BT=\(\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{12\left(3-\sqrt{3}\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)

\(=\dfrac{2\left(\sqrt{3}-1\right)}{2}+\dfrac{2+\sqrt{3}}{4-3}+\dfrac{12\left(3-\sqrt{3}\right)}{9-3}\)

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

\(=\sqrt{3}-1+2+\sqrt{3}+6-2\sqrt{3}=7\)

\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)

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

\(=3\sqrt{2}\)

\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)

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

\(=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}\)

\(=\dfrac{3}{2}\)

lỗi

?

\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)

\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)

\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)

\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)

péo cày gòi đý è

a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)

c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)

d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)

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

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:
\(\frac{3-2\sqrt{3}}{\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3}-2-\sqrt{(\sqrt{3}-1)^2}\)

\(=\sqrt{3}-2-|\sqrt{3}-1|=(\sqrt{3}-2)-(\sqrt{3}-1)=-1\)

a: ĐKXĐ: x>=2

Ta có: \(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\frac{x-2}{4}}=26\)

=>\(4\sqrt{x-2}+3\sqrt{x-2}-\frac12\cdot\sqrt{x-2}=26\)

=>\(6,5\cdot\sqrt{x-2}=26\)

=>\(\sqrt{x-2}=4\)

=>x-2=16

=>x=18(nhận)

b: ĐKXĐ: x∈R

\(3x+\sqrt{4x^2-8x+4}=1\)

=>\(3x+\sqrt{\left(2x-2\right)^2}=1\)

=>3x+|2x-2|=1

=>3x-1+|2x-2|=0(1)

TH1: x>=1

(1) sẽ trở thành: 3x-1+2x-2=0

=>5x-3=0

=>5x=3

=>x=3/5(loại)

TH2: x<1

(1) sẽ trở thành: 3x-1-2x+2=0

=>x+1=0

=>x=-1(nhận)

c: ĐKXĐ: x>=0

Ta có: \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\)

=>\(2x-4\sqrt{x}+\sqrt{x}-2-7=0\)

=>\(2x-3\sqrt{x}-9=0\)

=>\(2x-6\sqrt{x}+3\sqrt{x}-9=0\)

=>\(\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\)

=>\(\sqrt{x}-3=0\)

=>\(\sqrt{x}=3\)

=>x=9(nhận)