Tìm x , y sao cho :
x/y = 4/9 và 3x - 2y = -12
Làm giúp mình với , mình đang cần gấp . Cảm ơn các bạn nhiều !!!
Tìm các sô nguyên x,y thoả mãn x>y>1 và 2x+2y+1 chia hết cho xy. Các bạn giúp mình với, mình đang cần gấp. Mình cảm ơn!!!
Trước hết ta thấy rằng nếu có một trong hai số xy chẵn còn 2x+2y+1 không thể chia hết cho
tìm số tự nhiên x,y biết x^2-3x+2=6
MK ĐANG CẦN GẤP !!!!GIÚP MÌNH VỚI!!
cảm ơn các bạn nhiều
x²-3x+2=6
=>x²-3x=4
=>x.(x-3)=4
=>x và x-3 thuộc Ư(4)
Làm nốt nhé. Bạn chia TH ra thì hai cái này cùng dấu và tính
Trường hợp. Bạn chia ra đi mình đg bận gợi ý vậy là làm đc rồi đó
Tìm x, y, z ϵ Q , biết x(x + y + z) = -5; y(x + y + z) = 9; z(x + y + z) = 5
Gíup mình với mình đang cần gấp!!! Cảm ơn các bạn nhiều!!!!
Ai giải được cho 100 like!!!!!!!!
Từ 3 phương trình trên
\(\left(x+y+z\right)=\dfrac{-5}{x}=\dfrac{9}{y}=\dfrac{5}{z}=\dfrac{-5+9+5}{x+y+z}=\dfrac{9}{x+y+z}\)
\(\Rightarrow\left(x+y+z\right)^2=9\Rightarrow\left(x+y+z\right)=\pm3\)
+ Với \(x+y+z=3\) Thay vào từng phương trình ta có
\(x=-\dfrac{5}{3};y=3;z=\dfrac{5}{3}\)
+ Với \(x+y+z=-3\) Thay vào từng phương trình có
\(x=\dfrac{5}{3};y=3;z=-\dfrac{5}{3}\)
Mình cần gấp:
Tìm x,y thuộc Z, biết: 8x+y = 2y-9
Bạn nào giúp, mình xin cảm ơn nhiều!
tìm phần số x/9 (x thuộc z) sao cho:
x/9<4/7<x+1/9
GIÚP MÌNH VỚI Ạ. MÌNH CẦN GẤP. CẢM ƠN CÁC BẠN NHIỀU!
CÔ NGUYỄN THỊ THƯƠNG HOÀI GIÚP EM VỚI Ạ
\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)
\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)
7\(x\) < 36 < 63\(x\) + 7
⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)
\(\dfrac{29}{63}\)< \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}
⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)
tìm phần số x/9 (x thuộc z) sao cho:
x/9<4/7<x+1/9
GIÚP MÌNH VỚI Ạ. MÌNH CẦN GẤP. CẢM ƠN CÁC BẠN NHIỀU!
CÔ NGUYỄN THỊ THƯƠNG HOÀI GIÚP EM VỚI Ạ
\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)
=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)
\(\Rightarrow7x< 36< 7x+7\)
\(\Rightarrow x< \dfrac{36}{7}< x+1\)
\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)
\(\Rightarrow x=5\)
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tìm các số nguyên x và y sao cho:xy+3x-2y=11
giúp mình với đang cần gấp hạn mai rồi bạn nào nhanh mình tick
xy+3x-2y=11
\(\Rightarrow x.\left(y+3\right)-2.\left(y+3\right)=17\)
\(\Rightarrow\left(x-2\right).\left(y+3\right)=17\)
\(\Rightarrow17⋮x-2\)
\(\Rightarrow x-2\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
+)Ta có bảng:
x-2 | -1 | 1 | -7 | 7 |
y+3 | -7 | 7 | -1 | 1 |
x | 1\(\in Z\) | 3\(\in Z\) | -5\(\in Z\) | 9\(\in Z\) |
y | -10\(\in Z\) | 4\(\in Z\) | -4\(\in Z\) | -2\(\in Z\) |
Vậy \(\left(x,y\right)\in\left\{\left(1;-10\right);\left(3;4\right);\left(-5;-4\right);\left(9;-2\right)\right\}\)
Chúc bn học tốt
Ban kia sai r ! vì trừ VT thì phải trừ VP chứ ? sao lại trừ VT mà cộng VP ?
\(xy+3x-2y=11\)
\(=>x.\left(y+3\right)-2.\left(y+3\right)=5\)
\(=>\left(x-2\right).\left(y+3\right)=5\)
\(Do:x;y\inℤ=>x-2;y+3\inℤ\)
\(=>x-2;y+3\inƯ\left(5\right)\)
Nên ta có bảng sau :
x-2 | 1 | 5 | -1 | -5 |
y+3 | 5 | 1 | -5 | -1 |
x | 3 | 7 | 1 | -3 |
y | 2 | -2 | -8 | -4 |
Quân sai rồi 2.(y+3)=2y+6 chứ nhề xem lại đi bn ơi
Chúc bn học tốt
Tìm x:
-x - \(\dfrac{1}{9}\) = -\(\dfrac{2}{45}\)
Giúp mình với nha. Mình đang cần gấp. Cảm ơn các bạn nhiều!!!
x=2/45+1/9
x=7/45
Mình tưởng -\(\frac{2}{45}\)
Các bạn ơi giúp mình với mình đang cần gấp
Chứng minh rằng:
\(\frac{2x^3+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}=\frac{1}{x-y}\)
Cảm ơn nhiều
Sửa lại đề : \(\frac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}\)
Ta có : \(\frac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}\) \(=\) \(\frac{2x^2+3xy+y^2}{\left(x-y\right)\left(2x^2+3xy+y^2\right)}\)
\(=\frac{1}{x-y}\) ( Chia cả tử và mẫu cho \(2x^2+3xy+y^2\))