\(a,=\sqrt{3}+4\sqrt{3}+20\sqrt{3}-10\sqrt{3}=15\sqrt{3}\\ b,=4\sqrt{5}+\sqrt{5}-1-\dfrac{20\left(\sqrt{5}-1\right)}{4}\\ =5\sqrt{5}-1-5\sqrt{5}+5=4\\ c,=\dfrac{6\sqrt{13}+6+6\sqrt{13}-6}{\left(\sqrt{13}-1\right)\left(\sqrt{13}+1\right)}=\dfrac{12\sqrt{13}}{12}=\sqrt{13}\\ d,=\left(\sin^238^0+\cos^238^0\right)+\left(\tan67^0-\tan67^0\right)=1+0=1\)

a: \(=\sqrt{3}+4\sqrt{3}+4\cdot5\sqrt{3}-10\sqrt{3}\)

\(=15\sqrt{3}\)

b: \(=2\cdot2\sqrt{5}+\sqrt{5}-1-5+5\sqrt{5}\)

=-6

Lời giải:
Gọi biểu thức là $A$

\(A=\frac{-\sqrt{3}(1-\sqrt{2})}{1-\sqrt{2}}+\frac{\sqrt{3}(\sqrt{3}+6)}{\sqrt{3}}-\frac{13(\sqrt{3}+4)}{(\sqrt{3}+4)(\sqrt{3}-4)}\)

\(=-\sqrt{3}+\sqrt{3}+6-\frac{13(\sqrt{3}+4)}{3-16}=6-(-\sqrt{3}-4)=10+\sqrt{3}\)

a) Ta có: \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)

\(=10\sqrt{5}\)

b) Ta có: \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=3\left(2-\sqrt{3}\right)+4+\sqrt{3}+2\sqrt{3}\)

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

\(=10+\sqrt{3}\)

c) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

=7-5=2

d) Ta có: \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)

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

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

a. \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)

\(=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)

\(=10\sqrt{5}\)

b. \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\dfrac{6\sqrt{3}}{\sqrt{3}.\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{4-3}+\dfrac{13\left(4+\sqrt{3}\right)}{16-3}+\dfrac{6\sqrt{3}}{3}\)

\(=3\left(2-\sqrt{3}\right)+\dfrac{13\left(4+\sqrt{3}\right)}{13}+2\sqrt{3}\)

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

\(=10\)

c. \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=7-5=2\)

d. \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)

\(=\left|2+\sqrt{3}\right|-\sqrt{5^2-2.5.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

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

\(=2+\sqrt{3}-\left(5-\sqrt{3}\right)\) (vì \(\left|2+\sqrt{3}\right|\ge0,\left|5-\sqrt{3}\right|\ge0\))

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

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

Lời giải:
a.

\(=2\sqrt{4^2.5}+3\sqrt{3^2.5}-\sqrt{7^2.5}=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}=10\sqrt{5}\)

b.

\(=\frac{3(2-\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{13(4+\sqrt{3})}{(4-\sqrt{3})(4+\sqrt{3})}+\frac{6\sqrt{3}}{3}\)

\(=\frac{6-3\sqrt{3}}{1}+\frac{13(4+\sqrt{3})}{13}+2\sqrt{3}=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)

\(=10\)

c.

\(=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{\sqrt{2}-1}+\frac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{3}-1}\right].(\sqrt{7}-\sqrt{5})\)

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

d.

\(=|2+\sqrt{3}|-\sqrt{5^2-2.5\sqrt{3}+3}=|2+\sqrt{3}|-\sqrt{(5-\sqrt{3})^2}\)

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

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

2: \(\Leftrightarrow D=\dfrac{4\sqrt{x}+12-x+\sqrt{x}-13}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

\(\Leftrightarrow D=\dfrac{-x+5\sqrt{x}-1}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

\(\Leftrightarrow\dfrac{-x+5\sqrt{x}-1}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}=1\)

\(\Leftrightarrow\left(-x+5\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)=\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\)

\(\Leftrightarrow-2x\sqrt{x}-x+10x+5\sqrt{x}-2\sqrt{x}-1=x\sqrt{x}+3x+x+3\sqrt{x}+\sqrt{x}+3\)

\(\Leftrightarrow-2x\sqrt{x}+9x-3\sqrt{x}-1=x\sqrt{x}+4x+4\sqrt{x}+3\)

\(\Leftrightarrow-3x\sqrt{x}+5x-7\sqrt{x}-4=0\)

Bạn xem lại đề nhé, nghiệm rất xấu

a: \(=2\cdot\sqrt{\dfrac{18-2\sqrt{77}}{4}}-\sqrt{20+6\sqrt{11}}\)

\(=\sqrt{11}-\sqrt{7}-\sqrt{11}-3=-\sqrt{7}-3\)

b: B=\(=\left(\sqrt{13}-1\right)\cdot\sqrt{\dfrac{7+\sqrt{13}}{18}}+\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

Đặt \(C=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(\Leftrightarrow C^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{16-10-2\sqrt{5}}\)

\(=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)

=>\(C=\sqrt{5}+1\)

\(B=\left(\sqrt{13}-1\right)\cdot\sqrt{\dfrac{14+2\sqrt{13}}{36}}+\sqrt{5}+1\)

\(=\dfrac{\left(\sqrt{13}-1\right)\left(\sqrt{13}+1\right)}{6}+\sqrt{5}+1\)

=(13-1)/6+căn5+1

=3+căn5