a) \(B=3+3^2+3^3+..+3^{60}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)

\(=4\left(3+3^3+...+3^{59}\right)⋮4\)

=>đpcm

b) \(B=3+3^2+3^3+..+3^{60}\)

\(=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)

\(=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)=13\left(3+..+3^{58}\right)⋮13\)

=>đpcm

có :

     [2n-0] : 2 + 1= 2n +2 +1 = n+ 1 [phần tử]

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

\(\Leftrightarrow\left|2x-5\right|=3\)                 (1)

+)TH1: \(2x-5\ge0\Leftrightarrow x\ge\frac{5}{2}\) thì:

(1)<=> \(2x-5=3\Leftrightarrow2x=8\Leftrightarrow x=4\left(tm\right)\)

+) TH2: \(2x-5< 0\Leftrightarrow x< \frac{5}{2}\) thì

(1)<=> \(5-2x=3\Leftrightarrow-2x=-2\Leftrightarrow x=1\left(tm\right)\)

Vậy x={1;4}

1. \(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)

2. \(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\\ =3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)

3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\\ =2\sqrt{6}-3+3+\sqrt{6}=3\sqrt{6}\)

\(x^3+x^2+4=0\)

\(\Leftrightarrow\left(x^3-4x\right)+\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow x\left(x^2-4\right)+\left(x+2\right)^2=0\)

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

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

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

\(\Leftrightarrow x+2=0\) (Vì: \(x^2-x+2>0\) )

\(\Leftrightarrow x=-2\)

a) \(\frac{2}{17}+\frac{7}{21}+\frac{15}{17}-\frac{7}{5}+\frac{2}{3}=\left(\frac{2}{17}+\frac{15}{17}\right)+\left(\frac{1}{3}+\frac{2}{3}\right)-\frac{7}{5}=1+1-\frac{7}{5}=\frac{3}{5}\)

b) \(\frac{3}{4}\cdot\frac{11}{5}-\frac{3}{4}\cdot\frac{31}{5}+\sqrt{81}=\frac{3}{4}\left(\frac{11}{5}-\frac{31}{5}\right)+9=\frac{3}{4}\cdot\left(-4\right)+9=-3+9=6\)

c) \(\left(-2\right)^3\cdot\left(\frac{3}{4}-0,25\right):2\frac{1}{4}=-8\cdot\frac{1}{2}:\frac{9}{4}=-8\cdot\frac{2}{9}=-\frac{16}{9}\)

Xét ΔAED và ΔACB có:

        AE=AC(gt)

  \(\widehat{EAD}=\widehat{CAB}\left(dd\right)\)

      AD=AB(gt)

=>ΔAED=ΔACB(c.g.c)

=>\(\widehat{ADE}=\widehat{ABC}\). Mà hai góc này ở vị trí sole trong)

=>BC//DE

b)Xét ΔAMD và ΔANB có:

  \(\widehat{ADM}=\widehat{ABN}\left(cmt\right)\)

     AD=AB(gt)

 \(\widehat{MAD}=\widehat{NAB}\left(dd\right)\)

=>ΔAMD=ΔANB(g.c.g)

=>AM=AN

viết luôn bài ra ik

Làm theo ABCD là ht cân

a)  Xét ΔADN và ΔBCN có:

  AD=BC(gt)

^D=^C(gt)

DN=CN(gt)

=> ΔADN =ΔBCN(c.g.c)

=> NA=NB

=>ΔABN cân tại N

b) ΔABN cân tại N(cmt)

Có: NM là đường trung gtuyeens uungs vs cạnh AB

=>NM cx là đg trung trực của AB