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1) \(x+\dfrac{30}{100}x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}x=-\dfrac{131}{100}\)
\(\Leftrightarrow100x+30x=-131\)
\(\Leftrightarrow130x=-131\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
Vậy \(x=-\dfrac{131}{130}\)
b) \(\left(4,5-2x\right)\cdot\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow\left(\dfrac{9}{2}-2x\right)\cdot\left(-\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow-\dfrac{18}{7}+\dfrac{8}{7}x=\dfrac{11}{4}\)
\(\Leftrightarrow-72+32x=77\)
\(\Leftrightarrow32x=77+72\)
\(\Leftrightarrow32x=149\)
\(\Leftrightarrow x=\dfrac{149}{32}\)
Vậy \(x=\dfrac{149}{32}\)
=>3/10(x-5)=2x+5
=>3/10x-3/2=2x+5
=>-17/10x=5+3/2=6/2+3/2=11/2
=>x=-55/17
1.
\(\dfrac{30}{100}.x+\dfrac{1}{4}=\dfrac{1}{5}.x-\dfrac{1}{2}\)
\(\dfrac{3}{10}.x=\dfrac{1}{5}.x-\dfrac{1}{2}-\dfrac{1}{4}\)
\(\dfrac{3}{10}.x=\dfrac{1}{5}.x-\dfrac{1}{4}\)
\(\dfrac{1}{5}.x-\dfrac{3}{10}.x=\dfrac{1}{4}\)
\(\left(\dfrac{1}{5}-\dfrac{3}{10}\right).x=\dfrac{1}{4}\)
\(\dfrac{-1}{10}.x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}:\dfrac{-1}{10}\)
x=\(\dfrac{5}{-2}\)=\(\dfrac{-5}{2}\)
2.
\(\left(\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{31.33}\right).x=\left(0,25-3,5\right).\dfrac{27}{3}\)
\(\dfrac{2}{2}.\left(\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{31.33}\right).x=-3,25.9\)
\(\dfrac{1}{2}.\left(\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{31.33}\right).x=-29,25\)
\(\dfrac{1}{2}.\left(\dfrac{1}{7}-\dfrac{1}{33}\right).x=-29,25\)
\(\dfrac{1}{2}.\dfrac{26}{231}.x=-29,25\)
\(\dfrac{13}{231}.x=-29,25\)
\(x=-29,25:\dfrac{13}{231}\)
\(x=\dfrac{-2079}{4}\)
tick mink nha :)
\(\Leftrightarrow\left(x-5\right)\cdot\dfrac{3}{10}=\dfrac{x}{5}+5\)
=>3/10x-3/2=1/5x+5
=>1/10x=5+3/2=10/2+3/2=13/2
=>x=13/2:1/10=13/2x10=65
a: =>3/2x=5/6
hay x=5/6:3/2=5/6x2/3=10/18=5/9
b: \(\Leftrightarrow219\cdot1-7\left(x+1\right)=100\)
=>7(x+1)=119
=>x+1=17
hay x=16
x-(60-x)=25x/100+25x/100
x-60+x=25x+25x/100
x-60+x=x(25+25)/100
x-60+x=x.50/100
x-60+x=x.50/50.2
x-60+x=x/2
.....
tới đây thui, mình ko biết làm nữa
ko biết là đúng hay sai đâu nhoa
a, \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
ta có: \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}=>\dfrac{3}{x}=\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{5-2y}{6}\)
=>\(\dfrac{3}{x}=\dfrac{5-2y}{6}=>x.\left(5-2y\right)=3.6=18\)
=> x và 5-2y thuộc Ư của 18={1,-1,2,-2,3,-3,6,-6}
vì 5-2y là số lẻ=> 5-2y= +-1 hoặc 5-2y=+-3
xét bảng
5-2y | 1 | -1 | 3 | -3 |
y | 2 | 3 | 1 | 4 |
x | 18 | -18 | 6 | -6 |
vậy giá trị x,y cần tìm là: {x=18.y=2}
{x=-18.y=3}
{x=6, y=1}Ư
{x=-6,y=4}
1a.Vì \(\left|x\right|\) là 1 số tự nhiên nên \(\left|x\right|+2017\ge2017\)(1)
Mà ta đã biết:\(\dfrac{a}{b}\ge\dfrac{a}{b+n}\)với n là một số tự nhiên.
