Help me
\(\dfrac{x+3}{5}-\dfrac{1}{y-1}=\dfrac{2}{5}\)
\(-1-\dfrac{7}{14}\le\dfrac{x}{3}\le1+\dfrac{12}{48}\)
\(\dfrac{2y+7}{y-2}\in Z\)
Bài 1 : Tìm GTNN
A = \(\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|+\dfrac{13}{14}\)
Bài 2 : Tìm x , y là số nguyên
a) \(\dfrac{x+3}{5}-\dfrac{1}{y-1}=\dfrac{2}{5}\)
b) \(-1-\dfrac{7}{14}\le\dfrac{x}{3}\le1+\dfrac{12}{48}\)
c) \(\dfrac{2y+7}{y-2}\in Z\)
giúp mk với sáng mai mk có tiết toán
Bài 1:
Ta có: \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2\ge0\\\left|y+\dfrac{1}{4}\right|\ge0\end{matrix}\right.\Rightarrow\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|\ge0\)
\(\Rightarrow A=\left(x-\dfrac{1}{4}\right)^2+\left|y+\dfrac{1}{4}\right|+\dfrac{13}{14}\ge\dfrac{13}{14}\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{4}\right)^2=0\\\left|y+\dfrac{1}{4}\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{4}=0\\y+\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{-1}{4}\end{matrix}\right.\)
Vậy \(MIN_A=\dfrac{13}{14}\) khi \(x=\dfrac{1}{4};y=-\dfrac{1}{4}\)
Nguyễn Huy TúAce LegonaNghiêm Gia Phương
Hoàng Thị Ngọc AnhHoang Hung QuanngonhuminhĐức Huy ABC
An NguyễnNguyễn Nhật MinhPhạm Nguyễn Tất Đạt
Đức Minh
tìm x, y, z khi
1) \(\dfrac{x}{7}=\dfrac{y}{3}\) và x-24=y
2) \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}\) và y-x=48
3) \(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}\) và x-y=4009
4) \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\) và x-y-z=28
5) \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\) và 2x+3y-z=-14
6) 3x=y; 5y=4z và 6x+7y+8z=456
a, Ta có:
\(x-24=y\\ x-y=24\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{24}{4}=6\)
+) \(\dfrac{x}{7}=6\Rightarrow x=6\cdot7=42\)
+) \(\dfrac{y}{3}=6\Rightarrow6\cdot3=18\)
Vậy \(x=42;y=18\)
b, Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}=\dfrac{y-z}{7-2}=\dfrac{48}{5}=9,6\)
+) \(\dfrac{x}{5}=9,6\Rightarrow x=9,6\cdot5=48\)
+) \(\dfrac{y}{7}=9,6\Rightarrow y=9,6\cdot7=67,2\)
+) \(\dfrac{z}{2}=9,6\Rightarrow z=9,6\cdot2=19,2\)
Vậy \(x=48;y=67,2;z=19,2\)
mk giải đc bao nhiêu thì bn làm bấy nhiêu nha
bài tìm x,y,z biết :a)\(\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{z}{7}\)và x+y-z=69
b)\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\)và 2x-3y+z=6
c)\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{6}\)và x+y=14
d)\(\dfrac{2}{3x}=\dfrac{1}{2y}=\dfrac{2}{z}\)và 3x+2y+z=1
a) \(\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{z}{7}\)và \(x+y-z=69\)
Theo đề bài, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{5}\times\dfrac{1}{8}=\dfrac{y}{6}\times\dfrac{1}{8}\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}\)(1)
\(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{8}\times\dfrac{1}{6}=\dfrac{z}{7}\times\dfrac{1}{6}\Rightarrow\dfrac{y}{48}=\dfrac{z}{42}\)(2)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}=\dfrac{x+y-z}{40+48-42}=\dfrac{69}{46}=\dfrac{3}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{40}=\dfrac{3}{2}\Rightarrow x=\dfrac{40\times3}{2}=60\\\dfrac{y}{48}=\dfrac{3}{2}\Rightarrow y=\dfrac{48\times3}{2}=72\\\dfrac{z}{42}=\dfrac{3}{2}\Rightarrow z=\dfrac{42\times3}{2}=63\end{matrix}\right.\)
Vậy \(\Rightarrow\left\{{}\begin{matrix}x=60\\y=72\\z=63\end{matrix}\right.\)
Ta có:\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\)(Nhân 2 vế với \(\dfrac{1}{4}\))
