\(a_1,=\left(x^3+x^2-2x^2-2x+3x+3\right):\left(x+1\right)\\ =\left(x+1\right)\left(x^2-2x+3\right):\left(x+1\right)\\ =x^2-2x+3\\ a_2,=\left(2x^2+3y^2\right)^2:\left(2x^2+3y^2\right)=2x^2+3y^2\\ b_1,=\left(x^3-7x^2+x^2-7x-2x+14\right):\left(x-7\right)\\ =\left(x-7\right)\left(x^2+x-2\right):\left(x-7\right)\\ =x^2+x-2\\ b_2,=\left(8ab-7m^2n\right)\left(8ab+7m^2n\right):\left(8ab+7m^2n\right)=8ab-7m^2n\\ c,=\left(3x-2y^2\right)\left(9x^2+6xy^2+4y^4\right):\left(3x-2y^2\right)\\ =9x^2+6xy^2+4y^4\\ d,=\left(3x+2y^2\right)\left(9x^2-6xy^2+4y^4\right):\left(9x^2-6xy^2+4y^4\right)\\ =3x+2y^2\)

Bài 2: 

\(A=2+2^2+2^3+...+2^{60}\)

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

\(A=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+...+2^{59}\cdot\left(1+2\right)\)

\(A=2\cdot3+2^3\cdot3+...+2^{59}\cdot3\)

\(A=3\cdot\left(2+2^3+...+2^{59}\right)\) ⋮ 3

Vậy: A ⋮ 3

_____________

\(A=2+2^2+...+2^{60}\)

\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(A=2\cdot\left(1+2+4\right)+2^4\cdot\left(1+2+4\right)+...+2^{58}\cdot\left(1+2+4\right)\)

\(A=2\cdot7+2^4\cdot7+...+2^{58}\cdot7\)

\(A=7\cdot\left(2+2^4+....+2^{58}\right)\) ⋮ 7

Vậy: A ⋮ 7

___________________

\(A=2+2^2+...+2^{60}\)

\(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)

\(A=2\cdot\left(1+4\right)+2^2\cdot\left(1+4\right)+...+2^{58}\cdot\left(1+4\right)\)

\(A=2\cdot5+2^2\cdot5+...+2^{58}\cdot5\)

\(A=5\cdot\left(2+2^2+...+2^{58}\right)\) ⋮ 5

Vậy: A ⋮ 5 

B

YRGFGYSTHRBHFYSVGSYG

GYE4F4GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

Dễ vậy mà ko làm đc àk

\(a_1.a_2=b_1.b_2\Rightarrow\frac{a_1}{b_1}=\frac{b_2}{a_2}\)

\(\Rightarrow\frac{a_1}{b_1}+\frac{a_2}{b_2}=\frac{b_2}{a_2}+\frac{a_2}{b_2}\ge2\sqrt{\frac{b_2}{a_2}.\frac{a_2}{b_2}}=2\) (AM - GM)

có a1.a2=b1.b2

=> a1/b1=b2/a2

có \(\frac{a1}{b1}+\frac{a2}{b2}=\frac{b2}{a2}+\frac{a2}{b2}\)

áp dụng bất đẳng thức cosi cho 2 số dương có

\(\frac{b2}{a2}+\frac{a2}{b2}\ge2\sqrt{\frac{b2}{a2}.\left(\frac{a2}{b2}\right)}=2\)(đpcm)

Chọn A

Bài 1:

a: \(\left|x-\dfrac{1}{2}\right|+\dfrac{1}{2}=x\)

=>\(\left|x-\dfrac{1}{2}\right|=x-\dfrac{1}{2}\)

=>\(x-\dfrac{1}{2}>=0\)

=>\(x>=\dfrac{1}{2}\)

b: \(\left|1-3x\right|+1=3x\)

=>\(\left|1-3x\right|=3x-1\)

=>\(1-3x< =0\)

=>3x-1>=0

=>3x>=1

=>\(x>=\dfrac{1}{3}\)

Bài 2:

a: \(C=\left|5-x\right|+x=\left|x-5\right|+x\)

TH1: x>=5

\(C=x-5+x=2x-5\)

TH2: x<5

C=5-x+x=5

b: D=|2x-1|-x

TH1: x>=1/2

\(D=2x-1-x=x-1\)

TH2: \(x< \dfrac{1}{2}\)

D=1-2x-x=1-3x

a) \(\approx\) 36,6

b) 130

c) 100

 a1/a2 = b1/b2 = c1/c2 = k

a1=k.a2, b1=k.b2, c1=k.c2

Biểu thức trở thành

√(k.a2 + k.b2 + k.c2).(a2 + b2 + c2)= √k.a2.a2 + √k.b2.b2 + √k.c2.c2

√k.(a2+b2+c2)² = a2. √k + b2. √k + c2. √k

(a2+b2+c2). √k = (a2+b2+c2). √k (hiển nhiên đúng)

Suy ra điều phải chứng minh