Tìm x:

$\sqrt{4x-4}$+$\sqrt{25x-25}$+$\sqrt{81x-81}$=$-$1

Rút gọn:

A=(x -2).9(x&#x2212;2)2" 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">√9(x−2)2$\sqrt{{\displaystyle \frac{9}{{\left(x-2\right)}^2}}}$+3 với x<0

B=a6" 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">√a6$\sqrt{{\displaystyle \frac{a}{6}}}$+2a3" 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">√2a3$\sqrt{{\displaystyle \frac{2a}{3}}}$+3a2" 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">√3a2$\sqrt{{\displaystyle \frac{3a}{2}}}$(với a≥0)

E=$\sqrt{9a^2}$+$\sqrt{4a^2}$+$\sqrt{{\left(1-a\right)}^2}$+$\sqrt{16a^2}$. Với a≤0

F=$\left|x-2\right|$+x2x" 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">√x2x${\displaystyle \frac{\sqrt{x^2}}{x}}$ với x<0

H=x2+23.x+3x2&#x2212;3" 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">x2+2√3.x+3x2−3${\displaystyle \frac{x^2+2\sqrt{3}.x+3}{x^2-3}}$

I=$\left|x-\sqrt{{\left(x-1\right)}^2}\right|$$-$2x (x<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 biểu thức P=(x&#x2212;y1+xy" 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−√y1+√xy${\displaystyle \frac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}}$+x+y1&#x2212;xy" 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+√y1−√xy${\displaystyle \frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}}$):(x+y+2xy1&#x2212;xy" 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+y+2xy1−xy${\displaystyle \frac{x+y+2xy}{1-xy}}$+1)

a) Rút gọn P

b) Tính giá trị của P tại x=22+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">22+√3${\displaystyle \frac{2}{2+\sqrt{3}}}$

c) Chứng minh: P≤1

1.Cho &#x0394;ABC" 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">ΔABC$\mathrm{\Delta }ABC$ , AB = 5cm, AC = 9cm, B&#x005E;" 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">ˆB$\hat{B}$ = 45⁰. Tính các góc và các cạnh còn lại ( bằng 2 cách )

2. Cho &#x0394;ABC" 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">ΔABC$\mathrm{\Delta }ABC$ có AB = 3cm, BC = 4cm, CA = 5cm. Tính góc A, B, C ? ( bằng 2 cách )

3 Cho &#x0394;ABC" 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">ΔABC$\mathrm{\Delta }ABC$ có A&#x005E;" 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">ˆA$\hat{A}$ = 45^o , cạnh AC = 3cm, cạnh AB = 5cm. Tính cạnh BC, góc B và góc C ? ( bằng hai cách )

Chứng minh P ≥ 0 biết P = a - 2 $\sqrt{a-1}$ ( a ≥1)

Cho x>14" 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">14${\displaystyle \frac{1}{4}}$ Tìm giá trị nhỏ nhất của iểu thức 2x&#x2212;x+82x&#x2212;1" 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=2x−√x+82√x−1

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 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$DE$.