\(\dfrac{y}{8}=\dfrac{x}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)(Nhân 2 vế với \(\dfrac{1}{3}\))
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)và x+y-z=6
Áp dụng tính chất dãy tỉ số bằng nhau. Ta có:
\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y-z}{20+24-21}=\dfrac{69}{23}=3\)
Vì \(\dfrac{x}{20}=3\Rightarrow x=20.3=60\)
\(\dfrac{y}{24}=3\Rightarrow y=24.3=72\)
\(\dfrac{z}{21}=3\Rightarrow z=3.21=63\)
Vậy x=60; y=72; z=63
a) \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\)và \(2x-3y+z=6\)
Theo đề bài, ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{3}\times\dfrac{1}{3}=\dfrac{y}{4}\times\dfrac{1}{3}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\)(1)
\(\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{3}\times\dfrac{1}{4}=\dfrac{z}{5}\times\dfrac{1}{4}\Rightarrow\dfrac{y}{12}=\dfrac{z}{20}\)(2)
Từ (1) và (2), ta có: \(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
\(\Leftrightarrow\dfrac{x}{9}\Rightarrow\dfrac{2x}{18};\dfrac{y}{12}\Rightarrow\dfrac{3y}{36}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\Rightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x}{18}=3\Rightarrow x=\dfrac{18\times3}{2}=27\\\dfrac{3y}{36}=3\Rightarrow y=\dfrac{36\times3}{3}=36\\\dfrac{z}{20}=3\Rightarrow z=20\times3=60\end{matrix}\right.\)
Vậy \(\Rightarrow\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)
TÌM CÁC SỐ x,y,z BIẾT:
a. 5x = 2y = 3z và x + y - z = 33
b.2x = -3y = 4z và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) = 3
c.\(\dfrac{x-1}{2}=\dfrac{7-2}{3}=\dfrac{z-3}{4}\) và x - 2y +3z = 14
d.\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và \(^{x^2}\) +\(2y^2-z^2\) = -12
GIÚP MK NHA
a) Giải
Vì \(5x=2y=3z\)
\(\Rightarrow\dfrac{5x}{30}=\dfrac{2y}{30}=\dfrac{3z}{30}\)
\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{x+y-z}{6+15-10}=\dfrac{33}{11}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{6}=3\Rightarrow x=18\\\dfrac{y}{15}=3\Rightarrow y=45\\\dfrac{z}{10}=3\Rightarrow z=30\end{matrix}\right.\)
Vậy \(x=18,\) \(y=45\) hoặc \(z=30.\)
c) Giải
(Vì mk bt bạn bấm nhầm nên đề bị sai, mk sửa 7 \(\rightarrow\) y do trên bàn phím, 7 với y ở vị trí gần nhau mà 2 với y ở cách xa nhau nên sửa như vậy nhé)
Vì \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
\(\Rightarrow\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{\left(x-1\right)-\left(2y-4\right)+\left(3z-9\right)}{4-6+12}=\dfrac{x-1-2y+4+3z-9}{10}\)
\(=\dfrac{\left(x-2y+3z\right)-\left(1-4+9\right)}{10}=\dfrac{14-6}{10}=\dfrac{4}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{4}{5}\Rightarrow x=\dfrac{13}{5}\\\dfrac{y-2}{3}=\dfrac{4}{5}\Rightarrow y=\dfrac{22}{5}\\\dfrac{z-3}{4}=\dfrac{4}{5}\Rightarrow z=\dfrac{31}{5}\end{matrix}\right.\)
Vậy \(x=\dfrac{13}{5},\) \(y=\dfrac{22}{5}\) và \(z=\dfrac{31}{5}.\)
c) Giải
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
Mà \(x^2+2y^2-z^2=-12\)
\(\Rightarrow\left(2k\right)^2+2\left(3k\right)^2-\left(5k\right)^2=-12\)
\(\Rightarrow4.k^2+18.k^2-25.k^2=-12\)
\(\Rightarrow\left(-3\right)k^2=-12\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=\pm2\)
\(\circledast k=-2\Rightarrow\left\{{}\begin{matrix}x=-4\\y=-6\\z=-10\end{matrix}\right.\)
\(\circledast k=2\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=10\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-4;y=-6;z=-10\\x=4;y=6;z=10\end{matrix}\right..\)
