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$\frac{1}{a^3}+\frac{a^3}{b^3}+b^3\ge \frac{1}{a}+\frac{a}{b}+b$

Chọn đáp án đúng để nêu định nghĩa của các tỉ số lượng giác? (Viết tắt : cạnh kề (k), cạnh đối(đ), cạnh huyền (h))

1. A đh,tan=đk,cot=kđ" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:25.99px; 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">sin=kh,cos=đh,tan=đk,cot=kđ$sin=\frac{k}{h},cos=\frac{đ}{h},tan=\frac{đ}{k},cot=\frac{k}{đ}$
2. B đh,cos=kh,tan=kđ,cot=đk" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:25.99px; 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">sin=đh,cos=kh,tan=kđ,cot=đk$sin=\frac{đ}{h},cos=\frac{k}{h},tan=\frac{k}{đ},cot=\frac{đ}{k}$
3. C đh,cos=kh,tan=đk,cot=kđ" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:25.99px; 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">sin=đh,cos=kh,tan=đk,cot=kđ$sin=\frac{đ}{h},cos=\frac{k}{h},tan=\frac{đ}{k},cot=\frac{k}{đ}$
4. D đk,cos=kđ,tan=đh,cot=kh" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:25.99px; 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">sin=đk,cos=kđ,tan=đh,cot=kh

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 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

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 tam giác ABC có AB=c, BC=a, AC=b và $P=\frac{a+b+c}{2}$

CMR: a) (P-a)(P-b)(P-c) $\le$$\frac{1}{8}abc$

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

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