Tìm x/y biết x d ) x y x 4 = 7 9
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1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
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áp dung tc cua day ti so bang nhau co
\(\frac{x}{2}=\frac{y}{3}=\frac{x+y}{2+3}=\frac{-15}{5}=-3\)
x=-6;y=-9
y b lam tuong tuu nhung thay cong bang tru
y c
co \(\frac{x}{y}=\frac{7}{-9}\Rightarrow\frac{x}{7}=\frac{y}{-9}\Rightarrow\frac{2x}{14}=\frac{3y}{-27}\)
lam tuong tuu y a
d,
h cheo
7 ( x + 4 ) = 4 ( 7 + y )
7x + 28 = 4y + 28
7x = 4y
\(\Rightarrow\frac{x}{4}=\frac{y}{7}\)
ap dung tc cua day ti so bang nhau va lam tuong tuu y a
t i c k nha
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a, \(x-\frac{1}{9}=\frac{8}{3}\Rightarrow x=\frac{8}{3}+\frac{1}{9}=\frac{25}{9}\)
\(-\frac{x}{4}=-\frac{9}{x}\Rightarrow x^2=-9.-4=36\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
\(\frac{x}{4}=\frac{18}{x+1}\Rightarrow x\left(x+1\right)=18.4\Rightarrow x\left(x+1\right)=72\Rightarrow x=8\)
\(\frac{x}{7}=\frac{9}{y}\Rightarrow xy=63.\) Bạn tự làm tiếp là ra nhé
x-1/9=8/3
x=8/3+1/9
x=25/9
b)-x/4=-9/x
=>x/4=9/x
=>x.x=9.4
=>x2=36
=>x\(\in\){-6;6}
c)x/4=18/x+1
=>x(x+1)=18.4
=>x(x+1)=72=8.9
=>x=8
d) x/7=9/y
=>x.y=9.7=63
Mà x>9 =>y<63:9=7
=>y=1 hoặc y=3
Với y=1, ta có x=63
Với y=3 ta có x=21
e) -2/x=y/5
=> x.y=-2.5=-10
Vì x<0<y nên ta có bảng sau
x | -1 | -2 | -5 | -10 |
y | 10 | 5 | 2 | 1 |
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a: =>4/x=y/21=4/7
=>x=7; y=21*4/7=12
b: x/7=9/y
=>xy=63
mà x>y
nên \(\left(x,y\right)\in\left\{\left(63;1\right);\left(21;3\right);\left(9;7\right);\left(-7;-9\right);\left(-3;-21\right);\left(-1;-63\right)\right\}\)
c: x/15=3/y
=>xy=45
mà x<y<0
nên \(\left(x,y\right)\in\left\{\left(-45;-1\right);\left(-15;-3\right);\left(-9;-5\right)\right\}\)
d: x/y=21/28=3/4
=>x/3=y/4=k
=>x=3k; y=4k(k\(\in Z\))
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a,\(\frac{-\chi}{4}=\frac{-9}{\chi}\Rightarrow-\chi.\chi=4.\left(-9\right)\)
\(\Rightarrow-2\chi=-36\Rightarrow\chi=-36:\left(-2\right)\)
\(\Rightarrow\chi=18\)
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a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
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1.a.
\(\left(x+3\right)\left(x-2\right)< 0\)
\(TH1:\hept{\begin{cases}x+3< 0\\x-2>0\end{cases}}\Rightarrow\hept{\begin{cases}x< -3\\x>2\end{cases}}\)
\(TH2:\hept{\begin{cases}x+3>0\\x-2< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-3\\x< 2\end{cases}}}\)
không biết có đúng không nữa!
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d: x+y=5
nên x=5-y
Ta có: xy=6
=>y(5-y)=6
=>y2-5y+6=0
=>(y-2)(y-3)=0
=>y=2 hoặc y=3
=>x=3 hoặc x=2
a: \(\Leftrightarrow\left(x-3;y+4\right)\in\left\{\left(1;-7\right);\left(-1;7\right);\left(-7;1\right);\left(7;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(4;-11\right);\left(2;3\right);\left(-4;-3\right);\left(10;-5\right)\right\}\)
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