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15S+1=15+15.42+15.44+...+15.420+1
=16+15.42+15.44+...+15.420
=42+15.42+15.44+...+15.420
=16.42+15.44+...+15.420 =44+15.44+...+15.420=16.44+...+15.420=16.418+15.420=16.420=422
vậy x-5=22 <=>x=27
A=4+(22+23+24+...+220)
A-4=22+23+24+...+220
2(A-4)=23+24+25+...+221
A-4=2(A-4)-(A-4)=(23+24+25+...+221)-(22+23+24+...+220)
A-4=(23-23)+(24-24)+(25-25)+...+(220-220)+(221-22)
A-4=221-4
A =221-4+4
A =221
Bạn làm tiếp nha .
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{99}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{98}\right)=0\)
\(\Rightarrow x+100=0\)
\(\Rightarrow x=-100\)
a) \(\left(4,5-2x\right)\cdot1\frac{4}{7}=\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)\cdot\frac{11}{7}=\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{7}\div\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=1\)
\(2x=\frac{9}{2}-1\)
\(x=\frac{7}{2}\div2\)
\(x=\frac{7}{4}\)
b) \(|\frac{3}{4}\cdot x-\frac{1}{2}|-1=\frac{1}{4}\)
\(|\frac{3}{4}\cdot x-\frac{1}{2}|=\frac{1}{4}+1\)
\(|\frac{3}{4}\cdot x|=\frac{5}{4}+\frac{1}{2}\)
\(x=\frac{7}{4}\div\frac{3}{4}\)
\(x=\frac{7}{3}\)
c) \(\frac{1}{4}-|3-x|=-\frac{3}{4}\)
\(|3-x|=\frac{1}{4}-\left(-\frac{3}{4}\right)\)
\(|3-x|=1\)
\(x=3-1\)
\(\Rightarrow x=2\)
d) \(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=1,4\)
\(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=\frac{7}{5}\)
\(4\cdot\left(x-\frac{6}{7}\right)=\frac{7}{5}+\frac{3}{5}\)
\(4\cdot\left(x-\frac{6}{7}\right)=2\)
\(\left(x-\frac{6}{7}\right)=2\div4\)
\(x=\frac{1}{2}+\frac{6}{7}\)
\(x=\frac{19}{14}\)
\(\)
a) \(3x-\frac{1}{5}=\frac{4+x}{2}\)
=> \(\frac{15x-1}{5}=\frac{4+x}{2}\)
=> \(\left(15x-1\right).2=\left(4+x\right).5\)
=> \(30x-2=20+5x\)
=> \(30x-5x=20+2\)
=> \(25x=22\)
=> \(x=\frac{22}{25}\)
b) \(\frac{4}{3}x-1=\frac{4\left(x+1\right)}{3}-\frac{1}{3}\)
=> \(\frac{1}{3}x=\frac{4x+4-1}{3}\)
=> \(\frac{1}{3}x=\frac{4x+3}{3}\)
=> \(3x=3\left(4x+3\right)\)
=> \(3x=12x+9\)
=> \(3x-12x=9\)
=> \(-9x=9\)
=> \(x=9:\left(-9\right)=-1\)
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
A = 1 + 4 + 42 + ... + 499
4A = 4 + 42 + ... + 4100
4A - A = 4100 - 1
3A = 4100 - 1
=> 4100 - 1 + 1 = 4x
=> 4100 = 4x
=> x = 100