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\(\left(\frac{2x-3}{4}\right)^{2014}+\left(\frac{3y+4}{5}\right)^{2016}=0\)
Có: \(\left(\frac{2x-3}{4}\right)^{2014}\ge0;\left(\frac{3y+4}{5}\right)^{2016}\ge0\)
Mà theo bài ra: \(\left(\frac{2x-3}{4}\right)^{2014}+\left(\frac{3y+4}{5}\right)^{2016}=0\)
\(\Rightarrow\hept{\begin{cases}\frac{2x-3}{4}=0\\\frac{3y+4}{5}=0\end{cases}}\Rightarrow\hept{\begin{cases}2x-3=0\\3y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}2x=3\\3y=-4\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-\frac{4}{3}\end{cases}}\)
Vậy: \(\hept{\begin{cases}x=\frac{3}{2}\\y=-\frac{4}{3}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{2x-3}{4}=0\\\frac{3y+4}{5}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-\frac{4}{3}\end{cases}}}\)
đầu bài trên tớ làm luôn nhá !!!
a, / 3x+1/= 5-3
/ 3x+1/= 2
3x+1=2
x+1 = 2:3
x+1 = 2 phần 3
x= 2/3 -1
x= -1/3
Vì \(\hept{\begin{cases}\left(x+2y-4\right)^2\ge0\\\left(2x-3y-1\right)^2\ge0\end{cases}}\)=> \(\left(x+2y-4\right)^2+\left(2x-3y-1\right)^2\ge0\)
\(\left(x+2y-4\right)^2+\left(2x-3y-1\right)^2=0\) <=> \(\left(x+2y-4\right)^2=\left(2x-3y-1\right)^2=0\)
<=>\(x+2y-4=2x-3y-1=0\)
\(x+2y-4=0\Leftrightarrow x+2y=4\Leftrightarrow2\left(x+2y\right)=8\Leftrightarrow2x+4y=8\)
\(2x-3y-1=0\Leftrightarrow2x-3y=1\)
=>\(\left(2x-3y\right)-\left(2x+4y\right)=1-8\)
=>\(2x-3y-2x-4y=-7\)
=>\(-7y=-7\)=>\(y=1\)=>\(x=2\)
Vậy .............................
Ta có:\(C=2\left(x-y\right)+13x^3y^2\left(x-y\right)+15xy\left(y-x\right)+1\)Thế \(x-y=0\) vào C ta được:
\(C=0+0+0+1\)
C = 0
1,
Vì \(\left|2x-27\right|^{2007}\ge0;\left(3y+10\right)^{2008}\ge0\)
\(\Rightarrow\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}\ge0\)
Mà \(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}=0\)
\(\Rightarrow\hept{\begin{cases}\left|2x-27\right|^{2007}=0\\\left(3y+10\right)^{2008}=0\end{cases}\Rightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{27}{2}\\y=\frac{-10}{3}\end{cases}}}\)
2,
TH1: \(x\ge\frac{3}{5}\)
<=> 2(5x-3)-2x=14
<=> 10x-6-2x=14
<=>8x-6=14
<=>8x=20
<=>x=5/2 (thỏa mãn)
TH2: x < 3/5
<=> 2(3-5x)-2x=14
<=>6-10x-2x=14
<=>6-12x=14
<=>12x=-8
<=>x=-2/3 (thỏa mãn)
Vậy \(x\in\left\{\frac{5}{2};\frac{-2}{3}\right\}\)
A=2(x+y)+3xy(x+y)+5x2y2(x+y)+2
A=2.0+3xy.0+5x2y2.0+2
A=2
B=xy(x+y)+2x2y (x+y)+5
B=xy.0+2x2y.0+5=5
a,Ta có 2(x+y)+3xy(x+y)+5x2y2(x+y)+4
Xg thay x+y=0 vào là dc bn nhó
Chúc bn hok tốt
Vì \(\left(x+2y-3\right)^{2016}\ge0;\left|2x+3y-5\right|\ge0\forall x;y\)
\(\Rightarrow\left(x+2y-3\right)^{2016}+\left|2x+3y-5\right|\ge0\forall x;y\)
Mà \(\left(x+2y-3\right)^{2016}+\left|2x+3y-5\right|=0\) \(\Leftrightarrow\left(x+2y-3\right)^{2016}=0\) ; \(\left|2x+3y-5\right|=0\)
\(\Rightarrow x+2y-3=0;2x+3y-5=0\)
\(\Leftrightarrow x+2y=3;2x+3y=5\)
\(\Rightarrow x=3-2y\)
\(\Rightarrow2\left(3-2y\right)+3y=5\Leftrightarrow6-4y+3y=5\Leftrightarrow6-y=5\Rightarrow y=1\)
\(\Rightarrow x=3-2.1=1\)
Vậy \(x=1;y=1\)