Khoảng cách giữa hai điểm 

A

A 

1

​

 (x 

1

​

 ,y 

1

​

 )

 và 

A_2\left(x_2,y_2\right)

A 

2

​

 (x 

2

​

 ,y 

2

​

 )

 là:

A_1A_2=\sqrt{\left(x_1+x_2\right)^2+\left(y_1+y_2\right)^2}

A 

1

​

 A 

2

​

 = 

(x 

1

​

 +x 

2

​

 ) 

2

 +(y 

1

​

 +y 

2

​

 ) 

2

​

A_1A_2=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}

A 

1

​

 A 

2

​

 = 

(x 

1

​

 −x 

2

​

 ) 

2

 +(y 

1

​

 −y 

2

​

 ) 

2

​

A_1A_2=\sqrt{\left(x_1+x_2\right)^2}+\sqrt{\left(y_1+y_2\right)^2}

A 

1

​

 A 

2

​

 = 

(x 

1

​

 +x 

2

​

 ) 

2

​

 + 

(y 

1

​

 +y 

2

​

 ) 

2

​

A_1A_2=\sqrt{\left(x_1-x_2\right)^2}+\sqrt{\left(y_1-y_2\right)^2}

A 

1

​

 A 

2

​

 = 

(x 

1

​

 −x 

2

​

 ) 

2

​

 + 

(y 

1

​

 −y 

2

​

 ) 

2

​

{x3+x−2=y3+3y2+4yx5+y3+1=0

Chứng minh rằng với mọi số dương a,b,c,d>0" role="presentation" style="border:0px; color:rgb(0, 0, 128); direction:ltr; display:inline-block; float:none; font-family:times new roman; font-size:18.18px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">a,b,c,d>0$a,b,c,d>0$ thì
a−db+d+d−bb+c+b−cc+a+c−aa+d≥0" role="presentation" style="border:0px; color:rgb(0, 0, 128); direction:ltr; display:inline-block; float:none; font-family:times new roman; font-size:18.18px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">a−db+d+d−bb+c+b−cc+a+c−aa+d≥0$\frac{a-d}{b+d}+\frac{d-b}{b+c}+\frac{b-c}{c+a}+\frac{c-a}{a+d}\ge 0$

Cho tam giác AB có AB=c, BC=a, CA=b

CMR: a) (b+c-a) (c+a-b) (a+b-c) $\le$abc (1)

b) $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ge 3\left(2\right)$

c) $\frac{a^2}{b+c-1}+\frac{b^2}{c+a-b}+\frac{c^2}{a+b-c}\ge a+b+c\left(3\right)$

Giải hệ phương trình:\(\hept{ }\)
{x+10+y−1=11y+10+x−1=11" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">√x+10+√y−1=11√y+10+√x−1=11

CMR: Với n lẻ: 
an+bn=(a+b)(an−1−an−2b+an−3b2−...+a2bn−3−abn−2+bn−1)" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">an+bn=(a+b)(an−1−an−2b+an−3b2−...+a2bn−3−abn−2+bn−1)${\displaystyle a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}{b}^2-...+a^2{b}^{n-3}-a{b}^{n-2}+b^{n-1})}$ 

Chứng minh bằng phương pháp quy nạp nhé!

Cho 2 số thực dương x,y thỏa mãn

$x^3+y^3-3xy\left(x^2+y^2\right)+4x^2{y}^2\left(x+y\right)-4x^3{y}^3=0$

Tìm giá trị nhỏ nhất của biểu thức M=x+y

a,b>0" role="presentation" style="border:0px; color:rgb(0, 128, 128); direction:ltr; display:inline-block; float:none; font-family:fixedsys; font-size:18.18px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">a,b>0$a,b>0$ chứng minh :
1a3+a3b3+b3≥1a+ab+b" role="presentation" style="border:0px; color:rgb(0, 128, 128); direction:ltr; display:inline-block; float:none; font-family:fixedsys; font-size:18.18px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">1a3+a3b3+b3≥1a+ab+b$\frac{1}{a^3}+\frac{a^3}{b^3}+b^3\ge \frac{1}{a}+\frac{a}{b}+b$

Hai số dương x2+y2=1" role="presentation" style="border:0px; color:rgb(0, 0, 128); direction:ltr; display:inline-block; float:none; font-family:times new roman; font-size:18.18px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">x2+y2=1$x^2+y^2=1$ có tổng bằng 1" role="presentation" style="border:0px; color:rgb(0, 0, 128); direction:ltr; display:inline-block; float:none; font-family:times new roman; font-size:18.18px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">1$1$.Chứng minh rằng 
12≤x3+y3≤1" role="presentation" style="border:0px; color:rgb(0, 0, 128); direction:ltr; display:inline-block; float:none; font-family:times new roman; font-size:18.18px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">1√2≤x3+y3≤1

1, A= $\frac{\sqrt{x}+4}{\sqrt{x}-1}$         B= $\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\left(x\ge 0,x\ne 1\right)$

Tìm x để $\frac{A}{B}\ge \frac{x}{4}+5$biết B= $\frac{1}{\sqrt{x}-1}$

2, A= $\frac{4\left(\sqrt{x}+1\right)}{25-x}$ B= $\left(\frac{15-5x}{x-25}+\frac{2}{\sqrt{x}+5}\right):\frac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge 0,x\ne 25\right)$

Tìm giá trị nguyên của x để P= A.B đặt giá trị nguyên lớn nhất

GIÚP MK VỚI! THANKS