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1) \(\left|2x+5\right|\ge21\Rightarrow2x+5\ge21\)hoặc \(2x+5<-21\)<=> \(x\ge8\) hoặc \(x<-13\)
2)
a) |2x-3|>=0 => A>=0-5=-5 => Min A=-5 <=> x=3/2
b) \(\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|=\left|2\right|=2\Rightarrow B\ge2+5=7\)=> MinB=7 <=>x=1
3)
\(\left|2x-1\right|\ge0\Rightarrow-\left|2x-1\right|\le0\Leftrightarrow A\le0+7=7\Rightarrow MaxA=7\Leftrightarrow x=-\frac{1}{2}\)
b)
th1: nếu x<-3/2 => B=-2x-3+2x+2=-1
th2: nếu \(-\frac{3}{2}\le x\le-1\)=> B=2x+3+2x+2=4x+5
ta có:\(-\frac{3}{2}\le x\le-1\Rightarrow-6\le4x\le-4\Leftrightarrow-1\le4x+5\le1\Rightarrow-1\le B\le1\)
th3: nếu x>-1 => B=2x+3-2x-2=1=>
Max B=1 <=> x>-1 hoặc \(-\frac{3}{2}\le x\le-1\)
2b) Áp dụng bất đẳng thức giá trị tuyệt đối: |a| + |b| \(\ge\) |a + b|. Dấu "=" xảy ra khi tích a.b \(\ge\) 0
Ta có: B = |2x - 1| + |3 - 2x| + 5 \(\ge\) |2x - 1+3 - 2x| + 5 = |2| + 5 = 7
=> Min B = 7 khi
(2x - 1)( 3 - 2x) \(\ge\) 0 => (2x - 1)(2x - 3) \(\le\) 0
Mà 2x - 1 > 2x - 3 nên 2x - 1 \(\ge\) 0 và 2x - 3 \(\le\) 0
=> x \(\ge\) 1/2 và x \(\le\) 3/2
Nguyễn Trà My
Phần a)
\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(32-3x+13=76-x\)
\(116-3x=76-x\)
\(116-76=3x-x\)
\(46=2x\)
\(x=46\div2\)
\(x=13\)
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
Sửa đề : a) Tìm GTNN A
a) \(A=\left|x-5\right|+3\)có : \(\left|x-5\right|\ge0\Rightarrow\left|x-5\right|+3\ge0\)
\(\Leftrightarrow A\ge3\)dấu "=" xảy ra khi : \(\left|x-5\right|=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy GTNN A = 3 khi x = 5.
b) \(C=-\left|x+1\right|+5\)có : \(-\left|x+1\right|\le0\Rightarrow-\left|x+1\right|+5\le5\)
\(\Leftrightarrow C\le5\)dấu "=" xảy ra khi : \(-\left|x+1\right|=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy GTLN C = 5 khi x = -1.
\(D=5-\left|2x+3\right|\)có : \(-\left|2x+3\right|\le0\Rightarrow5-\left|2x+3\right|\le5\)
\(\Leftrightarrow D\le5\)dấu "=" xảy ra khi : \(-\left|2x+3\right|=0\Leftrightarrow2x+3=0\Leftrightarrow x=-\frac{3}{2}\)
Vậy GTLN D = 5 khi x = -3/2.
c) \(\left|x-3\right|+\left|y+1\right|=0\)có \(\left|x-3\right|\ge0;\left|y+1\right|\ge0\Rightarrow\left|x-3\right|+\left|y+1\right|\ge0\)
\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=0\\\left|y+1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}.\)
1.
a) \(\frac{-7}{9}.2\frac{3}{4}=\frac{-7}{9}.\frac{11}{4}=\frac{-77}{36}\)
b) \(\frac{2}{3}+\frac{1}{3}.\frac{-2}{5}=\frac{2}{3}+\frac{-2}{15}=\frac{8}{15}\)
c) \(\frac{3}{4}.15\frac{1}{3}-\frac{3}{4}.43\frac{1}{3}=\frac{3}{4}.\frac{46}{3}-\frac{3}{4}.\frac{130}{3}=\frac{23}{2}-\frac{65}{2}=-21\)
d) \(\left(-49,1\right).\frac{13}{27}-58,9.\frac{13}{27}=\frac{13}{27}.\left(-49,1-58,9\right)=\frac{13}{27}.\left(-108\right)=-52\)
e) \(0,375:\left(-4,5\right)=\frac{-1}{12}\)
f) \(3\frac{1}{7}:\left(-1\frac{3}{7}\right)=\frac{22}{7}:\frac{-10}{7}=\frac{-11}{5}\)
g) \(9\frac{1}{3}:4\frac{2}{3}-2=\frac{28}{3}:\frac{14}{3}-2=2-2=0\)
h) \(\left(7\frac{3}{4}:0,3125+4,5.2\frac{2}{45}\right):\left(-8,5\right)=\left(\frac{31}{4}:\frac{5}{16}+\frac{9}{2}.\frac{92}{45}\right):\frac{-17}{2}=\left(\frac{124}{5}+\frac{46}{5}\right):\frac{-17}{2}=34:\frac{-17}{2}=-4\)
Bài 1 : Tính:
a)
\(\frac{-7}{9}.2\frac{3}{4}=\frac{-7}{9}.\frac{11}{4}=\frac{-77}{36}\)
b)
\(\frac{2}{3}+\frac{1}{3}.\frac{-2}{5}=\frac{2}{3}+\frac{-2}{15}=\frac{10}{15}+\frac{-2}{15}=\frac{8}{15}\)
c)
\(\frac{3}{4}.15\frac{1}{3}-\frac{3}{4}.43\frac{1}{3}=\frac{3}{4}.\frac{46}{3}-\frac{3}{4}.\frac{130}{3}\)\(=\frac{23}{2}-\frac{65}{2}=\frac{-42}{2}=-21\)
....
Tự lm tiếp dạng như v
Bài 2 :
\(A=\frac{-6}{11}.\frac{7}{10}.\frac{11}{-6}.-20=\left(\frac{-6}{11}.\frac{11}{-6}\right).\left(\frac{7}{10}.-20\right)\)\(=1.\left(-14\right)=-14\)
.....
Bài 3 :
\(\frac{3}{7}.x-\frac{2}{5}.x=\frac{-17}{35}\)
\(\Leftrightarrow\frac{3}{7}-\frac{2}{5}.x=\frac{-17}{35}\)
\(\Leftrightarrow\frac{1}{35}x=\frac{-17}{35}\)
\(\Leftrightarrow x=\frac{-17}{35}:\frac{1}{35}\)
\(\Leftrightarrow x=\frac{-17}{35}.35=-17\)
a; \(x\) - \(\dfrac{3}{5}\) = 1 - \(\dfrac{4}{5}\) + \(\dfrac{1}{6}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{30}{30}\) - \(\dfrac{24}{30}\) + \(\dfrac{5}{30}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{6}{30}\) + \(\dfrac{5}{30}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{11}{30}\) + \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{11}{30}\) + \(\dfrac{18}{30}\)
\(x\) = \(\dfrac{29}{30}\)
Vậy \(x\) = \(\dfrac{29}{30}\)
b; (- \(\dfrac{10}{4}\)) + \(\dfrac{1}{4}\) = \(\dfrac{3}{4}\) thế \(x\) của em đâu nhỉ???
c; - \(\dfrac{3}{2}\) + (\(x\) - \(\dfrac{1}{2}\)) = \(\dfrac{1}{2}\)
\(x\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\)
\(x\) - \(\dfrac{1}{2}\) = 2
\(x\) = 2 + \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{4}{2}\) + \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}\)