Nên từ (1)suy ra\(\dfrac{2016}{\left|x\right|+2017}\le\dfrac{2016}{2017}\)
Vậy để \(\dfrac{2016}{\left|x\right|+2017}\)lớn nhất thì \(\dfrac{2016}{\left|x\right|+2017}=\dfrac{2016}{2017}\)
1b.Ta thấy:
\(\dfrac{\left|x\right|+2016}{-2017}=\dfrac{-\left(\left|x\right|+2016\right)}{2017}\)
Để \(\dfrac{-\left(\left|x\right|+2016\right)}{2017}\)lớn nhất thì \(-\left(\left|x\right|+2016\right)\)lớn nhất
Mà theo câu a,ta có:\(\left|x\right|\)+2016 là một số tự nhiên nên \(-\left(\left|x\right|+2016\right)\)mang dấu âm hay \(-\left(\left|x\right|+2016\right)\le0\)( chú ý \(-0=0\))
Vậy để \(-\left(\left|x\right|+2016\right)\)lớn nhất hay \(\dfrac{\left|x\right|+2016}{-2017}\)lớn nhất thì \(\left|x\right|+2016=0\)
\(\Rightarrow\)Để \(\dfrac{\left|x\right|+2016}{-2017}\)lớn nhất thì nó bằng \(\dfrac{0}{-2017}\)hay nó bằng 0
2)
a)Để \(\dfrac{\left|x\right|+1945}{1975}\)nhỏ nhất thì \(\left|x\right|+1945\) nhỏ nhất
Vì \(\left|x\right|\ge0\) nên \(\left|x\right|+1945\ge1945\)
\(\Rightarrow\)Để \(\left|x\right|+1945\) nhỏ nhất thì \(\left|x\right|+1945\) = 1945
\(\Rightarrow\)Để \(\dfrac{\left|x\right|+1945}{1975}\)bé nhất thì nó phải bằng \(\dfrac{1945}{1975}\)hay\(\dfrac{389}{395}\)
b)Để \(\dfrac{-1}{\left|x\right|+1}\)thì \(\left|x\right|+1\)bé nhất
Vì \(\left|x\right|\ge0\) nên \(\left|x\right|+1\ge1\)
\(\Rightarrow\)Để \(\left|x\right|+1\)bé nhất thì \(\left|x\right|+1\)\(=1\)
\(\Rightarrow\)GTNN của \(\dfrac{-1}{\left|x\right|+1}\)là \(\dfrac{-1}{1}\) hay -1
c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)
\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)
\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)
\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)
\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)
\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)
\(\Rightarrow x=-36\)
Vậy ...
a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)
\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)
\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)
\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)
Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)
Chúc bạn học tốt!!!
\(x+\dfrac{30}{100}.x=-1,31\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x.\left(1+\dfrac{30}{100}\right)=\dfrac{-131}{100}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x.\left(\dfrac{100}{100}+\dfrac{30}{100}\right)=\dfrac{-131}{100}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ x.\dfrac{26}{20}=\dfrac{-131}{100}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\\\ \\ x=\dfrac{-131}{100}:\dfrac{26}{20}=\dfrac{-131}{100}.\dfrac{20}{26}\\ \\ \\ \\ \\ \\ \\ \\ \\\\ \\ x=\dfrac{-131}{130}\)
coi chừng mình làm sai nhưng trinh bày mình đúng nhưng cẩn thận kết quả. nhé bye
x+\(\dfrac{30}{100}\).x=-1,31
x+0,3.x=-1,31
x.(0,3+1)=-1,31
x.1,3=-1,31
x=-1,31:1,3
x=\(\dfrac{-131}{130}\)
Vậy x=\(\dfrac{-131}{130}\)
Chúc bạn học tốt nha