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\(\Rightarrow x\in\left\{-9;-8;...;16\right\}\)
Tổng là: \(\dfrac{\left(16-9\right)\left(\dfrac{16+9}{1}+1\right)}{2}=91\)
1. \(\frac{-7}{12}\)< \(\frac{x-1}{4}\)< \(\frac{2}{3}\)
=> \(\frac{-7}{12}\)< \(\frac{3.\left(x-1\right)}{12}\)< \(\frac{8}{12}\)
=> 3 . ( x - 1 ) thuộc { - 6 ; - 5 ; - 4 ; - 3 ; - 2 ; - 1 ; 0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7}
Lập bảng tính giá trị x , cái này dễ lên bạn tự làm nha
1/ \(-\frac{7}{12}< \frac{x-1}{4}< \frac{2}{3}\)
hay \(\frac{-7}{12}< \frac{3.\left(x-1\right)}{12}< \frac{8}{12}\)
Vậy \(-7< 3.\left(x-1\right)< 8\)
Vậy \(3.\left(x-1\right)\in\left\{-6;-5;-4;...;7\right\}\)
mà \(x\in Z\)nên \(3.\left(x-1\right)⋮3\)
Vậy \(3.\left(x-1\right)\in\left\{-6;-3;0;3;6\right\}\)
hay \(x-1\in\left\{-2;-1;0;1;2\right\}\)
tới đây dễ rồi thì làm nốt nhé, để thời gian làm mấy câu sau!
Ta có : \(\frac{3}{x+2}=\frac{x+2}{3}\) <=> \(\left(x+2\right)^2=3^2\)
=> \(\orbr{\begin{cases}x+2=3\\x+2=-3\end{cases}}\)
=>\(\orbr{\begin{cases}x=3-2=1\\x=-3-2=-5\end{cases}}\)
Vậy tập hợp các số nguyên x thỏa mãn là { 1 ; -5 }
\(\frac{3}{x+2}=\frac{x+2}{3}\Rightarrow3^2=\left(x+2\right)^2=9\)
\(\Rightarrow x+2=3\)hoặc \(-3\)
Với \(x+2=3\Rightarrow x=1\)
Với \(x+2=-3\Rightarrow x=-5\)
1) \(-4< x< 3\)
\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2\right\}\)
Tổng:
\(\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2\)
\(=\left(-2+2\right)+\left(-1+1\right)+0-3\)
\(=-3\)
2) \(-5< x< 5\)
\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)
Tổng:
\(\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+3\)
\(=\left(-4+4\right)+\left(-3+3\right)+\left(-2+2\right)+\left(-1+1\right)+0\)
\(=0\)
3) \(-10< x< 6\)
\(\Rightarrow x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
Tổng:
\(\left(-9\right)+\left(-8\right)+\left(-7\right)++\left(-6\right)+\left(-5\right)+\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+4+5\)
\(=-24\)
4) \(-6< x< 5\)
\(\Rightarrow x\in\left\{-5;-4;-3;-2;-1;0;1;2;3;4\right\}\)
Tổng:
\(\left(-5\right)+\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+4\)
\(=\left(-4+4\right)+\left(-3+3\right)+\left(-2+2\right)+\left(-1+1\right)+0-5\)
\(=-5\)
5) \(-5< x< 2\)
\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1\right\}\)
Tổng:
\(\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1\)
\(=\left(-1+1\right)+0+\left(-4-3-2\right)\)
\(=-6\)
\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)
\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)
\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)
\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)
=> x - 29 = 0
=> x = 29.
a) \(\frac{x}{7}+\frac{1}{14}=-\frac{1}{y}\)
\(\Rightarrow\frac{2x}{14}+\frac{1}{14}=\frac{-1}{y}\)
\(\Rightarrow\frac{2x+1}{14}=\frac{-1}{y}\)
\(\Rightarrow\left(2x+1\right).y=\left(-1\right).14=\left(-14\right)\)
Ta có bảng sau :
2x + 1 | 1 | -1 | 14 | -14 | 2 | -2 | 7 | -7 |
2x | 0 | -2 | 13 | -15 | 1 | -3 | 6 | -8 |
x | 0 | -1 | \(\frac{13}{2}\) | \(\frac{-15}{2}\) | \(\frac{1}{2}\) | \(\frac{-3}{2}\) | 3 | -4 |
y | -14 | 14 | -1 | 1 | -7 | 7 | -2 | 2 |
Vậy \(\left(x;y\right)\in\left\{\left(-1;14\right),\left(3;-2\right),\left(0;-14\right),\left(-4;2\right)\right\}\)
b) \(\frac{x}{9}+-\frac{1}{6}=-\frac{1}{y}\)
\(\Rightarrow\frac{2x}{18}+\frac{-3}{18}=\frac{-1}{y}\)
\(\Rightarrow\frac{2x-3}{18}=\frac{-1}{y}\)
\(\Rightarrow\left(2x-3\right).y=\left(-1\right).18=\left(-18\right)\)
Ta có bảng :
2x - 3 | 1 | -1 | 18 | -18 | 3 | -3 | 6 | -6 | 9 | -9 | -2 | 2 | ||||
2x | 4 | 2 | 21 | -15 | 6 | 0 | 9 | -3 | 12 | -6 | 1 | 5 | ||||
x | 2 | 1 | \(\frac{21}{2}\) | \(\frac{-15}{2}\) | 3 | 0 | \(\frac{9}{2}\) | \(\frac{-3}{2}\) | 6 | -3 | \(\frac{1}{2}\) | \(\frac{5}{2}\) | ||||
y | -18 | 18 | -1 | 1 | -6 | 6 | -3 | 3 | -2 | 2 | 9 | -9 |
Vậy \(\left(x;y\right)\in\left\{\left(2;-18\right),\left(1;18\right),\left(3;-6\right),\left(0;6\right),\left(6;-2\right),\left(-3,2\right)\right\}\)
\(\frac{10+x}{17+x}=\frac{3}{4}\)=>3.(17+x)=4.(10+x)
= 51+3x=40+4x
=>51-40=4x-3x
=>11=x
vậy x=11