tim cac gia tri cua x de bieu thuc sau nhan gia tri am
a)x mu 2 + 5x b)x mu 2 - 5/2x c)3(2x+3)(3x-5) d)x - 2/x-6 e) x mu 2 -1/x mu 2
Tim cac gia tri cua X de biet thuc sau nhan gia tri am
a)x^2+5x
b)3x(2x+3)x(4y-3)
tim cac gia tri cua x de bieu thuc sau nhan gia tri duong
a) x mu 2 + 4x b)2x mu 2 - 4x c)(x-3)(x+7) d)5(3x +1 )(4x-3) e)(1/2-x)(1/3-x)
Tim cac gia tri cua Y de bieu thuc sau nhan gia tri duong
a)2y^2-4y
b)5x(3y+1)x(4y-3)
tim cac gia tri nguyen cua x de bieu thuc sau nhan gia tri duong (x+1).(x-2)/x+6
\(\Rightarrow\)(x + 1) . (x - 2)\(⋮\)(x + 6)
\(\Rightarrow\)(x + 1) . (x -2)\(⋮\)x + 6
(x - 2) . (x+1) \(⋮\)x+ 6
(x - 2) . (x + 6 - 5)\(⋮\)x+ 6
x + 6 \(⋮\)x + 6
5\(⋮\)x + 6
( x -2 ) \(⋮\)6
6+x\(\in\)Ư (5) = ( 1 , 5) Vì biểu thức trên dương nên 6 + x cũng dương.
x + 6 = 1 x + 6 =5
x=-5 x=-1
Vậy x\(\in\)(-5, -1)
tim cac so nguyen x de cac bieu thuc sau co gia tri nguyen:
a.-24/x+18/x
b.2x-5/x+1
c.3x+2/x-1-x-5/x-1
d.5x-4/2x+3 - 2x-5/2x+3
a)\(-\frac{21}{x}+\frac{18}{x}=\frac{-21+18}{x}=\frac{-3}{x}\in Z\)
=>-3 chia hết x
=>x thuộc Ư(-3)
=>x thuộc {1;-1;3;-3}
b)\(\frac{2x-5}{x+1}=\frac{2\left(x+1\right)-7}{x+1}=\frac{2\left(x+1\right)}{x+1}-\frac{7}{x+1}=2-\frac{7}{x+1}\in Z\)
=>7 chia hết x+1
=>x+1 thuộc Ư(7)
=>x+1 thuộc {1;-1;7;-7}
=>x thuộc {0;-2;6;-8}
c)\(\frac{3x+2}{x-1}-\frac{x-5}{x-1}=\frac{3x+2-\left(x-5\right)}{x-1}=\frac{2x+7}{x-1}=\frac{2\left(x-1\right)+9}{x-1}=\frac{2\left(x-1\right)}{x-1}+\frac{9}{x-1}\)\(=2+\frac{9}{x-1}\in Z\)
=>9 chia hết x-1
=>x-1 thuộc Ư(9)
=>....
Còn lại bạn tự làm típ nha khi nào ko làm đc thì nhắn vs mk :)
tim cac gia tri nguyen cua x de gia tri bieu thuc y=2x^2+1/2x-1 la so nguyen
Bai 14 :Tim cac gia tri cua x de bieu thuc sau nhan gia tri duong:
a) 3x-2 b)x2-5x
\(3x-2>0\)
\(\Leftrightarrow3x>2\)
\(x>0\)
\(x^2-5x>0\)
\(x\left(x-5\right)>0\)
\(x>5\)
Tinh gia tri cac bieu thuc sau
a) A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
