Có: \(\left\{{}\begin{matrix}\left|x-3\right|\ge0\forall x\\\left|y-1\right|\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left|x-3\right|+\left|y-1\right|\ge0\forall x;y\)

Mà: \(\left|x-3\right|+\left|y-1\right|=0\)

nên: \(\left\{{}\begin{matrix}x-3=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)

\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)

\(-18x+13=0\)

\(x=\dfrac{13}{18}\)

Vậy \(S=\left\{\dfrac{13}{18}\right\}\)

\(b.\left(x-1\right)^3-125=0\)

\(\left(x-1\right)^3=125\)

\(x-1=5\)

\(x=6\)

Vậy \(S=\left\{6\right\}\)

\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)

\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)

Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy \(S=\left\{1;-2\right\}\)

\(d.x^2-4x+4+x^2-2xy+y^2=0\)

\(\left(x-2\right)^2+\left(x-y\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

Vậy \(S=\left\{2;2\right\}\)

Bài 1:

b) \(2x+6⋮x-3\)

\(\Leftrightarrow2\left(x-3\right)+12⋮x-3\)

Mà \(2\left(x-3\right)⋮x-3\)

\(\Rightarrow12⋮x-3\)

làm nốt

d) \(x-1⋮2x+1\)

\(\Leftrightarrow2x-2⋮2x+1\)

\(\Leftrightarrow2x+1-3⋮2x+1\)

Mà \(2x+1⋮2x+1\)

\(\Rightarrow3⋮2x+1\)

Làm nốt

Bài 2:

\(x479y⋮36\)

\(\Leftrightarrow x479y⋮4;9\)

+) Để \(x479y⋮4\)

\(\Leftrightarrow9y⋮4\)mà y là chữ số nên y<10

\(\Rightarrow y\in\left\{2;6\right\}\)

TH1:\(y=2\)Thay vào ta được :

\(x4792⋮9\)

\(\Leftrightarrow x+4+7+9+2⋮9\)

\(\Leftrightarrow x+22⋮9\)Mà x là chữ số

\(\Rightarrow x=5\)

TH2: \(y=6\)thay vào ta được :

\(x4796⋮9\)

\(\Leftrightarrow x+4+7+9+6⋮9\)

\(\Leftrightarrow x+26⋮9\)Mà x là chữ số

\(\Rightarrow x=1\)

Vậy \(\left(x;y\right)=\left\{\left(5;2\right);\left(1;6\right)\right\}\)

a) (y + 5) x 2020 = 205 x 2020

(y + 5) x 2020 = 414100

y + 5 = 414100 : 2020 = 205

y = 205 - 5 = 200

a) y= 200

b) y-45600 = 1600 × 4 ×25

y-45600=160000

y= 114400

b) y - 45600 = (2019 - 419) x 4 x 25

y - 45600 = 1600 x 100

y - 45600 = 160000

y = 160000 + 45600 = 205 600

\(a,y\times2,8+5,2\times y=48\\ y\times\left(2,8+5,2\right)=48\\ y\times8=48\\ y=\dfrac{48}{8}=6\\ ---\\ b,y\times12,25-y+y\times2,75=1050\\ y\times\left(12,25-1+2,75\right)=1050\\ y\times14=1050\\ y=\dfrac{1050}{14}=75\)

a: \(y\cdot2,8+y\cdot5,2=48\)

=>\(y\left(2,8+5,2\right)=48\)

=>\(8y=48\)

=>\(y=\dfrac{48}{8}=6\)

b: \(y\cdot12,25-y+y\cdot2,75=1050\)

=>\(y\left(12,25-1+2,75\right)=1050\)

=>\(y\cdot14=1050\)

=>\(y=\dfrac{1050}{14}=75\)

\(a,y\times27+y\times30+y\times44-y=10500\\\Rightarrow y\times\left(27+30+44-1\right)=10500\\ \Rightarrow y\times100=10500\\ \Rightarrow y=10500:100\\ \Rightarrow y=105\\ b,y\times285+115\times y=400\\ \Rightarrow y\times\left(285+115\right)=40\\ \Rightarrow y\times400=400\\ y=400:400\\ \Rightarrow y=1\)

\(a,\Rightarrow y\times\left(27+30+44-1\right)=10500\\ \Rightarrow y\times100=10500\\ \Rightarrow y=105\\ b,\Rightarrow y\times\left(285+115\right)=400\\ \Rightarrow y\times400=400\\ \Rightarrow y=1\)

Tìm y:

\(a,y+847\times2=1953-74\\ y+847\times2=1879\\ y+1694=1897\\ y=1897-1694\\ y=185\\ b,y:\left(7\times18\right)=5839+8591\\ y:126=14430\\ y=14430\times126\\ y=1818180.\)

a/ y + 847 x 2 = 1953 - 74
y + 1694 = 1879
y = 1879 - 1694
y = 185
b/y : (7 x 18) = 5839 + 8591
y : 126 = 14430
y = 14430 x 126
y = 1818180
#AvoidMe

 a. y + 847 x 2 = 1953 - 74

y + 1694 = 1879                              

            y = 1879 - 1694

            y = 185

 b. y : (7 x 18) = 5839 + 8591

y : 126 = 14430

              y = 14430 x 126

              y = 1818180

a: \(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

2x+1.3y=12x => 2x+1.3y=4x.3x

=> 2x+1.3y=22x.3x => x + 1 = 2x và y = x

=> x = 1 và y = x = 1

 Vậy x=y=1