a) Xét \(\Delta ABC\)có

\(AB=AC\left(gt\right)\)

\(\Rightarrow\Delta ABC\)cân tại A

\(\Rightarrow\widehat{B}=\widehat{C}\)

b) Vì M là trung điểm của BC 

=> AM là đường trung tuyến của \(\Delta ABC\)

Trong tam giác cân đường trung tuyến cũng là đường cao

\(\Rightarrow AM\perp BC\)

                                           A B M C 1 2

a) Xét \(\Delta ABC\)có : AB = BC ( gt )

\(\Rightarrow\Delta ABC\)cân tại A

\(\Rightarrow\widehat{B}=\widehat{C}\)

b) Xét \(\Delta ABM\)và \(\Delta ACM\)có :

                     \(AB=AC\left(gt\right)\)

                    \(BM=MC\)( M là trung điểm của BC )

                     AM chung

\(\Rightarrow\Delta ABM=\Delta ACM\left(c.c.c\right)\)

\(\Rightarrow\widehat{M_1}=\widehat{M_2}\)( 2 góc tương ứng )

mà \(\widehat{M_1}+\widehat{M_2}=180^o\)( kề bù )

\(\Rightarrow\widehat{M_1}=90^o\)

\(\Rightarrow AM\perp BC\)

a) \(ĐKXĐ:x>0;x\ne4\)

Ta có : \(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4x}{2\sqrt{x}-x}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}\right)\)

\(=\left[\frac{\sqrt{x}.\sqrt{x}-4x}{\sqrt{x}.\left(\sqrt{x}-2\right)}\right]\cdot\frac{\sqrt{x}-2}{\sqrt{x}+3}\)

\(=\frac{-3x}{\sqrt{x}.\left(\sqrt{x}+3\right)}\)

b) Ta có : \(x-1=10-4\sqrt{6}=\left(\sqrt{6}-2\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{6}-2\right)^2+1}\)

......

câu 1 sai đề :>>>

1) \(\left(x-2\right)^6=\left(x-2\right)^8\)

\(\Leftrightarrow\left(x-2\right)^8-\left(x-2\right)^6=0\)

\(\Leftrightarrow\left(x-2\right)^6.\left[\left(x-2\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=2\\x=3,x=1\end{cases}}\)

2) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

\(\Leftrightarrow2^x.\left(1+2+2^2+2^3\right)=480\)

\(\Leftrightarrow2^x.15=480\)

\(\Leftrightarrow2^x=32=2^5\)

\(\Leftrightarrow x=5\)

a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-\frac{13}{6}\end{cases}}\)

Đặt \(A=\left(\frac{1+2x}{4+2x}-\frac{x}{3x-6}+\frac{2x^2}{12-3x^2}\right)\cdot\frac{24-12x}{6+13x}\)

\(\Leftrightarrow A=\left(\frac{1+2x}{2\left(x+2\right)}-\frac{x}{3\left(x-2\right)}-\frac{2x^2}{3\left(x^2-4\right)}\right)\cdot\frac{12\left(2-x\right)}{6+13x}\)

\(\Leftrightarrow A=\frac{3\left(2x^2-3x-2\right)-2\left(x^2+2x\right)-4x^2}{6\left(x-2\right)\left(x+2\right)}\cdot\frac{12\left(2-x\right)}{6+13x}\)

\(\Leftrightarrow A=\frac{-2\left(6x^2-9x-6-2x^2-4x-4x^2\right)}{\left(x+2\right)\left(6+13x\right)}\)

\(\Leftrightarrow A=\frac{-2\left(-6-13x\right)}{\left(x+2\right)\left(6+13x\right)}\)

\(\Leftrightarrow A=\frac{2}{x+2}\)

b) Để biểu thức nhận giá trị dương

\(\Leftrightarrow\frac{2}{x+2}>0\)

\(\Leftrightarrow x+2>0\)

\(\Leftrightarrow x>-2\)

Vậy để biểu thức có giá trị dương thì \(x>-2\)

1. Where do you study ? - Currently, I'm studying at Co Nhue 2 Secondary School.

2. What subject do you ? Why ? 

- I to study Literature the most because they aim to educate about morality, personality and ideology through literary works rich in emotions and images.

3. How many people in your family ? Who are they ?

- My family consists of a total of four people : dad, mom, me anh my younger brother. 

1,I study at cam vu primary school

2,I english best because I want to talk with foreign friends

3,there are five people in my family.It is a grandmother,father,mother,sister and me

Vì \(\overline{x216y}\)chia 2 dư 1 \(\Rightarrow y\)là số lẻ (1)

mà \(\overline{x216y}\)chia 5 dư 3 \(\Rightarrow y\in\left\{3;8\right\}\)(2)

Từ (1) và (2) \(\Rightarrow y=3\)

Ta có: \(\overline{x2163}\)chia 9 dư 5 \(\Rightarrow\overline{x2163}-5=\overline{x2158}⋮9\)

\(\Rightarrow x+2+1+5+8=16+x⋮9\)

\(\Rightarrow x=2\)

Vậy số cần tìm là \(22163\)

Ta có :

\(\overline{x216y}\)chia 2 dư 1 \(\Rightarrow y\)là 1 số lẻ ( * )

\(\overline{x216y}\)chia 5 dư 3 \(\Rightarrow y\in\left\{3;8\right\}\)( ** )

Từ ( * ) và ( ** ) \(\Rightarrow y=3\)

Ta lại có :

\(\overline{x2163}\)chia 9 dư 5 \(\Rightarrow\overline{x2163}-5=\overline{x2158}⋮9\)

\(\Rightarrow x+2+1+6+3⋮9\)

\(\Rightarrow x+16⋮9\)

\(\Rightarrow x=2\)

Vậy số cần tìm là : \(22163\)

Bài 11:

1) Sửa lại đề là: \(A=127^2+146.127+73^2\)

\(\Rightarrow A=127^2+2.127.73+73^2\)

\(\Rightarrow A=\left(127+73\right)^2\)

\(\Rightarrow A=200^2\)

\(\Rightarrow A=40000\)

Vậy \(A=40000.\)

2) Sửa lại đề là: \(B=9^8.2^8-\left(18^4-1\right).\left(18^4+1\right)\)

\(\Rightarrow B=\left(9.2\right)^8-\left[\left(18^4\right)^2-1^2\right]\)

\(\Rightarrow B=18^8-\left(18^8-1\right)\)

\(\Rightarrow B=18^8-18^8+1\)

\(\Rightarrow B=0+1\)

\(\Rightarrow B=1\)

Vậy \(B=1.\)

4) \(D=\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(\Rightarrow2D=\left(3-1\right).\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(=3^{32}-1\)

\(\Rightarrow D=\frac{3^{32}-1}{2}\)