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

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

. Cho P=xxy+x+3+yyz+y+1+3zxz+3z+3" 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; 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">√x√xy+√x+3+√y√yz+√y+1+3√z√xz+3√z+3${\displaystyle \frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+3}}+{\displaystyle \frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}}+{\displaystyle \frac{3\sqrt{z}}{\sqrt{xz}+3\sqrt{z}+3}}$ và xyz =9. Tính $\sqrt{10P-1}$

Cho P=xxy+x+3+yyz+y+1+3zxz+3z+3" 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; 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">√x√xy+√x+3+√y√yz+√y+1+3√z√xz+3√z+3${\displaystyle \frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+3}}+{\displaystyle \frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}}+{\displaystyle \frac{3\sqrt{z}}{\sqrt{xz}+3\sqrt{z}+3}}$ và xyz =9. Tính $\sqrt{10P-1}$

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

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 !

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