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a: \(81^7-27^9-9^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}\left(3^2-3-1\right)=3^{26}\cdot5\)
\(=3^{22}\cdot3^4\cdot5=3^{22}\cdot405⋮405\)
b: Để A<0 thì \(\dfrac{x+3}{x-1}< 0\)
=>-3<x<1
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
Bài 1:
a) Ta có: 2x + 2x+3 = 144
2x.(1+23) = 144
2x.9 = 144
2x = 16
x = 4
để A âm thì x2 - 1 + x2 - 9 = 0
<=> 2x2 - 10 = 0
<=> 2x2 = 10 <=> x2 = 5
mà x nguyên => vô lí, vậy ko tìm đc giá trị của x
Ta có: \(A.B=\left(-\frac{4}{15}x^3y\right).\left(\frac{3}{7}x^5y^3\right)\)
\(\Leftrightarrow A.B=\left(-\frac{4}{15}.\frac{3}{7}\right).\left(x^3yy^3x^5\right)\)
\(\Leftrightarrow A.B=\frac{-12}{105}x^8y^4\)
Mà \(\hept{\begin{cases}x^8\ge0\\y^4\ge0\end{cases}}\)\(\Leftrightarrow A.B\le0\)
Suy ra A và B không thể cùng giá trị âm
\(81^7=3^{28};27^9=3^{27};9^{13}=3^{26}\)
=\(3^{28}-3^{27}-3^{26}=3^{26}-\left(3^2-3-1\right)\)
=\(3^{26}.5=3^{13}.3^2.5=3^{13}.45⋮45\)
Mà 405=45.9
\(\Rightarrow\)dpcm
Ta có x+3=(x-1)+4
Nên 4\(\inƯ\left(x-1\right)=\left\{\pm1,\pm2,\pm4\right\}\)
x-1=1 \(\Rightarrow\) x=2 x-1=-2\(\Rightarrow\)x=-1
x-1=-1 \(\Rightarrow\) x=0 x-1=4\(\Rightarrow\)x=5
x-1=2\(\Rightarrow\)x=3 x-1=-4 \(\Rightarrow\) x=-3
\(\Rightarrow\) x\(\in\left\{2,-1,0,5,3,-3\right\}\)