Bài 13:

\(1)A=x^2-x+1\\ =\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\\ 2)B=x^2+x+1\\ =\left(x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ =\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\\ 3)C=x^2+2x+2\\ =\left(x^2+2x+1\right)+1\\ =\left(x+1\right)^2+1\ge1>0\forall x\)

\(4)A=x^2-5x+10\\ =\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}\right)+\dfrac{15}{4}\\ =\left(x-\dfrac{5}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}>0\forall x\\ 5)B=x^2-8x+20\\ =\left(x^2-8x+16\right)+4\\ =\left(x-4\right)^2+4\ge4>0\forall x\\ 6)C=x^2-8x+17\\ =\left(x^2-8x+16\right)+1\\ =\left(x-4\right)^2+1\ge1>0\forall x\) 

\(7)A=x^2-6x+10\\ =\left(x^2-6x+9\right)+1\\ =\left(x-3\right)^2+1\ge1>0\forall x\\ 8)B=9x^2-6x+2\\ =\left(9x^2-6x+1\right)+1\\ =\left(3x-1\right)^2+1\ge1>0\forall x\\ 9)C=2x^2+8x+15\\ =\left(2x^2+8x+8\right)+7\\ =2\left(x^2+4x+4\right)+7\\ =2\left(x+2\right)^2+7\ge7>0\forall x\)

Gọi số đó có dạng \(\overline{ab}\)

Khi thêm số 0 vào giữa thì ta có số mới là: \(\overline{a0b}=100a+b\)

Mà số mới gấp 7 lần số cũ nên ta có: 

\(\overline{a0b}=7\overline{ab}\\ 100a+b=7\left(10a+b\right)\\ 100a+b=70a+7b\\ 100a-70a=7b-b\\ a\left(100-70\right)=b\left(7-1\right)\\ 30a=6b\\ \dfrac{a}{b}=\dfrac{6}{30}=\dfrac{1}{5}\)

`=> a=1;b=5`

Vậy sso cần tìm là 15

cần mua vip k vậy

\(a)\left(x+5\right)^2=x^2+2\cdot x\cdot5+5^2=x^2+10x+25\\ b)\left(2-y\right)^2=2^2-2\cdot2\cdot y+y^2=4-4y+y^2\\ c)\left(5x-1\right)^2=\left(5x\right)^2-2\cdot5x\cdot1+1^2=25x^2-10x+1\\ d)\left(1+5x^3\right)^2=1^2+2\cdot1\cdot5x^3+\left(5x^3\right)^2=1+10x^3+25x^6\\ e)\left(7-a^2\right)\left(7+a^2\right)=7^2-\left(a^2\right)^2=49-a^4\\ \left(x-y\right)^2-\left(x+y\right)^2=\left(x-y+x+y\right)\left(x-y-x-y\right)=2x\cdot-2y=-4xy\\ g)\left(2x^3-\dfrac{1}{2}y\right)^2=\left(2x^3\right)^2-2\cdot2x^3\cdot\dfrac{1}{2}y+\left(\dfrac{1}{2}y^2\right)=4x^6-2x^3y+\dfrac{1}{4}y^2\\ h)\left(x^2+4y\right)^2=\left(x^2\right)^2+2\cdot x^2\cdot4y+\left(4y\right)^2=x^4+8x^2y+16y^2\\ i)\left(a+b+c\right)^2=\left[a+\left(b+c\right)\right]^2=a^2+2a\left(b+c\right)+\left(b+c\right)^2\\ =a^2+2ab+2ac+b^2+2bc+c^2=a^2+b^2+c^2+2ab+2bc+2ac\\ k)\left(a-b-c\right)^2=\left[a-\left(b+c\right)\right]^2=a^2-2a\left(b+c\right)+\left(b+c\right)^2\\ =a^2-2ab-2ac+b^2+2bc+c^2=a^2+b^2+c^2-2ab-2ac+2bc\)

Bài 9:

a; 23,4 : 0,5 và 23 x 2

   23,4 : 0,5 = 23,4 x 2 = 46,8

   23 x 2 = 46

   23,4 : 0,5 > 23 x 24

b; 18: 0,25 và 18 x 4

    18 : 0,25 = 18 x 4 = 73

    18 x 4 = 73

Vậy 18: 0,25  = 18 x 4

Bài 10:

a; 23,4 + 164 : 4,5

=  \(\dfrac{234}{10}\) + 164 : \(\dfrac{45}{10}\)

= \(\dfrac{234}{10}\) + 164 x \(\dfrac{10}{45}\)

= \(\dfrac{117}{5}\) + \(\dfrac{328}{9}\)

= \(\dfrac{1053}{45}\) + \(\dfrac{1640}{45}\)

= \(\dfrac{2693}{45}\) 

Bài 8:

\(34,8dm=348cm\\ 10,9m=1090cm\\ 0,189m=18,9\left(cm\right)\\ 6,27m=627cm\)

Bài 9:

100 can nhựa như thế đựng được số lít dầu là: 

\(100\times6,3=630\left(l\right)\) 

ĐS:...

còn bài 10 ở đâu rồi bn, mik hỏi 3 bài mà, bn trả lời mới 2 à!

\(0,5+3\dfrac{1}{2}-4\\ =0,5+\left(3+\dfrac{1}{2}\right)-4\\ =\dfrac{1}{2}+3+\dfrac{1}{2}-4\\ =\left(\dfrac{1}{2}+\dfrac{1}{2}+3\right)-4\\ =\left(1+3\right)-4\\ =4-4\\ =0\)

\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=0\\ \dfrac{1}{2}\times\left(x+1\right)+\dfrac{1}{3}\times\left(x+1\right)+\dfrac{1}{4}\times\left(x+1\right)=0\\ \left(x+1\right)\times\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)=0\\ x+1=0:\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)\\ x+1=0\\ x=-1\) 

câu b đề bài cho B < \(\dfrac{1}{3}\) thì cần gì chứng minh nữa em.

Đáng lẽ phải là: Cho B = \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) +...-\(\dfrac{1}{64}\)

Chứng minh B < \(\dfrac{1}{3}\)