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

a)
=(x-2)³
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
 

d)
=\(\left(x^2\right)^3-3.\left(x^2\right)^2.2+3.x^2.2^2-2^3\)
\(=x^5-6x^4+12x^2-8\)
e)
\(=\left(2x\right)^3-3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2-\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)

a)
\(=x^3+3.x^2.1+3.x.1^2+1^3\)
\(=x^3+3x^2+3x+1\)
b)
\(=\left(2x\right)^3+3.\left(2x\right)^2.3+3.2x.3^2+3^3\)
\(=8x^3+36x^2+54x+27\)
c)
\(x^3+3.x^2.\dfrac{1}{2}+3.x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)
\(=x^3+1,5x^2+0,75x+0,125\)

a)
\(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{3}+\sqrt{2}-\sqrt{2}+1=\sqrt{3}+1\)
b)
\(\sqrt{\left(\sqrt{9}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{16}+\sqrt{2}\right)^2}=\sqrt{9}+\sqrt{2}-\sqrt{16}-\sqrt{2}=3-4=-1\)
c)
\(=\sqrt{2\left(2-\sqrt{3}\right)}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)

1)
\(1001^2=\left(1000+1\right)^2=1000^2+2000+1^2=1000000+2000+1=1002001\)
2)
\(=\left(30-0,1\right)\left(30+0,1\right)=30^2-0,1^2=900-0,01=899.99\)
3)
\(201^2=\left(200+1\right)^2=200^2+400+1^2=40000+400+1=40401\)
4)
\(37.43=\left(40-3\right)\left(40+3\right)=40^2-3^2=1600-9=1591\)
5)
\(199^2=\left(200-1\right)^2=200^2-400+1=40000-400+1=39601\)

có AB=CD
AD=BC
=> ABCD là hình thoi

a)
\(=\left(4x-1\right)^2-3\left(4x-1\right)=\left(4x-1\right)\left(4x-1-3\right)=\left(4x-1\right)\left(4x-4\right)=4\left(x-1\right)\left(4x-1\right)\)
b)
\(=\left(3x+2\right)\left(9x^2-6x+4\right)\)
c)
\(=-x^4y^4\left(16y^2+24xy+9x^2\right)=-x^4y^4\left(4y+3x\right)^2\)
d)
\(=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\)
\(=\left[a\left(x-y\right)+b\left(y-x\right)\right]\left[a\left(x+y\right)+b\left(x+y\right)\right]\)
\(=\left[a\left(x-y\right)-b\left(x-y\right)\right]\left(a+b\right)\left(x+y\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(x-y\right)\left(x+y\right)=\left(a^2-b^2\right)\left(x^2-y^2\right)\)

có góc ABb = 180⁰-góc B1
    góc ABb = 180⁰-130⁰=50⁰
mà a//b
=>A1=góc ABb = 50^o(so le trong)
câu 5
đường thẳng c có vuông góc với đường thẳng a vì a//b mà b vuông góc với c
góc bBA = 180⁰-góc B
góc bBA=180⁰-125⁰=55⁰
mà a//b
=> góc BAD=góc bBA= 55 độ

C=a(b-c)-2(b-c)
C=(a-2)(b-c)
thay a=2  ;  b=1,007  ;  c=-0,006 vào C ta có
C=(2-2)(1,007-0,006)
C=0.1,001
C=0