y \(\times\) 2 +\(\dfrac{y}{\dfrac{1}{3}}\) = 20

\(y\times2+y\div\dfrac{1}{3}=20\)

\(y\times2+y\times3=20\)

\(y\times\left(2+3\right)=20\)

\(y\times5=20\)

\(y=20\div5\)

\(y=4\)

y $\times$ 2 +�1331​y​ = 20

�×2+�÷13=20y×2+y÷31​=20

$y\times 2+y\times 3=20$

$y\times \left(2+3\right)=20$

$y\times 5=20$

$y=20÷5$

$y=4$

a: \(\Leftrightarrow\left(x+1;y-4\right)\in\left\{\left(1;19\right);\left(19;1\right);\left(-1;-19\right);\left(-19;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;23\right);\left(18;5\right);\left(-2;-15\right);\left(-20;3\right)\right\}\)

b: \(\Leftrightarrow\left(2x+1;y-5\right)\in\left\{\left(1;23\right);\left(23;1\right);\left(-1;-23\right);\left(-23;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;28\right);\left(11;6\right);\left(-1;-18\right);\left(-12;4\right)\right\}\)

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Thế thì hỏi làm j 

1./ \(x+y=3\Rightarrow\left(x+y\right)^3=27\Rightarrow x^3+y^3+3xy\left(x+y\right)=27\Rightarrow x^3+y^3+3\cdot2\cdot3=27.\)

\(\Rightarrow x^3+y^3=9\)

2./ \(\left(x+3\right)\left(x^2-3x+3^2\right)-x^3-2x-4=0\)

\(\Leftrightarrow x^3+27-x^3-2x-4=0\Leftrightarrow2x=23\Leftrightarrow x=\frac{23}{2}\)

1/ \(x+y=3\)

\(\Rightarrow\left(x+y\right)^2=9\)

\(\Rightarrow x^2+2xy+y^2=9\)

\(\Rightarrow x^2+4+y^2=9\)

\(\Rightarrow x^2+y^2=5\)

\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3.1=3\)

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

Giải:

a) \(\left(x-4\right).\left(y+1\right)=8\) 

\(\Rightarrow\left(x-4\right)\) và \(\left(y+1\right)\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

Ta có bảng giá trị:

Vì \(\left(x;y\right)\in N\) nên \(\left(x;y\right)=\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)

Vậy \(\left(x;y\right)=\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\) 

b) \(\left(2x+3\right).\left(y-2\right)=15\) 

\(\Rightarrow\left(2x+3\right)\) và \(\left(y-2\right)\inƯ\left(15\right)=\left\{\pm1;\pm3;\pm5;\pm15\right\}\) 

Vì \(\left(x;y\right)\in N\) nên \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\) 

Vậy \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\) 

c) \(xy+2x+y=12\) 

\(\Rightarrow x.\left(y+2\right)+\left(y+2\right)=14\) 

\(\Rightarrow\left(x+1\right).\left(y+2\right)=14\) 

\(\Rightarrow\left(x+1\right)\) và \(\left(y+2\right)\inƯ\left(14\right)=\left\{1;2;7;14\right\}\) 

Vì \(\left(x;y\right)\in N\) nên \(\left(x;y\right)\in\left\{\left(0;12\right);\left(1;5\right);\left(6;0\right)\right\}\) 

Vậy \(\left(x;y\right)\in\left\{\left(0;12\right);\left(1;5\right);\left(6;0\right)\right\}\) 

d) \(xy-x-3y=4\) 

\(\Rightarrow y.\left(x-3\right)-\left(x-3\right)=7\) 

\(\Rightarrow\left(y-1\right).\left(x-3\right)=7\) 

\(\Rightarrow\left(y-1\right)\) và \(\left(x-3\right)\inƯ\left(7\right)=\left\{1;7\right\}\) 

Ta có bảng giá trị:

Vậy \(\left(x;y\right)\in\left\{\left(4;8\right);\left(10;2\right)\right\}\)

(x-y)(2y+1)= 11

=> x-y \(\in\)Ư(11) ={ 1;11; -1; -11}

Nếu x-y = 1 thì 2y+1= 11 => 2y= 10 => y=5 => x= 6

Nếu x-y= 11 thì 2y+1 = 1 => 2y=0 => y=0 => x= 11

Nếu x-y = -1 thì 2y+1= -11 => 2y= -12 => y= -6 => x= -7

.............................

Vậy....

(x-y)(2y+1)=11

x,y nguyên => x-y; 2y+1 = nguyên

=> x-y; 2y+1 thuộc Ư (11)={-11;-1;1;11}
Ta có bảng

Ta có: \(\frac{1}{x}+\frac{y}{2}=\frac{5}{8}\)

\(\Rightarrow\frac{2}{2x}+\frac{xy}{2x}=\frac{5}{8}\)

\(\Rightarrow\frac{2+xy}{2x}=\frac{5}{8}\)

\(\Rightarrow8.\left(2+xy\right)=5.2x\)

\(\Rightarrow16+8xy=10x\)

\(\Rightarrow10x-8xy=16\)

\(\Rightarrow2x.5-2x.4y=16\)

\(\Rightarrow2x.\left(5-4y\right)=16\)

Với \(x;y\inℕ^∗\Rightarrow\hept{\begin{cases}2x\inℕ^∗\\5-4y\inℕ^∗\end{cases}}\)

mà 16 = 1.16 = 2.8 = 4.4 

Lập bảng xét 6 trường hợp ta có :

Vậy x = 8 ; y = 1