viết có chắc chữ giải mà cũng đúng thật vô lý

7X-8=713

\(\Rightarrow=\frac{51\cdot52\cdot53\cdot...\cdot100\cdot\left(1\cdot2\cdot3\cdot...\cdot50\right)\cdot2\cdot2\cdot2\cdot....\cdot2}{51\cdot52\cdot53\cdot...\cdot100}\)

rút gọn còn lại:\(\frac{1\cdot2\cdot3\cdot...\cdot50\left(2\cdot2\cdot2\cdot..\cdot2\right)}{1}\)

\(=1\cdot2\cdot3\cdot....\cdot50\left(2\cdot2\cdot2\cdot2\cdot...\cdot2\right)\)(52 số 2)

ok!

phan gia huy:bn làm sai rùi!!!!

xim lỗi nhầm bài tương tự,cho mình giải lại

\(A=\frac{\left(1\cdot3\cdot\cdot5\cdot....\cdot99\right)\left(2\cdot2\cdot2\cdot2\cdot2...\cdot2\cdot2\right)}{51\cdot52\cdot53\cdot...\cdot100}\)52 SỐ 2

\(=\frac{\left(1\cdot2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot2\cdot2\cdot..\cdot2\right)}{2\cdot4\cdot6\cdot...\cdot100\left(51\cdot52\cdot53\cdot...\cdot100\right)}\)(52 SỐ 2)

\(=\frac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot100\right)\left(2\cdot2\cdot2\cdot...\cdot2\right)}{\left(1\cdot2\right)\left(2\cdot2\right)\left(3\cdot2\right)\cdot...\cdot\left(50\cdot2\right)\left(51\cdot52\cdot53\cdot...\cdot100\right)}\)(52 SỐ 2)

\(=\frac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot..\cdot100\right)\left(2\cdot2\cdot2\cdot2\cdot....\cdot2\right)}{1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot100\left(2\cdot2\cdot2\cdot2\cdot...\cdot2\right)}\)(TỬ CÓ 52 SỐ 2,MÃU CÓ 50 SỐ)

\(=4\)

a: \(\Leftrightarrow\dfrac{x-51}{9}-1+\dfrac{x-52}{8}-1=\dfrac{x-53}{7}-1+\dfrac{x-54}{6}-1\)

=>x-60=0

hay x=60

b: \(\Leftrightarrow\left(x-2\right)^2-3\left(x+2\right)=x-14\)

\(\Leftrightarrow x^2-4x+4-3x-6-x+14=0\)

\(\Leftrightarrow x^2-8x+12=0\)

=>(x-2)(x-6)=0

=>x=2(loại) hoặc x=6(nhận)

Đặt \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(\Rightarrow A-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0\)

\(\Rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0\)

Ta có 51/2.52/2...100/2

       = 1.2.3....100/1.2...50.2.2...2 (nhân cả tử và mẫu với 1.2.3...50)

      = 1.2.3...100/(1.2)(2.2)(3.2)...(50.2)

      = 1.2.3...100/2.4.6...100

      = 1.3.5...99 => đpcm nhớ giữ lời hứa đấy

a, 2 + 4 + 6 +...+ 2x = 210

=> 2(1 + 2 + 3 +...+ x) = 210

=> \(\frac{2x\left(x+1\right)}{2}=210\)

=> x(x + 1) = 210

=> x(x + 1) = 14.15

=> x = 14

b, Ta có: \(B=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}....\frac{100}{2}=\frac{51.52.53....100}{2^{50}}\)

\(=\frac{\left(51.52.53....100\right)\left(1.2.3.....50\right)}{2^{50}\left(1.2.3.....50\right)}\)

\(=\frac{1.2.3.....100}{\left(2.1\right)\left(2.2\right)\left(2.3\right)....\left(2.50\right)}\)

\(=\frac{\left(1.3.5....99\right)\left(2.4.6....100\right)}{2.4.6.....100}\)

\(=1.3.5.....99=B\)

Vậy A = B

\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

\(=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{99}-\dfrac{1}{100}\right)\)

=\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

\(\left(\dfrac{2}{1\cdot2}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\right)\cdot\dfrac{x^2+x+1945}{2}>1975\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\)=>\(2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\cdot\dfrac{x^2+x+1945}{2}>1975\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)

=>\(x^2+x+1945>1975\)

=>\(x^2+x-30>0\)

=>(x+6)(x-5)>0

TH1: \(\left\{{}\begin{matrix}x+6>0\\x-5>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>-6\\x>5\end{matrix}\right.\)

=>x>5

TH2: \(\left\{{}\begin{matrix}x+6< 0\\x-5< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< -6\\x< 5\end{matrix}\right.\)

=>x<-6