a. a.-4 là B(a-1)
b. 2a là Ư(-10)
c. 12 thuộc B(2a+1)
d. 2a-1 là Ư (3a+2)
e. (2x-1)*(y-4)
f. (5x+1)*(y-1)
g. 5xy+5x+y
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a, (x+3)(y+2) = 1
=> (x+3) \(\in\)Ư(1) = \(\left\{-1;1\right\}\)
Do (x+3)(y+2) là số dương
=> (x+3) và (y+2) cùng dấu
\(\Rightarrow\hept{\begin{cases}x+3=1\\y+2=1\end{cases}}\)hay \(\hept{\begin{cases}x+3=-1\\y+2=-1\end{cases}}\)
TH1:
\(\hept{\begin{cases}x+3=1\\y+2=1\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\y=-1\end{cases}}}\)
TH2:
\(\hept{\begin{cases}x+3=-1\\y+2=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\y=-3\end{cases}}}\)
Vậy ............
b, (2x - 5)(y-6) = 17
=> \(\left(2x-5\right)\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
Ta có bảng sau:
2x - 5 | -17 | -1 | 1 | 17 |
x | -6 | 2 | 3 | 11 |
y - 6 | -1 | -17 | 17 | 1 |
y | 5 | -11 | 23 | 7 |
Vậy \(\left(x,y\right)\in\left\{\left(-6,5\right);\left(2,-11\right);\left(3,23\right);\left(11,7\right)\right\}\)
c, Tương tự câu b
\(\left(3a-1\right)^2=9a^2-6a+1\)
\(\left(a-2\right)^2=a^2-4a+4\)
\(\left(1-5a\right)^2=1-10a+25a^2\)
\(\left(3a-2b\right)^2=9a^2-12ab+4a^2\)
\(\left(4-3a\right)^2=16-24a+9a^2\)
\(\left(5a-4b\right)^2=25a^2-40ab+16b^2\)
\(\left(5a-3b\right)\left(5a+3b\right)=25a^2-9b^2\)
\(\left(3x+1\right)\left(3x-1\right)=9x^2-1\)
\(\left(5x^2-2\right)\left(5x^2+2\right)=25x^4-4\)
\(\left(2a+\dfrac{1}{2}\right)\left(2a-\dfrac{1}{2}\right)=4a^2-\dfrac{1}{4}\)
\(\left(3x^2-y\right)\left(3x^2+y\right)=9x^4-y^2\)
\(\left(\dfrac{1}{2}x-1\right)\left(\dfrac{1}{2}x+1\right)=\dfrac{1}{4}x^2-1\)
\(\left(\dfrac{3}{4}x+2\right)\left(\dfrac{3}{4}x-2\right)=\dfrac{9}{16}x^2-4\)
\(\left(5x-\dfrac{3}{2}\right)\left(5x+\dfrac{3}{2}\right)=25x^2-\dfrac{9}{4}\)
\(\left(2a^2-7\right)\left(2a^2+7\right)=4a^2-49\)
a) A=(4-5x)2-(3+5x)2=(4-5x-3-5x)(4-5x+3+5x)=(-25x+1)1=-25x+1
B=(3x-1)(1+3x)-(3x+1)2=9x2-1-(3x+1)2=9x2-1-(9x2+6x+1)=9x2-1-9x2-6x-1=-6x-2=-2(3x+1)
\(a,=x^2+x+\dfrac{1}{4}\\ b,=4x^2+2x+\dfrac{1}{4}\\ c,=x^2-2+\dfrac{1}{x^2}\\ d,=4x^2+\dfrac{8}{3}x+\dfrac{4}{9}x^2\\ e,=a^2-1\\ f,=25x^4-4\)
\(a,\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)
\(b,\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)
\(c,\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)
\(d,\left(\dfrac{2x+2}{3x}\right)^2=\dfrac{\left(2x+2\right)^2}{9x^2}=\dfrac{4x^2+8x+4}{9x^2}\)
\(e,\left(a-1\right).\left(a+1\right)=a^2-1\)
\(f,\left(5x^2-2\right).\left(5x^2+2\right)=25x^4-4\)
a/ -4 là B( a- 1 )
Hay a - 1 là Ư(-4)
Ta có Ư(-4) = { -4; -2; -1; 1; 2 ; 4 }
Xét :
a - 1 = -4 => a = -3
a - 1 = -2 => a = -1
a - 1 = -1 => a = 0
a - 1 = 1 => a= 2
a - 1 = 2 => a = 3
a - 1 = 4 => a = 5