{1}{2}\sqrt{10^2.3}\)

\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)

\(A=3\sqrt{3}\)

vậy \(A=3\sqrt{3}\)

\(B=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)  \(ĐKXĐ:x>0;x\ne1\)

\(B=\left[1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\)

\(B=\left[1+\sqrt{x}\right]\left[1-\sqrt{x}\right]\)

\(B=1-x\)

vậy \(B=1-x\)

\(C=\sqrt[3]{64}-\sqrt[3]{-125}+\sqrt[3]{216}\)

\(C=\sqrt[3]{4^3}-\sqrt[3]{\left(-5\right)^3}+\sqrt[3]{6^3}\)

\(C=4+5+6\)

\(C=15\)

vậy \(C=15\)

Cho mk giải câu a:

\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)

\(A=18\sqrt{3}-10\sqrt{3}-\frac{1}{2}10\sqrt{3}\)

\(A=18\sqrt{3}-10\sqrt{3}-10:2\sqrt{3}\)

\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)

\(A=\left(18-10-5\right)\sqrt{3}\)

\(A=3\sqrt{3}\)

\(=3\sqrt{6}-2\sqrt{6}+\dfrac{3\sqrt{6}}{2}-\sqrt{6}=0+\dfrac{3\sqrt{6}}{2}=\dfrac{3\sqrt{6}}{2}\)