( x + 3 ) . ( y - 5 ) =-25
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\(D=4^2+4^3+...+4^{25}\)
\(\Rightarrow4D=4^3+4^4+...+4^{26}\)
\(\Rightarrow3D=4D-D=4^3+4^4+...+4^{26}-4^2-4^3-...-4^{25}=4^{26}-4^2\)
\(\Rightarrow D=\dfrac{4^{26}-4^2}{3}\)
2) Ta có: \(\left(2x+1\right).\left(3y-2\right)=-55=\left(-1\right).55=1.\left(-55\right)=\left(-5\right).11=5.\left(-11\right)\)
- Ta có bảng giá trị:
\(2x+1\) | \(-55\) | \(-11\) | \(-5\) | \(-1\) | \(1\) | \(5\) | \(11\) | \(55\) |
\(3y-2\) | \(1\) | \(5\) | \(11\) | \(55\) | \(-55\) | \(-11\) | \(-5\) | \(-1\) |
\(x\) | \(-28\) | \(-6\) | \(-3\) | \(-1\) | \(0\) | \(2\) | \(5\) | \(27\) |
\(y\) | \(1\) | \(\frac{7}{3}\) | \(\frac{13}{3}\) | \(19\) | \(-\frac{53}{3}\) | \(-3\) | \(-1\) | \(\frac{1}{3}\) |
\(\left(TM\right)\) | \(\left(L\right)\) | \(\left(L\right)\) | \(\left(TM\right)\) | \(\left(L\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(L\right)\) |
Vậy \(\left(x,y\right)\in\left\{\left(-28,1\right);\left(-1,19\right);\left(2,-3\right);\left(5,-1\right)\right\}\)
3) Ta có: \(\left(x-2\right).\left(y+3\right)=5=\left(-1\right).\left(-5\right)=1.5\)
- Ta có bảng giá trị:
\(x-2\) | \(-1\) | \(1\) | \(-5\) | \(5\) |
\(y+3\) | \(-5\) | \(5\) | \(-1\) | \(1\) |
\(x\) | \(1\) | \(3\) | \(-3\) | \(7\) |
\(y\) | \(-8\) | \(2\) | \(-4\) | \(-2\) |
\(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(\left(x,y\right)\in\left\{\left(1,-8\right);\left(3,2\right);\left(-3,-4\right);\left(7,-2\right)\right\}\)
4) Ta có: \(\left(2x+3\right).\left(y-5\right)=10=\left(-1\right).\left(-10\right)=1.10=\left(-2\right).\left(-5\right)=2.5\)
- Vì \(x\in Z\)mà \(2x+3\)là số lẻ \(\Rightarrow\)\(2x+3\in\left\{-1,1,-5,5\right\}\)
- Ta có bảng giá trị:
\(2x+3\) | \(-1\) | \(1\) | \(-5\) | \(5\) |
\(y-5\) | \(-10\) | \(11\) | \(-2\) | \(2\) |
\(x\) | \(-2\) | \(-1\) | \(-4\) | \(1\) |
\(y\) | \(-5\) | \(16\) | \(3\) | \(7\) |
\(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) | \(\left(TM\right)\) |
Vậy \(\left(x,y\right)\in\left\{\left(-2,-5\right);\left(-1,16\right);\left(-4,3\right);\left(1,7\right)\right\}\)
Theo đề: \(2x+y=0\Leftrightarrow y=-2x\) \(\left(1\right)\)
Ta có:
\(\dfrac{3-x}{y-4}=\dfrac{2}{5}\)
\(\Leftrightarrow5\left(3-x\right)=2\left(y-4\right)\)
\(\Leftrightarrow15-5x=2y-8\)
\(\Leftrightarrow15+8=2y+5x\)
\(\Leftrightarrow5x+2y=23\) \(\left(2\right)\)
Thế (1) vào (2), suy ra:
\(5x+2.\left(-2x\right)=23\)
\(\Leftrightarrow5x-4x=23\)
\(\Leftrightarrow x=23\)
\(\Rightarrow y=-2.23=-46\)
\(\dfrac{2}{5}:\dfrac{6}{25}\\ =\dfrac{2}{5}\times\dfrac{25}{6}\\ =\dfrac{2}{5}\times\dfrac{5\times5}{2\times3}=\dfrac{5}{3}\)
\(\dfrac{2}{5}:\dfrac{6}{25}\)
\(=\dfrac{2}{5}\times\dfrac{25}{6}\)
\(=\dfrac{2}{5}\times\dfrac{5\times5}{2\times3}\)
\(=\dfrac{5}{3}\)
a, Xét : \(\frac{x}{-30}=-\frac{12}{20}=-\frac{3}{5}\Leftrightarrow5x=90\Leftrightarrow x=18\)
Xét : \(\frac{-36}{y}=\frac{-3}{5}\Leftrightarrow3y=180\Leftrightarrow y=60\)
Vậy \(x=18;y=60\)
b, \(\frac{x-1}{7}=\frac{2y+5}{3}\)và \(x+2y=-16\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x-1}{7}=\frac{2y+5}{3}=\frac{x+2y-1+5}{7+3}=\frac{-16+4}{10}=\frac{-12}{10}=-\frac{6}{5}\)
\(\Leftrightarrow\frac{x-1}{7}=-\frac{6}{5}\Leftrightarrow5x-5=-42\Leftrightarrow5x=-37\Leftrightarrow x=-\frac{37}{5}\)
\(\Leftrightarrow\frac{2y+5}{3}=-\frac{6}{5}\Leftrightarrow10y+25=-18\Leftrightarrow10y=-43\Leftrightarrow y=-\frac{43}{10}\)
Ta có: -25 = -1 . 25 = 1 . -25
= 5 . -5
Ta có bảng
x + 3 1 -1 -25 25 5 -5
y - 5 -25 25 1 -1 -5 5
x -2 -4 -28 22 2 -8
y -20 30 6 4 0 10
Vậy các cặp số nguyên ( x ; y ) \(\in\){ (-2;-20) ; (-4;30) ; (-28;6); (22;4) ; ( 2 ; 0 ) ; (-8;10) }
# HOK TỐT #