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