(x-2)(2y+3)=26
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\(TH1,\left\{{}\begin{matrix}x-2=2\\2y+3=13\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ TH2,\left\{{}\begin{matrix}x-2=13\\2y+3=2\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=15\\y=-\dfrac{1}{2}\end{matrix}\right.\\ TH3,\left\{{}\begin{matrix}x-2=-2\\2y+3=-13\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=0\\y=-8\end{matrix}\right.\\ Th4,\left\{{}\begin{matrix}x-2=-13\\2y+3=-2\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=-11\\y=-\dfrac{5}{2}\end{matrix}\right.\)
26=1.26=(-1)*(-26)=2.13=(-2)*-13)
2y+3=1=>y=-1; x-2=26=>x=28....> (x,y)=28,-1)
2y+3=-1=>y=2; x-2=-26=>x=-24...>(x,y)=(-24,2)
2y+3=13=>y=5; x-2=2=>x=4.......>(x,y)=(4,5)
2y+3=-13=>y=-8; x-2=-2=>x=0....>(x,y)=(0,-8)
(x-2)(2y+3)=26
Ta thấy 2y+3 là số lẻ
x-2 26 2
2y+3 1 13
x 28 4
y -1 5
Vậy x=4 ; y= 5
loại y=-1 Vì -1 l;à số nguyên âm
(x-2)(2y+3)=26=1.26=2.13
\(\Rightarrow\hept{\begin{cases}x-2=1\\2y+3=26\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=\frac{23}{2}\end{cases}}}\)(loại)
\(\Rightarrow\hept{\begin{cases}x-2=2\\2y+3=13\end{cases}\Rightarrow\hept{\begin{cases}x=4\\y=5\end{cases}}}\)(nhận)
Vậy \(\left(x;y\right)=\left(4;5\right)\)
( x - 2 ) . ( 2y + 3 ) = 26
x, y \(\in\) N => x - 2 \(\in\) Ư(26)
Mà x - 2 \(\ge\) 2
=> x - 2 \(\in\) \(\left\{1;2;13;26\right\}\)
Ta có bảng:
x - 2 | 1 | 2 | 13 | 26 |
2y + 3 | 26 | 13 | 2 | 1 |
x | 3 | 4 | 15 | 28 |
y | loại | 5 | loại | loại |
Vậy x = 4; y = 5
\(1,=-\left(y^2+12y+36\right)=-y^2-12y-36\)
\(2,=-\left(16-8y+y^2\right)=-16+8y-y^2\)
\(3,=-\left(\dfrac{4}{9}+\dfrac{4}{3}x+x^2\right)=-\dfrac{4}{9}-\dfrac{4}{3}x-x^2\)
\(4,=-\left(x^2-3x+\dfrac{9}{4}\right)=-x^2+3x-\dfrac{9}{4}\)
\(5,-\left(2+3y\right)^2=-\left(4+12y+9y^2\right)=-4-12y-9y^2\)
.... mấy ý còn lại bn tự lm nhé, tương tự thhooi
1) \(-\left(y+6\right)^2=-y^2-12y-36\)
2) \(-\left(4-y\right)^2=-y^2+8y-16\)
3) \(-\left(x+\dfrac{2}{3}\right)^2=-x^2-\dfrac{4}{3}x-\dfrac{4}{9}\)
4) \(-\left(x-\dfrac{3}{2}\right)^2=-x^2+3x-\dfrac{9}{4}\)
5) \(-\left(3y+2\right)^2=-9y^2-12y-4\)
6) \(-\left(2y-3\right)^2=-4y^2+12y-9\)
7) \(-\left(5x+2y\right)^2=-25x^2-20xy-4y^2\)
8) \(-\left(2x-\dfrac{3}{2}\right)^2=-4x^2+6x-\dfrac{9}{4}\)
(x -2) .(2y +3)= 26
=>x -2 ;2y +3\(\in\)U(26)
U(26)={\(\pm1;\pm2;\pm13;\pm26\)}
=>x - 2 ;2y +3\(\in\){\(\pm1;\pm2;\pm13;\pm26\)}
ta có bảng sau:
x-2 | -1 | -26 | 1 | 26 | 2 | 13 | -13 | -2 |
2y+3 | -26 | -1 | 26 | 1 | 13 | 2 | -2 |
-13 |
x | 1 | -24 | 3 | 28 | 4 | 15 | -11 | 0 |
y | \(\varnothing\) | -2 | \(\varnothing\) | -1 | 5 | \(\varnothing\) | \(\varnothing\) | -8 |
=>x=-24 thì y=-2
x=28 thì y=-1
x=4 thì y=5
x=0 thì y=-8
(x.2).(2y+3)=26
2x(2y+3)=26
x(2y+3)=13
TH1<=>x=1;2y+3=13
<=>x=1; y=5
TH2 <=> x=13 ;2y+3=1
x=13 ; y= -1
P/s tham khảo nha
(x-2)(2y+3)=26
=>\(\left(x-2\right)\left(2y+3\right)=1\cdot26=26\cdot1=\left(-1\right)\cdot\left(-26\right)=\left(-26\right)\cdot\left(-1\right)=2\cdot13=13\cdot2=\left(-2\right)\cdot\left(-13\right)=\left(-13\right)\cdot\left(-2\right)\)
=>\(\left(x-2;2y+3\right)\in\left\{\left(1;26\right);\left(26;1\right);\left(-1;-26\right);\left(-26;-1\right);\left(2;13\right);\left(13;2\right);\left(-2;-13\right);\left(-13;-2\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(3;\dfrac{23}{2}\right);\left(28;-1\right);\left(1;-\dfrac{29}{2}\right);\left(-24;-2\right);\left(4;5\right);\left(15;-\dfrac{1}{2}\right);\left(0;-8\right);\left(-11;-\dfrac{5}{2}\right)\right\}\)