Tìm các số nguyên x,y biết:
a)\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
b) \(\dfrac{24}{7x-3}=\dfrac{-4}{25}\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
d) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
f) \(y\dfrac{5}{y}=\dfrac{86}{y}\) ( \(x\dfrac{2}{5};y\dfrac{5}{y}\) là các hỗn số)
a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)
⇒\(2x+1=21\)
\(2x=21-1\)
\(2x=20\)
⇒\(x=10\)
Tìm x,y,z khi :
\(\dfrac{x}{7}=\dfrac{y}{3}\) và x - 24 = y
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}\) và y - x = 48
\(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}\) và x - y = 4009
\(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\) và z - y - z =28
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\) và 2x + 3y - z = -14
3x = y ; 5y = 4z và 6x + 7y + 8z = 456
Ta có : x - 24 = y
=> x - y = 24
Lại có : \(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{24}{4}=6\)
( theo tính chất của dãy tỉ số bằng nhau )
Nên \(\dfrac{x}{7}=6\) => x = 42
\(\dfrac{y}{3}=6\) => y = 18
Vậy x = 42, y = 18
Ta có :\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}=\dfrac{y-x}{7-5}=\dfrac{48}{2}=24\)
( theo tính chất dãy tỉ số bằng nhau )
Nên \(\dfrac{x}{5}=24\) => x = 120
\(\dfrac{y}{7}=24\) => y = 168
\(\dfrac{z}{2}=24\) => z = 48
Vậy x = 120, y = 168, z = 48
Ta có : \(\dfrac{x}{2}=\dfrac{y}{3}\) => \(\dfrac{x}{8}=\dfrac{y}{12}\)
\(\dfrac{y}{4}=\dfrac{z}{5}\) => \(\dfrac{y}{12}=\dfrac{z}{15}\)
Nên ...
(BẠN TỰ LÀM NHÉ!!!!!!!!!
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)
\(\dfrac{1}{4}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)
Vậy \(x\in\left\{22;23;24;...\right\}\)
a) \(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11}{55}+\dfrac{10}{55}< \dfrac{x}{55}< \dfrac{22}{55}+\dfrac{1}{55}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{23}{55}\)
\(\Rightarrow\) \(x=22\)
b) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)
\(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}< x\le\dfrac{26}{8}+\dfrac{14}{8}\)
\(1< x\le5\)
\(\Rightarrow\) \(x\in\) {\(2;3;4;5\)}
c) \(\dfrac{1}{3}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\)
Ko biết làm
d) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)
\(\dfrac{79}{15}+\dfrac{21}{15}+\dfrac{-40}{15}\le x\le\dfrac{40}{12}+\dfrac{45}{12}+\dfrac{23}{12}\)
\(4\le x\le9\)
\(\Rightarrow\) \(x\in\) {\(4;5;6;7;8;9\)}
Tìm x,y,z trong dãy tỉ số bằng nhau
1)\(\dfrac{3x}{8}=\dfrac{3y}{64}=\dfrac{3z}{216}\)và \(2x^2+2y^2.z^2=1\)
2) \(\dfrac{2x+1}{5}=\dfrac{4y-5}{9}=\dfrac{2x+4y-4}{7x}\)
3) \(\dfrac{x^3+y^3}{6}=\dfrac{x^3-2y^3}{4}\)và x6 . y6 =14
4) \(\dfrac{x+4}{6}=\dfrac{3y-1}{8}=\dfrac{3y-x-5}{x}\)
5) \(\dfrac{3}{x-1}=\dfrac{4}{y-2}=\dfrac{5}{z-3}\)và x.y.z=192
6)\(\dfrac{x-y}{3}=\dfrac{x+y}{13}=\dfrac{x.y}{200}\)
7)\(\dfrac{x+1}{2}=\dfrac{y-1}{3}=\dfrac{z+2}{4}=\dfrac{x+y+z+2}{2x+5}\)
8) \(\dfrac{15}{x-9}=\dfrac{20}{y-12}=\dfrac{40}{z-24}\)và x.y = 1200
9)\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\) và x.y.z = 22400
10)15x = -10y =6z và x.y.z = -30000
11) Cho\(\dfrac{x+1}{3}=\dfrac{y-2}{5}=\dfrac{2z+14}{9}\)và x+z=y
12) Cho \(\dfrac{x}{3}=\dfrac{y}{4}\)và \(\dfrac{y}{5}=\dfrac{z}{6}\).Tính M=\(\dfrac{2x+3y+4z}{3x+4y+5z}\)
\(\dfrac{-29}{12}+1+\dfrac{19}{-12}\)\(\le\)x\(\le\dfrac{-1}{3}+\dfrac{3}{4}+\dfrac{7}{12}\)
\(\dfrac{-5}{3}+1+\dfrac{1}{-3}\le x\le\dfrac{8}{10}+\dfrac{1}{5}+2\)