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a, 1,5 +|2x - 2/3| = 3/2
|2x - 2/3| = 3/2 - 1,5
|2x - 2/3| = 0
<=> 2x - 2/3 = 0
<=> 2x = 0 + 2/3
<=> 2x = 2/3
<=> x = 2/3 : 2
<=> x = 1/3
Vậy x = 1/3
b, 3/4 - |1/4 - x| = 5/8
|1/4 - x| = 3/4 - 5/8
|1/4 - x| = 1/8
<=> 1/4 - x = 1/8
1/4 - x = /1/8
<=> x = 1/4 - 1/8
x = 1/4 - ( -1/8)
<=> x = 1/8
x = 3/8
Vậy x thuộc { 1/8 ; 3/8 }
\(3.M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{38}}\)
=> \(3M-M=2M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{38}}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{39}}\)
=> \(2M=1-\frac{1}{3^{39}}\)
=> \(M=\frac{1}{2}\left(1-\frac{1}{3^{39}}\right)\)
do \(1-\frac{1}{3^{39}}< 1\)
=> \(\frac{1}{2}\left(1-\frac{1}{3^{39}}\right)< \frac{1}{2}.1=\frac{1}{2}\)
Vay \(M< \frac{1}{2}\)
Chuc bn hoc tot !
1/4×2/6×3/8×4/10×...×14/30×15/32=1/2^x
<=>1/(2×2)×2/(2×3)×...×14/(2×15)×15/2^5=1/2^x
<=>1/2×1/2×...×1/2×1/(2^5)=1/2^x
<=>1/2^19=1/2^x=>x=19
Đề mình không ghi lại nhé.
\(\Rightarrow\frac{1\times2\times3\times4\times...\times14\times15}{4\times6\times10\times...\times30\times32}=\frac{1}{2^x}\)\(\frac{1}{2^x}\)
\(\Rightarrow\frac{1\times2\times3\times4\times...\times14\times15}{2\times4\times6\times8\times10\times...\times30\times32}\)\(=\frac{1}{2^{x+1}}\)
\(\Rightarrow\frac{1}{2^{15}\times32}=\)\(\frac{1}{2^{x+1}}\)
\(\Rightarrow2^{15}\times2^5=2^{x+1}\)
\(\Rightarrow2^{20}=2^{x+1}\)
\(\Rightarrow x+1=20\Rightarrow x=19\)
Vậy \(x=1\)
Học tốt nhaaa!
\(-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-3\right)^2\)
\(=-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-\left(4x^2-12x+9\right)\)
\(=-3x^3-12x^2-12x+x^3-x+3x^2-3-4x^2+12x-9\)
\(=-2x^3-13x^2-x-12\)
(x+1)+(x+2)+(x+3)=4x
x+1+x+2+x+3=4x
(x+x+x)+(1+2+3)=4x
x*3+6=4x
6=1*x(bớt cả hai vế đi 3*x)
x=6/1(Tìm thừa số)
x=6
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Leftrightarrow-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
=> x \(\in\) {-1;0;1;2;3;4;5;6}
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(\Leftrightarrow\)\(\frac{9-10}{12}\le\frac{x}{12}< 1-\left(\frac{8-3}{12}\right)\)
\(\Leftrightarrow\)\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
\(\Leftrightarrow-1\le x< 7\)
Mà x nguyên
=>x={-1;0;1;2;3;4;5;6}
Ta có : \(\left|2x-1\right|-\frac{1}{2}=\frac{1}{3}\)
=> \(\left|2x-1\right|=\frac{5}{6}\)
TH1 : \(2x-1\ge0\left(x\ge\frac{1}{2}\right)\)
=> \(\left|2x-1\right|=2x-1=\frac{5}{6}\)
=> \(x=\frac{\frac{5}{6}+1}{2}=\frac{11}{12}\)( TM )
TH2 : \(2x-1< 0\left(x< \frac{1}{2}\right)\)
=> \(\left|2x-1\right|=1-2x=\frac{5}{6}\)
=> \(x=\frac{\frac{5}{6}-1}{-2}=\frac{1}{12}\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{\frac{1}{12};\frac{11}{12}\right\}\)
\(|2x-1|-\frac{1}{2}=\frac{1}{3}\)
\(|2x-1|=\frac{1}{3}+\frac{1}{2}\)
\(|2x-1|=\frac{2+3}{6}\)
\(|2x-1|=\frac{5}{6}\)
\(\Rightarrow2x-1=\frac{5}{6}\) hoặc \(2x-1=-\frac{5}{6}\)
\(TH1:2x-1=\frac{5}{6}\)
\(2x=\frac{5}{6}+1\)
\(2x=\frac{5+6}{6}\)
\(2x=\frac{11}{6}\)
\(x=\frac{11}{6}:2\)
\(x=\frac{11}{6}.\frac{1}{2}\)
\(x=\frac{11}{12}\)
\(TH2:2x-1=-\frac{5}{6}\)
\(2x=-\frac{5}{6}+1\)
\(2x=\frac{-5+6}{6}\)
\(2x=\frac{1}{6}\)
\(x=\frac{1}{6}:2\)
\(x=\frac{1}{6}.\frac{1}{2}\)
\(x=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{11}{12};\frac{1}{12}\right\}\)