\(B=\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{2}{\sqrt{3}}-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)

\(\Rightarrow B=\dfrac{\sqrt{1}}{\sqrt{3}}+\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\dfrac{2}{\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\)

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

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

\(\Rightarrow B=\sqrt{3}+\dfrac{4-2\sqrt{3}}{\sqrt{3^2}-1^2}-\sqrt{3}\)

\(\Rightarrow B=\dfrac{4-2\sqrt{3}}{2}\)

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

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

⇔ \(B=\dfrac{\text{3}}{\sqrt{3}}+\dfrac{3-2\sqrt{3}+1}{3-1}-\sqrt{3}\)

⇔ \(B=\sqrt{3}+\dfrac{\text{4}-2\sqrt{3}}{\text{2}}-\sqrt{3}\)

⇔ \(B=\dfrac{2\left(2-\sqrt{3}\right)}{2}=2-\sqrt{3}\)

a: ĐKXĐ: \(x\le0\)

b: ĐKXĐ: \(x\le2\)

\(a,ĐK:-3x\ge0\Leftrightarrow x\le0\left(-3< 0\right)\\ b,ĐK:4-2x\ge0\Leftrightarrow-2x\ge-4\Leftrightarrow x\le2\\ c,ĐK:\dfrac{1}{2x-5}\ge0\Leftrightarrow2x-5>0\left(1>0;2x-5\ne0\right)\\ \Leftrightarrow x>\dfrac{5}{2}\\ d,ĐK:\dfrac{4x+7}{-3}\ge0\Leftrightarrow4x+7\le0\left(-3< 0\right)\Leftrightarrow x\le-\dfrac{7}{4}\)

a: Ta có: \(A=x^2-20x+101\)

\(=x^2-20x+100+1\)

\(=\left(x-10\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x=10

1: \(\begin{cases}x-2y+z=2\\ x+2y-3z=-4\\ x-3y+2z=-11\end{cases}\)

=>\(\begin{cases}x-2y+z-x-2y+3z=2+4=6\\ x-2y+z-x+3y-2z=2-\left(-11\right)=2+11=13\\ x-2y+z=2\end{cases}\)

=>\(\begin{cases}-4y+4z=6\\ y-z=13\\ x-2y+z=2\end{cases}\Rightarrow\begin{cases}y-z=-1,5\\ y-z=13\\ x-2y+z=2\end{cases}\)

=>(x;y;z)∈∅

a: Xét ΔADE có \(\frac{AB}{BD}=\frac{AC}{CE}\)

nên BC//DE
b: Ta có: \(\hat{DBM}=\hat{ABC}\) (hai góc đối đỉnh)

\(\hat{ECN}=\hat{ACB}\) (hai góc đối đỉnh)

mà \(\hat{ABC}=\hat{ACB}\) (ΔABC cân tại A)

nên \(\hat{DBM}=\hat{ECN}\)

Xét ΔDBM vuông tại M và ΔECN vuông tại N có

DB=EC

\(\hat{DBM}=\hat{ECN}\)

Do đó: ΔDBM=ΔECN

=>DM=EN

c: ΔDBM=ΔECN

=>\(\hat{BDN}=\hat{CEN}\)

AD=AB+BD

AE=AC+CE
mà AB=AC và BD=CE

nên AD=AE

Xét ΔADM và ΔAEN có

AD=AE

\(\hat{ADM}=\hat{AEN}\)

DM=EN

Do đó: ΔADM=ΔAEN

=>AM=AN

=>ΔAMN cân tại A