tìm các số nguyên x, y biết
-5^x+y^16=-18^72
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{-5}{x}=\frac{y}{16}=-\frac{18}{72}\)
\(\Rightarrow\)\(x=\frac{\left(-5\right).72}{-18}=-20\)
\(y=\frac{\left(-18\right).16}{72}=4\)
Vậy....
\(\frac{-5}{x}\)=\(\frac{y}{16}\)=\(\frac{-18}{72}\)
Ta có: \(\frac{-18}{72}\)= \(\frac{\left(-18\right):18}{72:18}\)=\(\frac{-1}{4}\)
=>\(\frac{-5}{x}\)= \(\frac{y}{16}\)= \(\frac{-1}{4}\)
=> \(\frac{-5}{x}\)= \(\frac{-1}{4}\): \(\frac{y}{16}\)= \(\frac{-1}{4}\)
+ \(\frac{-5}{x}\)= \(\frac{-1}{4}\)
=> (-5). 4= x. (-1)
x. (-1) = (-5). 4
x. (-1)= -20
x= (-20): (-1)
x= 20. \(\left(1\right)\)
+ \(\frac{y}{16}\)= \(\frac{-1}{4}\)
=> y. 4= 16. (-1)
y.4= -16
y= (-16): 4
y= -4. \(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)ta được giá trị x, y thỏa mãn bài toán là: x= 20 và y= -4.
Mik ko chắc chắn lắm nên bạn kiểm tra lại và cho mik ý kiến nhak!
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
Xét \(\frac{-5}{x}=\frac{-18}{72}\)
\(\Rightarrow\)(-5) . 72 = x . (-18)
\(\Rightarrow\)-360 = x . (-18)
\(\Rightarrow\)x = (-360) : (-18) = 20
Xét \(\frac{y}{16}=\frac{-18}{72}\)
\(\Rightarrow\)72y = 16 . (-18)
\(\Rightarrow\)72y = -288
\(\Rightarrow\)y = (-288) : 72 = -4
Vậy x = 20 ; y = -4
Ta có :
\(\frac{-5}{x}=\frac{-18}{72}\Rightarrow\frac{-5}{x}=\frac{-1}{4}\)
\(\Rightarrow-5.4=x.-1\)
\(\Rightarrow-20=-x\)
\(\Rightarrow x=20\)
Thay x = 20 vào \(\frac{-5}{x}\)ta được :
\(\frac{-5}{20}=\frac{y}{16}\)\(\Rightarrow\frac{-1}{4}=\frac{y}{16}\Rightarrow-1.16=y.4\)
\(\Rightarrow-16=y.4\)
\(\Rightarrow y=-4\)
Vậy x = 20 , y = -4
Tk mk nha !!!
Ta có : 2x + xy - 3y = 18
=> x(y + 2) - 3y = 18
=> x(y + 2) - 3y - 6 = 18 - 6
=> x(y + 2) - 3(x + 2) = 12
=> (x - 3)(y + 2) = 12
Vì \(x;y\inℤ\Rightarrow\hept{\begin{cases}x-3\inℤ\\y+2\inℤ\end{cases}}\)
Lại có : 12 = 1.12 = 3.4 = 2.6 = (-1).(-12) = (-3).(-4) = (-2).(-6)
Lập bảng xét 12 trường hợp
x - 3 | 1 | 12 | -1 | -12 | 3 | 4 | -3 | -4 | 2 | 6 | -2 | -6 |
y + 2 | 12 | 1 | -12 | -1 | 4 | 3 | -4 | -3 | 6 | 2 | -6 | -2 |
x | 4 | 15 | 2 | -9 | 6 | 7 | 0 | -1 | 5 | 9 | 1 | -3 |
y | 10 | -1 | -14 | -3 | 2 | 1 | -6 | -5 | 4 | 0 | -8 | -4 |
Vậy các cặp số (x;y) nguyên thỏa mãn là : (4 ; 10) ; (15 ; - 1) ; (2 ; -14) ; (-9 ; -3) ; (6 ; 2) ; (7 ; 1) ; (0 ; -6) ; (-1 ' 5) ; (5 ; 4) ; (9 ; 0) ;
(1 ; -8) ; (-3 ; -4)
b) \(\left(x^2-5\right)\left(x^2-25\right)< 0\)
TH1 : \(\hept{\begin{cases}x^2-5>0\\x^2-25< 0\end{cases}\Rightarrow\hept{\begin{cases}x^2>5\\x^2< 25\end{cases}}\Rightarrow5< x^2< 25\Rightarrow x^2\in\left\{9;16\right\}}\)(vì x là số nguyên)
=> \(x\in\left\{\pm3;\pm4\right\}\)
TH2 : \(\hept{\begin{cases}x^2-5< 0\\x^2-25>0\end{cases}}\Rightarrow\hept{\begin{cases}x^2< 5\\x^2>25\end{cases}}\Rightarrow x\in\varnothing\)
Vậy \(x\in\left\{\pm3;\pm4\right\}\)
2x + xy - 3y = 18
<=> 2x + xy - 6 - 3y = 12
<=> ( 2x + xy ) - ( 6 + 3y ) = 12
<=> x( 2 + y ) - 3( 2 + y ) = 12
<=> ( x - 3 )( 2 + y ) = 12
Lập bảng :
x-3 | 1 | -1 | 2 | -2 | 3 | -3 | 4 | -4 | 6 | -6 | 12 | -12 |
x | 4 | 2 | 5 | 1 | 6 | 0 | 7 | -1 | 9 | -3 | 15 | -9 |
2+y | 12 | -12 | 6 | -6 | 4 | -4 | 3 | -3 | 2 | -2 | 1 | -1 |
y | 10 | -14 | 4 | -8 | 2 | -6 | 1 | -5 | 0 | -4 | -1 | -3 |
Vậy ta có 12 cặp ( x ; y ) thỏa mãn
( 4 ; 10 ) , ( 2 ; -14 ) , ( 5 ; 4 ) , ( 1 ; -8 ) , ( 6 ; 2 ) , ( 0 ; -6 ) , ( 7 ; 1 ) , ( -1 ; -5 ) , ( 9 ; 0 ) , ( -3 ; -4 ) , ( 15 ; -1 ) , ( -9 ; -3 )
a) \(\dfrac{x}{-6}=\dfrac{-15}{45}\)
\(\dfrac{-x}{6}=\dfrac{-15}{45}\)
\(\dfrac{x}{6}=\dfrac{15}{45}\)
\(x=\dfrac{\left(15\cdot6\right)}{45}\)
\(x=2\)
b) \(\dfrac{x}{5}=\dfrac{16}{25}\)
\(x=\dfrac{\left(16\cdot5\right)}{25}\)
\(x=\dfrac{80}{25}\)
\(x=\dfrac{16}{5}\)
c) \(\dfrac{5}{x-3}=\dfrac{20}{-12}\)
\(x-3=\dfrac{\left(5\cdot-12\right)}{20}\)
\(x-3=-3\)
\(x=\left(-3\right)+3\)
\(x=0\)
d) \(\dfrac{2}{5}\cdot x=\dfrac{6}{35}\)
\(x=\dfrac{6}{35}\div\dfrac{2}{5}\)
\(x=\dfrac{3}{7}\)