` Y = ( 3x^2 - 3x - 3 )/(x^2+x-2) - (x+1)/(x+2) + (x-2)/(x).( (1)/(1-x) - 1)`
a) Rút gọn Y ( Đáp số Y = ` (x-2)/(x+2) ` )
b) Tìm x để Y = 2
c) Tìm x ∈ Z để Y ∈ Z
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)
\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)
\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)
\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)
=a, (x-3)(x+3)-(x-7)(x+7)= x2 - 9 - x2 + 7
= -2
b, (4x-5)2+(3x-2)2-2(4x+5)(3x-2)= (4x-5)2 - 2(4x+5)(3x-2) + (3x-2)2
= ( 4x - 5 - 3x + 2 )2
= ( x - 3 )2
c, 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2= 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2
= (3x-y)2+ 2(3x-y)(3x+y)+ (3x+y)2
= ( 3x - y + 3x + y )2
= ( 6x )2
= 36x2
d, (x-y+z)2+(z-y)2+2(x-y+z+2(x-y+z)(y-z-y+z)(y-z)
1, rút gọn
a, (x-3)(x+3)-(x-7)(x+7)
= x^2 - 9 - (x^2 - 49)
= x^2 - 9 - x^2 + 49
= 40
b, (4x-5)2+(3x-2)2-2(4x+5)(3x-2)
= 16x^2 - 40x + 25 + 9x^2 - 12x + 4 - 2(12x^2 - 8x + 15x - 10)
= 25x^2 - 52x + 29 - 24x^2 + 16x - 30x + 20
= x^2 - 66x + 49
c, 2(3x-y)(3x+y)+(3x-y)2+(3x+y)2
= 2(9x^2 - y^2) + 9x^2 - 6xy + y^2 + 9x^2 + 6xy + y^2
= 18x^2 - 2y^2 + 18x^2 + 2y^2
= 36x^2
d, (x-y+z)2+(z-y)2+2(x-y+z+2(x-y+z)(y-z-y+z)(y-z)
= dài vl
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
a/ \(\frac{3x^2-11x+8}{2x^2-9x+7}=\frac{\left(x-1\right)\left(3x-8\right)}{\left(x-1\right)\left(2x-7\right)}=\frac{3x-8}{2x-7}\)
câu b,c tương tự nha ^^
a: \(Y=\dfrac{3\left(x^2-x-1\right)-x^2+1}{\left(x+2\right)\left(x-1\right)}+\dfrac{x-2}{x}\cdot\dfrac{1-1+x}{1-x}\)
\(=\dfrac{2x^2-3x-2}{\left(x+2\right)\left(x-1\right)}+\dfrac{x-2}{x}\cdot\dfrac{-x}{x-1}\)
\(=\dfrac{2x^2-3x-2}{\left(x+2\right)\left(x-1\right)}-\dfrac{x-2}{x-1}\)
\(=\dfrac{2x^2-3x-2-x^2+4}{\left(x+2\right)\left(x-1\right)}=\dfrac{x^2-3x+2}{\left(x+2\right)\left(x-1\right)}=\dfrac{x-2}{x+2}\)
b: Y=2
=>2x+4=x-2
=>x=-6(nhận)
c; Y nguyên
=>x+2-4 chia hết cho x+2
=>x+2 thuộc {1;-1;2;-2;4;-4}
Kết hợp ĐKXĐ, ta được: x thuộc {-1;-3;-4;-6}