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${\displaystyle a}$.

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!

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 $ABC$ đều nội tiếp đường tròn tâm $O$, gọi $H$ là trung điểm $BC$. Trên các cạnh $AB,AC$ lần lượt lấy hai điểm $D,E$ sao cho $\widehat{DHE}={60}^{}$ độ . Lấy $M$ bất kì trên cung nhỏ $AB$.

a) Chứng minh ba đường phân giác của ba góc $\widehat{BAC},\widehat{BDE},\widehat{DEC}$ đồng quy.

b) Cho $AB$ có độ dài $1$ đơn vị. Chứng minh: $MA+MB<\frac{4}{\mathrm{/3}}$

a,b&gt;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&#x2265;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&#x2264;x3+y3&#x2264;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

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

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

Cho a,b,c >0 ,thõa mãn : a+b+c =1

CM: $\frac{3}{ab+bc+ac}+\frac{2}{a^2+b^2+c^2}>14$

Áp dụng bunhia nha các bạn

Chứng minh các đẳng thức sau:

a)$326+223&#x2212;432=66" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:20.32px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; text-align:left; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">$32√6+2√23−4√32=√66$\$\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{6}$

b) x6x+2x3+6x):6x=213" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:20.32px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; text-align:left; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">(x√6x+√2x3+√6x):√6x=213$\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2\mathrm{x}}{3}}+\sqrt{6\mathrm{x}}\right):\sqrt{6\mathrm{x}}=2\frac{1}{3}$ với x > 0.