a) |x| + |x + 1| = 1

Nếu x \(\le\) - 1

=> |x| = -x

=> |x + 1| = -(x + 1) = -x - 1

Khi đó  |x| + |x + 1| = 1 (1)

<=> -x - x - 1 = 1

=> -2x = 2

=> x = -1(tm)

Nếu -1 < x < 0

=> |x| = -x

=> |x + 1| = x + 1

Khi đó (1) <=> -x + x + 1 = 1

=> 0x = 0

=> \(x\in\varnothing\)

Nếu x \(\ge\) 0

=> |x| = x 

=> |x + 1| = x + 1

Khi đó (1) <=> x + x + 1 = 1

=> 2x = 0

=> x = 0 (tm)

Vậy \(x\in\left\{-1;0\right\}\)

b)  |x| + |x + 1| = 2020

Nếu x \(\le\) - 1

=> |x| = -x

=> |x + 1| = -(x + 1) = -x - 1

Khi đó  |x| + |x + 1| = 1 (1)

<=> -x - x - 1 = 2020

=> -2x = 2021

=> x = -1010,5(tm)

Nếu -1 < x < 0

=> |x| = -x

=> |x + 1| = x + 1

Khi đó (1) <=> -x + x + 1 = 2020

=> 0x = 2019

=> \(x\in\varnothing\)

Nếu x \(\ge\) 0

=> |x| = x 

=> |x + 1| = x + 1

Khi đó (1) <=> x + x + 1 = 2020

=> 2x = 2019

=> x = 1009,5 (tm)

Vậy \(x\in\left\{-1010,5;1009,5\right\}\)

c)\(\frac{x+1}{18}+\frac{x+2}{17}=\frac{x+3}{16}+\frac{x+4}{15}\)

=> \(\left(\frac{x+1}{18}+1\right)+\left(\frac{x+2}{17}+1\right)=\left(\frac{x+3}{16}+1\right)+\left(\frac{x+4}{15}+1\right)\)

=> \(\frac{x+19}{18}+\frac{x+19}{17}=\frac{x+19}{16}+\frac{x+19}{15}\)

=> \(\frac{x+19}{18}+\frac{x+19}{17}-\frac{x+19}{16}-\frac{x+19}{15}=0\)

=> \(\left(x+19\right)\left(\frac{1}{18}+\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)=0\)

=> x + 19 = 0 (Vì \(\frac{1}{18}+\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\ne0\)

=> x = -19

Vậy x =-19

a) | x | + | x + 1 | = 1 (*)

+) Với x < -1

(*) <=> -x - ( x + 1 ) = 1

     <=> -x - x - 1 = 1

     <=> -2x - 1 = 1

     <=> -2x = 2

     <=> x = -1 ( không thỏa mãn )

+) Với -1 ≤ x < 0 

(*) <=> -x + ( x + 1 ) = 1

     <=> -x + x + 1 = 1

     <=> 0 + 1 = 1 ( luôn đúng với mọi x ) (1)

+) Với ≥ 0 

(*) <=> x + ( x + 1 ) = 1

     <=> x + x + 1 = 1

     <=> 2x + 1 = 1

     <=> 2x = 0

     <=> x = 0 ( thỏa mãn ) (2)

Từ (1) và (2) => Với -1 ≤ x ≤ 0 thì thỏa mãn đề bài

b) | x | + | x + 1 | = 2020 (*)

+) Với x < -1

(*) <=> - x - ( x + 1 ) = 2020

     <=> -x - x - 1 = 2020

     <=> -2x - 1 = 2020

     <=> -2x = 2021

     <=> x = -2021/2 ( thỏa mãn )

+) Với -1 ≤ x < 0

(*) <=> -x + ( x + 1 ) = 2020

     <=> -x + x + 1 = 2020

     <=> 0 + 1 = 2020 ( vô lí )

+) Với x ≥ 0

(*) M <=> x + ( x + 1 ) = 2020

         <=> x + x + 1 = 2020

         <=> 2x + 1 = 2020

         <=> 2x = 2019

         <=> x = 2019/2 ( thỏa mãn )

Vậy x = -2021/2 hoặc x = 2019/2

c) \(\frac{x+1}{18}+\frac{x+2}{17}=\frac{x+3}{16}+\frac{x+4}{15}\)

\(\Leftrightarrow\left(\frac{x+1}{18}+1\right)+\left(\frac{x+2}{17}+1\right)=\left(\frac{x+3}{16}+1\right)+\left(\frac{x+4}{15}+1\right)\)

