Tìm GTNN
a. \(\mid x-7\mid\) + \(\mid x+5\mid\)
b. \((2x-1)^2 -3\mid2x-1\mid +2\)
c.\(\mid x^2 + x + 1\mid + \mid x^2 +x -12\mid\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
HX
0
NN
0
12 tháng 4 2017
a, Ta có: \(A=\left|x-1\right|+\left|x-2017\right|=\left|x-1\right|+\left|2017-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A\ge\left|x-1+2017-x\right|=\left|-2016\right|=2016\)
Dấu " = " khi \(\left\{{}\begin{matrix}x-1\ge0\\2017-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge1\\x\le2017\end{matrix}\right.\Rightarrow1\le x\le2017\)
Vậy \(MIN_A=2016\) khi \(1\le x\le2017\)
b, Ta có: \(\left\{{}\begin{matrix}\left(x-5\right)^2\ge0\\\left|x-5\right|\ge0\end{matrix}\right.\Rightarrow\left(x-5\right)^2+\left|x-5\right|\ge0\)
\(\Rightarrow B=\left(x-5\right)^2+\left|x-5\right|+2014\ge2014\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-5\right)^2=0\\\left|x-5\right|=0\end{matrix}\right.\Rightarrow x=5\)
Vậy \(MIN_B=2014\) khi x = 5
12 tháng 4 2017
b may cho chú là chung nghiệm là x=5 nếu (x-6)^2+|x-5| thì sao? cần phải nhớ (x-6)^2=|x-6|^2 sau đó áp dụng |a|+|b|>=|a+b|
MV
2 tháng 5 2017
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
NB
4
17 tháng 2 2019
\(\left(x+4\right).11=x.14\\ \Leftrightarrow11x+44=14x\\ \Leftrightarrow11x-14x=-44\\ \Leftrightarrow-3x=-44\\ \Leftrightarrow x=\dfrac{44}{3}\)
Vậy \(x=\dfrac{44}{3}\)
TT
17 tháng 2 2019
\(\left(x+4\right).11=x.14\)
\(11x+44-14x=0\)
\(-3x-44=0\)
\(-3x=44\)
\(x=\dfrac{44}{3}\)
TT
1 tháng 3 2020
a) Rút gọn :
ĐKXĐ : \(x\ne4,x\ne3\)
Ta có : \(Q=\frac{12x-45}{x^2-7x+12}-\frac{x+5}{x-4}+\frac{2x-3}{3-x}\)
\(=\frac{3\left(4x-15\right)}{\left(x-4\right)\left(x-3\right)}-\frac{\left(x+5\right)\left(x-3\right)}{\left(x-4\right)\left(x-3\right)}-\frac{\left(2x-3\right)\left(x-4\right)}{\left(x-4\right)\left(x-3\right)}\)
\(=\frac{12x-45-x^2-2x+15-2x^2+11x-12}{\left(x-4\right)\left(x-3\right)}\)
\(=\frac{-3x^2+21x-42}{\left(x-4\right)\left(x-3\right)}\)
... Chắc tui rút gọn sai òi :))
x-2015 
a) Bạn adct \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\)
Ta cóA= \(\left|x-7\right|+\left|x+5\right|=\left|7-x\right|+\left|x+5\right|\ge\left|7-x+x+5\right|\)
=> \(\left|7-x\right|+\left|x+5\right|\ge12\) vậy minA=12
b)Ta có \(\left(2x-1\right)^2-3\left|2x-1\right|+2=\left|2x-1\right|^2-2\left|2x-1\right|.\frac{3}{2}+\frac{9}{4}-\frac{1}{4}=\left(\left|2x-1\right|-\frac{3}{2}\right)^2-\frac{1}{4}\)=>minA=-1/4