A B H M C 60

Ta thấy AM là trung tuyến mà vuống tại A

=> A= 90 độ 

Vì là trung tuyến 

=> CAM= 45

chi mk nha 

ko biết đúng ko nữa mik chỉ mới hc lp 6

A B C H M

\(\Delta ABC\)có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

             Mà góc A=90⁰ góc B=60⁰

\(\Rightarrow\widehat{C}=30^0\)

\(\Delta ABC\)vuông tại A\(\Rightarrow\)AM=\(\frac{BC}{2}\)

Mà BM=MC=\(\frac{BC}{2}\)

\(\Rightarrow\Delta AMC\)cân tại M

\(\Rightarrow\widehat{MAC}=\widehat{MCA}\)

Mà \(\widehat{MCA}=30^0\)

\(\Rightarrow\widehat{MAC}=30^0\)

\(1+\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=2-\frac{1}{100}\)

\(=\frac{199}{100}\)

Gọi biểu thức là A

A=1+1/2+1/2.3+1/3.4+...+1/98.99+1/99.100

A-1=1/2+1/2.3+1/3.4+...+1/98.99+1/99.100

A-1=1-1/2+1/2-1/3+1/3-1/4+...+/198-1/99+1/99-1/100

A-1=1-1/100

A-1=99/100

A=99/100+1

A=199/100

\(\frac{6x^3y^4}{8x^2y^5}=\frac{3x}{4y}\)

\(\frac{6x^3y^4}{8^2y^5}=\frac{3x}{by}\)

\(\Rightarrow\frac{3x}{4y}=\frac{3x}{by}\Rightarrow b=4\)

\(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3\)

\(\Leftrightarrow\frac{201-x}{99}+1+\frac{203-x}{97}+1-\frac{205-x}{95}-1=4\)

\(\Leftrightarrow\frac{200-x}{99}+\frac{200-x}{97}-\frac{200-x}{95}=4\)

\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{99}+\frac{1}{97}-\frac{1}{95}\right)=4\)

Bạn tự làm tiếp.

X = -104,695575 

   Đáp số ra lẻ quá bạn nhỉ 

=1(2+1)(2^2+1)...(2^641)-2^128

=(2-1)(2+1)(2^2+1)(2^4+1)...(2^64+1)-2^128

=(2^2-1)(2^2+1)(2^4+1)...(2^64+1)-2^128.......

=(2^64-1)(2^64+1)-2^128

=2^128-1-2^128

=-1

\(=1.\left(2+1\right)\left(2^2+1\right)...\left(2^{64}+1\right)-2^{128}\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)-2^{128}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)-2^{128}\)

\(...\)

\(=\left(2^{64}-1\right)\left(2^{64}+1\right)-2^{128}\)

\(=2^{128}-1-2^{128}=-1\)

Thêm (-1) vào từng số hạng=> tử số các số hạng là:  \(\left(x^2-10x-2000\right)\)

\(\Leftrightarrow x^2-10x-2000=0\Leftrightarrow\left(x-5\right)^2=2025=45^2\)

\(\orbr{\begin{cases}x=50\\x=-40\end{cases}}\)

\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\\ \)
Cộng từng hạng tử của hai vế với 1
\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Rightarrow\frac{x+1+2004}{2004}+\frac{x+2+2003}{2003}=\frac{x+3+2002}{2002}+\frac{x+4+2001}{2001}\)
\(\Rightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2002}=0\)
\(\Rightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Vì \(\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)\ne0\)nên \(x+2005=0\Rightarrow x=-2005\)
Phương trình có nghiệm duy nhất: x=2005

(x+1)/2004+(x+2)/2003=(x+3)/2002+(x+4)/2001

(x+1)/2004+1  +(x+2)/2003 +1=(x+3)/2002+1 (x+4)/2001+1

=> x+2005/2004+(x+2005)/2003-(x+2005)/2002-(x+2005)/2002=0

(x+2005)(1/2004+1/2003-1/2002-1/2001)=0

=>x+2005=0

=>x=-2005