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$\frac{a-d}{b+d}+\frac{d-b}{b+c}+\frac{b-c}{c+a}+\frac{c-a}{a+d}\ge 0$

Bài toán: Cho ba số x,y,z" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; 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-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">x,y,z${\displaystyle x,y,z}$ thỏa mãn x+y+z=0" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; 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-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">x+y+z=0${\displaystyle x+y+z=0}$ và x2+y2+z2=a2" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; 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-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">x2+y2+z2=a2${\displaystyle x^2+y^2+z^2=a^2}$. Tính x4+y4+z4" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; 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-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">x4+y4+z4${\displaystyle x^4+y^4+z^4}$ theo a" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; 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-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">a${\displaystyle a}$.

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

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

Cho tam giác đều ABC$ABC$, O$O$ là trung điểm của BC$BC$. Trên các cạnh AB,AC$AB,AC$ lần lượt lấy các điểm di động D$D$ và E$E$ sao cho góc ˆDOE=600$\widehat{DOE}={60}^0$.

a) Chứng minh tích BD.CE$BD.CE$ không đổi.

b) Chứng minh ΔBOD$\Delta BOD$ đồng dạng ΔOED$\Delta OED$. Từ đó suy ra tia DO$DO$ là tia phân giác của góc BDE$BDE$.

c) Vẽ đường tròn tâm O$O$ tiếp xúc với AB$AB$. Chứng minh rằng đường tròn này luôn tiếp xúc với DE$DE$.

x−2+10−x=x2−12x+40" role="presentation" style="background-color:rgb(247, 247, 247); border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:14.04px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap" class="MathJax_CHTML mjx-chtml">√x−2+√10−x=x2−12x+40${\displaystyle \sqrt{x-2}+\sqrt{10-x}=x^2-12x+40}$

giúp  mình với!!

Cho a và b là các số thực dương. Chọn r là nghiệm có giá trị tuyệt đối nhỏ nhất của phương trình : x3&#x2212;ax+b=0" 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; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap" class="MathJax_CHTML mjx-chtml">x3−ax+b=0${\displaystyle x^3-ax+b=0}$ . CMR : ba&lt;r&#x2264;3b2a" 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; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap" class="MathJax_CHTML mjx-chtml">ba<r≤3b2a