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Ta có : \(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)
\(\Rightarrow\left|x+\frac{13}{14}\right|+\left|x-\frac{3}{7}\right|=0\)
Mà : \(\left|x+\frac{13}{14}\right|\ge0\forall x\)
\(\left|x-\frac{3}{7}\right|\ge0\forall x\)
Nên : \(\orbr{\begin{cases}\left|x+\frac{13}{14}\right|=0\\\left|x-\frac{3}{7}\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{13}{14}=0\\x-\frac{3}{7}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{14}\\x=\frac{3}{7}\end{cases}}\)
a) Ta có:
VT = |x + 1| + |x + 2| + |2x - 3| \(\ge\)|x + 1 + x + 2| + |3 - 2x| = |2x + 3| + |3 - 2x| \(\ge\)|2x + 3 + 3 - 2x| = 6
VP = 6
Dấu "=" xảy ra<=> \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\) => \(\orbr{\begin{cases}x\ge-1\\x\le-2\end{cases}}\)và \(-\frac{3}{2}\le x\le\frac{3}{2}\)
<=> \(-1\le x\le\frac{3}{2}\)
b) Ta có: VT = |x + 1| + |x - 2| + |x - 3| + |x - 5| = (|x + 1| + |5 - x|) + (|x - 2| + |3 - x|) \(\ge\)|x + 1 + 5 - x| + |x - 2 + 3 - x| = |6| + |1| = 7
VP = 7
Dấu "=" xảy ra<=> \(\hept{\begin{cases}\left(x+1\right)\left(5-x\right)\ge0\\\left(x-2\right)\left(3-x\right)\ge0\end{cases}}\) <=> \(\hept{\begin{cases}-1\le x\le5\\2\le x\le3\end{cases}}\) <=> \(2\le x\le3\)
Ta có bất đẳng thức giá trị tuyệt đối:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
Dấu \(=\)khi \(AB\ge0\).
d) \(\left|x+1\right|+\left|x+2\right|+\left|2x-3\right|\)
\(\ge\left|x+1+x+2\right|+\left|2x-3\right|\)
\(=\left|2x+3\right|+\left|3-2x\right|\)
\(\ge\left|2x+3+3-2x\right|=6\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le\frac{3}{2}\).
e) \(\left|x+1\right|+\left|x+2\right|+\left|x-3\right|+\left|x-5\right|\)
\(=\left(\left|x+1\right|+\left|3-x\right|\right)+\left(\left|x+2\right|+\left|5-x\right|\right)\)
\(\ge\left|x+1+3-x\right|+\left|x+2+5-x\right|\)
\(=4+7=11\)
Dấu \(=\)khi \(\hept{\begin{cases}\left(x+1\right)\left(3-x\right)\ge0\\\left(x+2\right)\left(5-x\right)\ge0\end{cases}}\Leftrightarrow-1\le x\le3\).
Do đó phương trình đã cho vô nghiệm.
a) (x + 2)2 = 81
=> (x + 2)2 = 92
=> \(\orbr{\begin{cases}x+2=-9\\x+2=9\end{cases}}\Rightarrow\orbr{\begin{cases}x=-11\\x=7\end{cases}}\)
b) 5x + 5x + 2 = 650
=> 5x + 5x . 52 = 650
=> 5x + 5x . 25 = 650
=> 5x (25 + 1) = 650
=> 5x . 26 = 650
=> 5x = 650 : 26
=> 5x = 25
=> 5x = 52
=> x = 2
d) (2x - 1)2 - 5 = 20
=> (2x - 1)2 = 25
=> (2x - 1)2 = 52
=> \(\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
g) (x - 1)3 = (x - 1)
=> (x - 1)3 - (x - 1) = 0
=> (x - 1) .[(x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x-1=\pm1\end{cases}}}\)
Nếu x - 1 = 1
=> x = 2
Nếu x - 1 = -1
=> x = 0
Vậy \(x\in\left\{0;1;2\right\}\)
a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)
=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
=> x + 1 = 0
=> x = -1
b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)
=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)
=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
=> x - 2021 = 0
=> x = 2021
c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
=> \(-\frac{1}{12}x+6=7\)
=> \(-\frac{1}{12}x=1\)
=> x = -12
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x^3-8x\left(x+2\right)=6\\ \Leftrightarrow\left(x^2+3x+2\right).\left(x+3\right)-x^3-8x^2-16x=6\\ \Leftrightarrow x^3+6x^2+11x+6-x^3-8x^2-16x-6=0\\ \Leftrightarrow-2x^2-5x=0\\ \Leftrightarrow x.\left(-2x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\-2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)