a) |x|+x=1/3
b) |x|-x=3/4
Làm gấp vs mn
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10 tháng 6 2021
Giải:
a) \(\left(x-4\right).\left(y+1\right)=8\)
\(\Rightarrow\left(x-4\right)\) và \(\left(y+1\right)\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Ta có bảng giá trị:
| x-4 | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
| y+1 | -1 | -2 | -4 | -8 | 8 | 4 | 2 | 1 |
| x | -4 | 0 | 2 | 3 | 5 | 6 | 8 | 12 |
| y | -2 | -3 | -5 | -9 | 7 | 3 | 1 | 0 |
Vì \(\left(x;y\right)\in N\) nên \(\left(x;y\right)=\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)
Vậy \(\left(x;y\right)=\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)
b) \(\left(2x+3\right).\left(y-2\right)=15\)
\(\Rightarrow\left(2x+3\right)\) và \(\left(y-2\right)\inƯ\left(15\right)=\left\{\pm1;\pm3;\pm5;\pm15\right\}\)
| 2x+3 | -15 | -5 | -3 | -1 | 1 | 3 | 5 | 15 |
| y-2 | -1 | -3 | -5 | -15 | 15 | 5 | 3 | 1 |
| x | -9 | -4 | -3 | -2 | -1 | 0 | 1 | 6 |
| y | 1 | -1 | -3 | -13 | 17 | 7 | 5 | 3 |
Vì \(\left(x;y\right)\in N\) nên \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)
Vậy \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)
c) \(xy+2x+y=12\)
\(\Rightarrow x.\left(y+2\right)+\left(y+2\right)=14\)
\(\Rightarrow\left(x+1\right).\left(y+2\right)=14\)
\(\Rightarrow\left(x+1\right)\) và \(\left(y+2\right)\inƯ\left(14\right)=\left\{1;2;7;14\right\}\)
| x+1 | 1 | 2 | 7 | 14 |
| y+2 | 14 | 7 | 2 | 1 |
| x | 0 | 1 | 6 | 13 |
| y | 12 | 5 | 0 | -1 |
Vì \(\left(x;y\right)\in N\) nên \(\left(x;y\right)\in\left\{\left(0;12\right);\left(1;5\right);\left(6;0\right)\right\}\)
Vậy \(\left(x;y\right)\in\left\{\left(0;12\right);\left(1;5\right);\left(6;0\right)\right\}\)
d) \(xy-x-3y=4\)
\(\Rightarrow y.\left(x-3\right)-\left(x-3\right)=7\)
\(\Rightarrow\left(y-1\right).\left(x-3\right)=7\)
\(\Rightarrow\left(y-1\right)\) và \(\left(x-3\right)\inƯ\left(7\right)=\left\{1;7\right\}\)
Ta có bảng giá trị:
| x-3 | 1 | 7 |
| y-1 | 7 | 1 |
| x | 4 | 10 |
| y | 8 | 2 |
Vậy \(\left(x;y\right)\in\left\{\left(4;8\right);\left(10;2\right)\right\}\)
AT
3
PT
2
23 tháng 1 2024
a: \(\dfrac{x}{6}=\dfrac{8}{3}\)
=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)
b: \(\dfrac{5}{x}=\dfrac{4}{9}\)
=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)
c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)
=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)
=>x=-1-3=-4
d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)
=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)
=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)
=>\(x=-\dfrac{69}{8}\)
f: ĐKXĐ: x<>1
\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)
=>\(\left(x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)
KL
2
22 tháng 10 2023
a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)
=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)
b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)
=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)
=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)
LQ
2
11 tháng 10 2021
\(a,\Rightarrow\left|x+\dfrac{4}{9}\right|=\dfrac{3}{2}+\dfrac{1}{2}=2\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{9}=2\\x+\dfrac{4}{9}=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{14}{9}\\x=-\dfrac{22}{9}\end{matrix}\right.\\ b,\Rightarrow\left\{{}\begin{matrix}x-\dfrac{4}{11}=0\\5+y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{11}\\y=-5\end{matrix}\right.\)
11 tháng 10 2021
a) \(\left|x+\dfrac{4}{9}\right|-\dfrac{1}{2}=\dfrac{3}{2}\)
\(\Rightarrow\left|x+\dfrac{4}{9}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{9}=2\\x+\dfrac{4}{9}=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{14}{9}\\x=-\dfrac{22}{9}\end{matrix}\right.\)
b) \(\left|x-\dfrac{4}{11}\right|+\left|5+y\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{4}{11}=0\\5+y=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{11}\\y=-5\end{matrix}\right.\)
H
1
18 tháng 10 2021
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=k\Rightarrow x=3k;y=5k\)
\(x^2-y^2=-4\\ \Rightarrow9k^2-25k^2=-4\\ \Rightarrow-16k^2=-4\Rightarrow k^2=4\\ \Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6;y=10\\x=-6;y=-10\end{matrix}\right.\)
29 tháng 6 2021
\(a,=27x^3+27x^2+9x+1\)
\(b,=\dfrac{x^3}{27}-\dfrac{x^2}{3}+x-1\)
\(c,=-\left(27x^3-27x^2y^2+9xy^4-y^6\right)\)
\(=-27x^3+27x^2y^2-9xy^4+y^6\)
\(d,=\dfrac{x^3}{y^3}-\dfrac{6x}{y}+\dfrac{12y}{x}-\dfrac{8y^3}{x^3}\)
29 tháng 6 2021
a) \(\left(3x+1\right)^3=27x^3+27x^2+9x+1\)
b) \(\left(\dfrac{x}{3}-1\right)^3=\dfrac{x^3}{27}-\dfrac{x^2}{3}\)
c) \(\left(-y^2+3x\right)^3=27x^3-27x^2y^2+9xy^4-y^6\)
d) \(\left(\dfrac{x}{y}-\dfrac{2y}{x}\right)^3=\dfrac{x^3}{y^3}-\dfrac{6x}{y}+\dfrac{12y}{x}-\dfrac{8y^3}{x^3}\)
gấp ạ
\(a,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.nghiệm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{1}{6}\)
\(b,\Leftrightarrow\left|x\right|=\dfrac{3}{4}+x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}+x\left(x\ge0\right)\\x=-\dfrac{3}{4}-x\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=\dfrac{3}{4}\left(vô.nghiệm\right)\\x=-\dfrac{3}{8}\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow x=-\dfrac{3}{8}\)