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NN
3 tháng 5 2020
1. \(\left(3x-5\right)^{2010}+\left(y-1\right)^{2012}+\left(x-z\right)^{2014}=0\)
Vì \(\left(3x-5\right)^{2010}\ge0\forall x\); \(\left(y-1\right)^{2012}\ge0\forall y\); \(\left(x-z\right)^{2014}\ge0\forall x,z\)
\(\Rightarrow\left(3x-5\right)^{2010}+\left(y-1\right)^{2012}+\left(x-z\right)^{2014}\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}3x-5=0\\y-1=0\\x-z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=5\\y=1\\x=z\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=1\\z=\frac{5}{3}\end{cases}}\)
Vậy \(x=z=\frac{5}{3}\)và \(y=1\)
PT
1
22 tháng 10 2015
1)
Từ: \(\frac{3}{y}=\frac{7}{x}\)=>\(\frac{x}{7}=\frac{y}{3}\)
x+16=y =>x-y=-16
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{7}=\frac{y}{3}=\frac{x-y}{7-3}=\frac{-16}{4}=-4\)(vì x-y=-16)
=>\(\frac{x}{7}=-4=>x=-28\)
=>\(\frac{y}{3}=-4=>y=-12\)
Vậy x=-28 ;y=-12
2)
=>x2-3x+5 chia hết cho x-3
mà (x-3)2 chia hết cho x-3
=>x2-3x+5 -(x-3)2 chia hết cho x-3
=> x2-3x+5 -x2-9 chia hết cho x-3
=>-3x+(-4) chia hết cho x-3
lại có : 3.(x-3) chia hết cho x-3
=>-3x-(-4)+3.(x-3) chia hết cho x-3
=>-3x+(-4)+3x-9 chia hết cho x-3
=>-13 chia hết cho x-3
=>x-3 \(\in\)Ư(13)={-1;1;-13;13}
=>x\(\in\){2;4;-9;16}
27 tháng 5 2022
Câu 2:
a: Ta có: \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{378}{395}\\z=2004\end{matrix}\right.\)
b: \(\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{3}{2}\right|+\left|x-y-z-\dfrac{1}{2}\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+\dfrac{3}{2}=0\\x-y-z-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{3}{2}\\z=\dfrac{3}{2}\end{matrix}\right.\)
MV
1 tháng 8 2017
a,
\(\left|x+\dfrac{9}{2}\right|\ge0\forall x\\ \left|y+\dfrac{4}{3}\right|\ge0\forall y\\ \left|z+\dfrac{7}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,z\)
Mà
\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{2}\\y=\dfrac{-4}{3}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy \(x=\dfrac{-9}{2};y=\dfrac{-4}{3};z=\dfrac{-7}{2}\)
d,
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{1}{5}\right|\ge0\forall y\\ \left|x+y+z\right|\ge0\forall x,y,z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)
Mà
\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-3}{4}+\dfrac{1}{5}+z=0\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-11}{20}+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)
9 tháng 12 2016
a)\(2^{x-1}=16\)
\(\Rightarrow2^{x-1}=2^4\)
\(\Rightarrow x-1=4\Rightarrow x=5\)
b)\(\left(x-1\right)^2=25\)
\(\Rightarrow\left(x-1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow\left(x-1\right)^2=5^2\) hoặc \(\left(x-1\right)^2=\left(-5\right)^2\)
\(\Rightarrow x-1=5\) hoặc \(x-1=-5\)
\(\Rightarrow x=6\) hoặc \(x=-4\)
JB
2
5 tháng 8 2018
a)Có \(\left(x-2\right)^2\ge0;\left(y-3\right)^2=0\)
Mà \(\left(x-2\right)^2+\left(y-3\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x-2=0\\y-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}}\)
b)\(\left(x-1\right)^{x+2}=0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
JB
7 tháng 8 2018
a) \(\left(x-2\right)^2+\left(y-3\right)^2=0\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\left(y-3\right)^2\ge0\forall y\)
\(\Rightarrow\)\(\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x-2=0\\y-3=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)
b) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)
\(\left(x-1\right)^{x+2}\times1-\left(x-1\right)^{x+2}\times\left(x-1\right)^4=0\)
\(\left(x-1\right)^{x+2}\times[1-\left(x-1^4\right)]=0\)
TH 1: \(\left(x-1\right)^{x+2}=0\) TH 2: \(1-\left(x-1\right)^4=0\)
\(\Rightarrow x-1=0\) \(\left(x-1\right)^4=1\)
\(\Rightarrow x=1\) \(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)
Vậy \(x\in[0;1;2]\)
HT
0