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1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow\)\(x+329=0\) (vì 1/327 + 1/326 + 1/325 + 1/324 + 1/5 khác 0 )
\(\Leftrightarrow\)\(x=-329\)
Bài 1 :
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
\(\Rightarrow\)\(x+329=0\)
\(\Rightarrow\)\(x=-329\)
Vậy \(x=-329\)
a) \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(A=\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}...\frac{-9999}{100^2}\)
\(A=-\left(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{9999}{100^2}\right)\) (vì A là tích của 99 thừa số âm nên kết quả là âm)
\(A=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\right)\)
\(A=-\left(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\right)\)
\(A=-\left(\frac{1}{100}.\frac{101}{2}\right)=\frac{-101}{200}\)
b) 2x + 2y = 2x+y
=> 2x = 2x.2y - 2y
=> 2x = 2y.(2x - 1)
\(\Rightarrow2^x⋮2^x-1\)
Mà (2x; 2x - 1) = 1
\(\Rightarrow\begin{cases}2^x-1=1\\2^y=2^x\end{cases}\)\(\Rightarrow\begin{cases}2^x=2=2^1\\x=y\end{cases}\)=> x = y = 1
Vậy x = y = 1
Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
a) (1-1/2)(1-1/3)...(1-1/100)=lx-1 99/100l
=> (1-1/2)(1-1/3)...(1-1/100)=1/2.2/3.3/4...99/100
=> (1-1/2)(1-1/3)...(1-1/100)=1.2.3.4....99/2.3.4....100
=>(1-1/2)(1-1/3)...(1-1/100)=1/100 (1)
từ (1)=>1/100= l x-1 99/100 l
TH1:x-1 99/100 =1/100 TH2 : x-1 99/100= -1/100
=>x- 199/100 =1/100 =>x- 199/100= -1/100
=>x=1/100+199/100 =>x=-1/100+199/100
=>x=200/100 =>x=198/100
=>x=2 =>x=99/50
Vậy x=2 hoặc x=99/50
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
a/ \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{100}\right)=\frac{3}{2}\times\frac{4}{3}\times....\times\frac{101}{100}=\frac{101}{2}\)
b/ Tự chép đề nha\(B=\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)....\left(1-\frac{1}{100}\right)\left(1+\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\frac{3}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{99}{100}\times\frac{101}{100}=\frac{1}{2}\times\frac{101}{100}=\frac{101}{200}\)
Đề a) (1+1/2) (1+1/3) (1+1/4)...(1+1/100)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)....\left(1+\frac{1}{100}\right)\)
\(=\frac{3}{2}.\frac{4}{3}....\frac{101}{100}=\frac{3.4...101}{2.3...100}=\frac{101}{2}\)
Học tốt
\(\left|5x-3\right|-3x=12\)
\(\Leftrightarrow\left|5x-3\right|=12+3x\)
\(\Leftrightarrow\hept{\begin{cases}5x-3=12+3x\\-\left(5x-3\right)=12+3x\end{cases}\Rightarrow\hept{\begin{cases}5x-3x=12+3\\-5x+3=12+3x\end{cases}\Rightarrow}\hept{\begin{cases}2x=15\\-5x-3x=12-3\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{15}{2}\\-8x=9\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{15}{2}\\x=\frac{-9}{8}\end{cases}}}\)
A=1+\(\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+\frac{1}{4}\cdot\frac{4\cdot5}{2}+....+\frac{1}{100}+\frac{100\cdot101}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)
\(=1+\left(\frac{101\cdot2}{2}-3\right)\cdot\frac{1}{2}=1+98\cdot\frac{1}{2}=49+1=50\)