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

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

J = $\left(1+\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)$ . $\left(1-\frac{2-\sqrt{2}}{1-\sqrt{2}}\right)$

M = $\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\frac{1}{\sqrt{6}}$

N = $\frac{6}{1+\sqrt{7}}+\frac{1}{\sqrt{7}}$

O = $\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{2-\sqrt{3}}$

Q = $\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)$

S = $\frac{2}{\sqrt{5}+1}-\sqrt{\frac{2}{3-\sqrt{5}}}$

Các thầy cô, các bạn giúp em với ạ. Em cảm ơn !

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

tìm giá trị nhỏ nhất của biểu thức

(x−1)2z+(y−1)2x+(z−1)2y" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:20.34px; font-style:normal; font-weight:normal; letter-spacing:normal; 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; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">A =(x−1)2z+(y−1)2x+(z−1)2y

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

Cho các số thực dương a,b,c thỏa mãn a+b+c = 1

CMR: a2a+b+b2b+c+c2c+a≥12" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:20.34px; font-style:normal; font-weight:normal; letter-spacing:normal; 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; text-align:left; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">a2a+b+b2b+c+c2c+a≥12

x−2x−3+x+1x+3+x−4x−99−x" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-table; float:none; font-size:20.34px; font-style:normal; font-weight:normal; letter-spacing:normal; 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; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">P=√x−2√x−3+√x+1√x+3+x−4√x−99−x$P={\displaystyle \frac{\sqrt{x}-2}{\sqrt{x}-3}}+{\displaystyle \frac{\sqrt{x}+1}{\sqrt{x}+3}}+{\displaystyle \frac{x-4\sqrt{x}-9}{9-x}}$và x+53−x" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:20.34px; font-style:normal; font-weight:normal; letter-spacing:normal; 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; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">Q=√x+53−√x$Q={\displaystyle \frac{\sqrt{x}+5}{3-\sqrt{x}}}$(với$x\ge 0;x\ne 9$)

1, Rút gọn P

2, Tìm x để P=3

3, Tìm $M=P:Q$. Tìm x để 12" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size:20.34px; font-style:normal; font-weight:normal; letter-spacing:normal; 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; text-indent:0px; text-transform:none; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">|M|<12