Chứng minh rằng với mọi số nguyên dương k" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; 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-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">k${\displaystyle k}$ ; ta có đẳng thức :

$$\frac{1}{{\sin }^2\frac{\pi }{4k+2}}+\frac{1}{{\sin }^2\frac{3\pi }{4k+2}}+\frac{1}{{\sin }^2\frac{5\pi }{4k+2}}+\cdots +\frac{1}{{\sin }^2\frac{(2k-1)\pi }{4k+2}}=2k(k+1$$ 

Cho a,b∈R;n∈N∗" 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">a,b∈R;n∈N∗${\displaystyle a,b\in R;n\in N^{\ast }}$. Chứng minh rằng:

Cho hình 56, trong đó //HK,AH//BK." role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.72px; 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; word-spacing:normal" class="MathJax_CHTML mjx-chtml">AB//HK,AH//BK.$AB//HK,AH//BK.$ Chứng minh rằng $AB=HK,AH=BK.$

Hàm số (x)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.72px; 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; word-spacing:normal" class="MathJax_CHTML mjx-chtml">y=f(x)$y=f(x)$ được cho bởi công thức x2−5" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.72px; 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; word-spacing:normal" class="MathJax_CHTML mjx-chtml">f(x)=2x2−5$f\left(x\right)=2x^2-5$.

Hãy tính: (1);f(−2);f(0);f(2)." role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:18.72px; 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; word-spacing:normal" class="MathJax_CHTML mjx-chtml">f(1);f(−2);f(0);f(2).

Cho ΔABC" id="MathJax-Element-1-Frame" role="presentation" style="box-sizing:border-box; display:none; font-size:19.98px; min-width:3.102em; text-align:center; width:10000em" class="MJXc-processed MathJax_CHTML mjx-full-width">ΔABCcân tại A. Kẻ AH là tia phân giác của BAC

a) CM: AH là đường cao của ΔABC" id="MathJax-Element-2-Frame" role="presentation" style="box-sizing:border-box; display:none; font-size:19.98px; min-width:3.102em; text-align:center; width:10000em" class="MJXc-processed MathJax_CHTML mjx-full-width">ΔABCΔABC, HB=HC

b)TỪ H kẻ ΔAC" id="MathJax-Element-3-Frame" role="presentation" style="box-sizing:border-box; display:none; font-size:19.98px; min-width:8.998em; text-align:center; width:10000em" class="MJXc-processed MathJax_CHTML mjx-full-width">HD⊥AB;HEΔACHD⊥AB;HEΔAC. CM: HD=HE

c) Trên tia đối của BC lấy điểm M, trên tia đối của CB lấy điểm N sao cho BM=CN. CM: AM=AN

d) CM: DE//BC

Cho tam giác $ABC$ có B^=C^" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-family:arial,tahoma,helvetica,sans-serif; font-size:18.72px; 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">ˆB=ˆC$\hat{B}=\hat{C}$. Tia phân giác góc $A$ cắt $BC$ tại $D.$ Chứng minh rằng $DB=DC,AB=AC.$

Tìm x,y,z \(\in\) N*

x7+y11+z13=0,(946053)${\displaystyle \frac{x}{7}+\frac{y}{11}+\frac{z}{13}=0,(946053)}$

Cho tam giác $ABC.$ Các tia phân giác của các góc $B$ và $C$ cắt nhau ở $O.$ Kẻ OD⊥AC" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-family:arial,tahoma,helvetica,sans-serif; font-size:18.72px; 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">OD⊥AC$\mathrm{O}\mathrm{D}\mathrm{\perp }AC$, kẻ OE⊥AB" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-family:arial,tahoma,helvetica,sans-serif; font-size:18.72px; 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">OE⊥AB$\mathrm{O}E\mathrm{\perp }AB$. Chứng minh rằng $OD=OE.$

Bài 1. Cho tam giác ABC, điểm D thuộc cạnh BC. Gọi M là trung điểm của AD. Trên tia đối của tia MB lấy điểm E sao cho ME = MB. Trên tia đối của tia MC lấy F sao cho MF = MC. Chứng minh:

a) AE = BD;

b) AF // BC.

c) Ba điểm A, E, F thẳng hàng.

Bài 2. Cho tam giác ABC cân tại A, M là trung điểm của BC.

a) Chứng minh AFE&#x005E;=ABC&#x005E;&#x21D2;EF//BC" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.08px; 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; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">ˆAFE=ˆABC⇒EF//BCAFE&#x005E;=ABC&#x005E;&#x21D2;EF//BC" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.08px; 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; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">$\stackrel{<u><em>^</em></u>}{\mathrm{<u><em>A</em></u>}\mathrm{<u><em>F</em></u>}\mathrm{<u><em>E</em></u>}}<u><em>=</em></u>\stackrel{<u><em>^</em></u>}{\mathrm{<u><em>A</em></u>}\mathrm{<u><em>B</em></u>}\mathrm{<u><em>C</em></u>}}<u><em>\Rightarrow </em></u>\mathrm{<u><em>E</em></u>}\mathrm{<u><em>F</em></u>}<u><em>/</em></u><u><em>/</em></u>\mathrm{<u><em>B</em></u>}\mathrm{<u><em>C</em></u>}$ &#x0394;ABM=&#x0394;ACM" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.08px; 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; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">ΔABM=ΔACM&#x0394;ABM=&#x0394;ACM" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.08px; 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; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">$\mathrm{<u><em>\Delta </em></u>}\mathrm{<u><em>A</em></u>}\mathrm{<u><em>B</em></u>}\mathrm{<u><em>M</em></u>}<u><em>=</em></u>\mathrm{<u><em>\Delta </em></u>}\mathrm{<u><em>A</em></u>}\mathrm{<u><em>C</em></u>}\mathrm{<u><em>M</em></u>}$.

b) Chứng minh &#x22A5;BC." role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.08px; 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; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">AM⊥BC.&#x22A5;BC." role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.08px; 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; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">$\mathrm{<u><em>A</em></u>}\mathrm{<u><em>M</em></u>}\mathrm{<u><em>\perp </em></u>}\mathrm{<u><em>B</em></u>}\mathrm{<u><em>C</em></u>}<u><em>.</em></u>$

c) Trên cạnh BA lấy  điểm E. Trên cạnh CA lấy điểm F sao cho BE = CF. Chứng minh &#x0394;EBC" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.08px; 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; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">ΔEBC&#x0394;EBC" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.08px; 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; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">$\mathrm{<u><em>\Delta </em></u>}\mathrm{<u><em>E</em></u>}\mathrm{<u><em>B</em></u>}\mathrm{<u><em>C</em></u>}$ và &#x0394;FCB" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.08px; 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; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">ΔFCB&#x0394;FCB" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:18.08px; 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; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">$\mathrm{<u><em>\Delta </em></u>}\mathrm{<u><em>F</em></u>}\mathrm{<u><em>C</em></u>}\mathrm{<u><em>B</em></u>}$bằng nhau.

d) Chứng minh EF // BC.

Cho hàm số x2&#x2212;2" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-family:arial,tahoma,helvetica,sans-serif; font-size:18.72px; 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">y=f(x)=x2−2$y=f\left(x\right)=x^2-2$ . Hãy tính: (2);f(1);f(0);f(&#x2212;1);f(&#x2212;2)" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-family:arial,tahoma,helvetica,sans-serif; font-size:18.72px; 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">f(2) ;f(1) ;f(0) ;f(−1) ;f(−2)