\(\Leftrightarrow\frac{x+1+18}{18}+\frac{x+2+17}{17}=\frac{x+3+16}{16}+\frac{x+4+15}{15}\)

\(\Leftrightarrow\frac{x+19}{18}+\frac{x+19}{17}=\frac{x+19}{16}+\frac{x+19}{15}\)

\(\Leftrightarrow\frac{x+19}{18}+\frac{x+19}{17}-\frac{x+19}{16}-\frac{x+19}{15}=0\)

\(\Leftrightarrow\left(x+19\right)\left(\frac{1}{18}+\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)=0\)

Vì \(\frac{1}{18}+\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\ne0\)

\(\Rightarrow x+19=0\)

\(\Rightarrow x=-19\)

\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)

\(\left|2-x\right|+\left|x+1\right|=5\)

TH1 : \(\left|2-x\right|=\pm5\)

+ ) \(2-x=5\)

            \(x=2-5\)

           \(x=-3\)

+ ) \(2-x=\left(-5\right)\)

           \(x=2-\left(-5\right)\)

          \(x=7\)

TH2 : \(\left|x+1\right|=\pm5\)

+ ) \(x+1=5\)

             \(x=5-1\)

            \(x=4\)

+ ) \(x+1=\left(-5\right)\)

       \(x=\left(-5\right)-1\)

      \(x=-6\)

2 ) \(\left|x+1\right|+\left|2x+1\right|=22\)

TH1 : \(\left|x+1\right|=\pm22\)

+ ) \(x+1=22\)

      \(x=22-1\)

      \(x=21\)

+ ) \(x+1=-22\)

      \(x=-22-1\)

     \(x=-23\)

 TH2: \(\left|2x+1\right|=\pm22\)

+ ) \(2x+1=22\)

     \(2x=21\)

      \(x=\frac{21}{2}\)

+ ) \(2x+1=-22\)

     \(2x=-23\)

     \(x=\frac{-23}{2}\) 

`|x+2|+|x+3|+|x-1|=4(x-1)`

`<=>|x+2|+|x+3|+|x-1|=4(x-1)`

Ta có nhận xét: Dễ thấy `|x+2|+|x+3|+|x-1|>=0AAx` suy ra `4(x-1)>=0` hay `x>=1`

Khi đó, pt trở thành:

`(x+2)+(x+3)+(x-1)=4(x-1)`

`<=>3x+4=4x-4`

`<=>4x-3x=4+4`

`<=>x=8(TM)`

Vậy `x=8` 

Giả sử cả 2 khí này cùng ở đktc:

\(n_{X_2}=\dfrac{2}{22,4}=\dfrac{5}{56}\left(mol\right)\)

\(\Rightarrow m_{X_2}=\dfrac{5}{56}.2X=\dfrac{5X}{28}\left(mol\right)\)

\(n_{H_2}=\dfrac{32}{22,4}=\dfrac{10}{7}\left(mol\right)\)

\(\Rightarrow m_{H_2}=\dfrac{10}{7}.2=\dfrac{20}{7}\left(g\right)\)

Theo đề bài ta có: 2 lít khí \(X_2\) có khối lượng bằng 32 lít \(H_2\)

\(\Rightarrow\dfrac{5X}{28}=\dfrac{20}{7}\)

\(\Rightarrow35X=560\)

\(\Rightarrow X=560:35=16\)

Vậy khí \(X_2\) là khí oxi có CTHH: O₂

Ta có:

\(V_{H_2}=n_{H_2}\cdot22,4\Leftrightarrow n_{H_2}=\dfrac{32}{22,4}\simeq1,43\left(mol\right)\\ m=n\cdot M\\ \Rightarrow m_{H_2}=1,43\cdot2\simeq2,9\left(g\right)\)

Theo bài ra, ta có: \(m_{X2}=m_{H2}\Rightarrow m_{X2}=2,9\left(g\right)\\ Lạicó:\\ V_{X2}=n_{X2}\cdot22,4\\ \Rightarrow n_{X2}=\dfrac{2}{22,4}\simeq0,09\left(mol\right)\\ m_{X2}=n_{X2}\cdot M_{X2}\\ \Rightarrow M_{X2}=\dfrac{2,9}{0,09}=32\left(g/mol\right)\\ \Rightarrow NguyêntửkhốicủanguyêntốXlà:\dfrac{32}{2}=16\left(đvC\right)\\ \Rightarrow CTHHcủaX_2là:O_2\)