Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
Ta có: \(3x=2y\)
nên \(\dfrac{x}{2}=\dfrac{y}{3}\)
mà x+y=-15
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{2}=-3\\\dfrac{y}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-9\end{matrix}\right.\)
Vậy: (x,y)=(-6;-9)
Bài 2:
a) Ta có: \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}\)
mà x+y-z=20
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y-z}{4+3-5}=\dfrac{20}{2}=10\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{4}=10\\\dfrac{y}{3}=10\\\dfrac{z}{5}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=40\\y=30\\z=50\end{matrix}\right.\)
Vậy: (x,y,z)=(40;30;50)
\(a,\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\left(-5< 0\right)\Leftrightarrow x>3\\ b,\dfrac{3-x}{x^2+1}\ge0\Leftrightarrow3-x\ge0\left(x^2+1>0\right)\Leftrightarrow x\le3\\ c,\dfrac{\left(x-1\right)^2}{x-2}< 0\Leftrightarrow x-2< 0\left[\left(x-1\right)^2\ge0\right]\Leftrightarrow x< 2\)
\(a,x-\dfrac{2}{3}=\dfrac{1}{6}\\ \Rightarrow x=\dfrac{5}{6}\\ b,2x+\dfrac{1}{2}=-\dfrac{5}{3}\\ \Rightarrow2x=-\dfrac{13}{6}\\ \Rightarrow x=-\dfrac{13}{12}\\ c,3x+\dfrac{3}{2}=x-\dfrac{5}{3}\\ \Rightarrow-4x+\dfrac{3}{2}=-\dfrac{5}{3}\\ \Rightarrow-4x=-\dfrac{19}{6}\\ \Rightarrow4x=\dfrac{19}{6}\\ \Rightarrow x=\dfrac{19}{24}.\)
a) x - 2/3 = 1/6
x = 1/6 + 2/3
x = 5/6
b) 2x + 1/2 = -5/3
2x = -5/3 - 1/2
2x = -13/6
x = -13/6 : 2
x = -13/12
c) 3x + 3/2 = x - 5/3
3x - x = -5/3 - 3/2
2x = -19/6
x = -19/6 : 2
x = -19/12
\(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
a. \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\dfrac{1}{3}\right|=-1+4=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=\dfrac{-10}{3}\end{matrix}\right.\)
Vậy..........
b. \(1\dfrac{3}{4}.x+1\dfrac{1}{2}=-\dfrac{4}{5}\)
\(\Rightarrow1\dfrac{3}{4}x=-\dfrac{4}{5}-1\dfrac{1}{2}=\dfrac{-23}{10}\)
\(\Rightarrow x=\dfrac{-23}{10}:1\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{-46}{35}\)
a) Ta có:
(x - 1)5 = - 243
=> (x - 1)5 = (-3)5
=> x - 1 = - 3
=> x = -3 + 1
=> x = -2
Vậy x = -2
b) Ta có:
\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)
\(\Rightarrow\left(x+2\right).\dfrac{1}{11}+\left(x+2\right).\dfrac{1}{12}+\left(x+2\right).\dfrac{1}{13}=\left(x+2\right).\dfrac{1}{14}+\left(x+2\right).\dfrac{1}{15}\)
=> \(\left(x+2\right).\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}\right)=\left(x+2\right).\left(\dfrac{1}{14}+\dfrac{1}{15}\right)\)
=> \(\left(x+2\right).\dfrac{431}{1716}=\left(x+2\right).\dfrac{29}{210}\)
=> \(\left(x+2\right).\dfrac{431}{1716}-\left(x+2\right).\dfrac{29}{210}=0\)
=> (x + 2).(\(\dfrac{431}{1716}-\dfrac{29}{210}\)) = 0
mà \(\dfrac{431}{1716}-\dfrac{29}{210}\) \(\ne\) 0
=> x + 2 = 0
=> x = -2
Vậy x = -2
c) Ta có :
\(\left|3x-2\right|+5x=4x-10\)
=> \(\left|3x-2\right|=4x-5x-10\)
=> \(\left|3x-2\right|=-x-10\)
=> 3x - 2 = -x - 10
hoặc 3x - 2 = -(-x -10)
*) Nếu 3x - 2 = -x - 10
=> 3x + x = -10 + 2
=> 4x = -8
=> x = -2
*) Nếu 3x - 2 = -(-x -10)
=> 3x - 2 = x +10
=> 3x - x = 10 + 2
=> 2x = 12
=> x = 6
Vậy x = -2 hoặc x = 6
a.Vì các số đó đều viết đc dưới dạng PS nên đó là SHT
b.=1/3+3/7=16/21
1. Vì : \(0.6=\dfrac{6}{10}=\dfrac{18}{30}=\dfrac{24}{40}=...\\ -1,25=\dfrac{-1}{25}=\dfrac{-2}{50}=\dfrac{-12}{150}=...\\ 1\dfrac{1}{3}=\dfrac{4}{3}=\dfrac{8}{6}=\dfrac{32}{24}=...\)
Bài 1:
a) \(\left|3x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{1}{3};x_2=3\)
b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
cho đáp án tự làm (vì cách lm của mik bị ném đá khá nhiều lần òi :D)
\(x=-1\)
c) như câu b nhé :D
\(x=-2004\)
a) \(4^{x+2}.3^x=16.12^5\)
\(\Leftrightarrow4^{x+2}.3^x=4^2.4^5.3^5\)
\(\Leftrightarrow4^{x-5}.3^{x-5}=1\)
\(\Leftrightarrow12^{x-5}=1\)
\(\Leftrightarrow x-5=0\Leftrightarrow x=5\)
\(a)4^{x+2}.3^x=4^2.\left(4.3\right)^5\)
\(4^{x+2}.3^x=4^7.3^5\)
\(\Rightarrow4^{x+2}=4^7;3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
\(3x-\dfrac{5}{3}=x-\dfrac{1}{4}\\ =>3x-x=\dfrac{5}{3}-\dfrac{1}{4}\\ =>2x=\dfrac{17}{12}\\ =>x=\dfrac{17}{24}\)