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7 tháng 1 2018
\(a,\dfrac{2x+2y}{a^2+2ab+b^2}.\dfrac{ax-ay+bx-by}{2x^2-2y^2}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{a+b}\)
\(b,\dfrac{a+b-c}{a^2+2ab+b^2-c^2}.\dfrac{a^2+2ab+b^2+ac+bc}{a^2-b^2}\)
\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{1}{a-b}\)
\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2-x+1\right)}\) \(=\dfrac{x-1}{2}\) \(d,\dfrac{x^8-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4\right)^2-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x+1}.\dfrac{1}{x^2+1}\) \(=\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\) \(=x-1\) \(e,\dfrac{x-y}{xy+y^2}-\dfrac{3x+y}{x^2-xy}.\dfrac{y-x}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x\left(x-y\right)}.\dfrac{-\left(x-y\right)}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x}.\dfrac{-1}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{-3x-y}{x\left(x+y\right)}\) \(=\dfrac{x\left(x-y\right)+y\left(3x+y\right)}{xy\left(x+y\right)}\) \(=\dfrac{x^2-xy+3xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{x^2+2xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}=\dfrac{x+y}{xy}\)19 tháng 2 2018
tìm giá trị của m để pt 2x-m=1-x nhận giá trị x=-2 là nghiệm
giải hộ e với :)
CM
20 tháng 8 2017
T a c ó : x 2 + y 2 = a 2 + b 2 ⇔ x 2 - a 2 = b 2 - y 2 ⇔ x - a x + a = b - y b + y M à x + y = a + b ⇔ x - a = b - y n ê n t a c ó x - a x + a = x - a b + y ⇔ x - a x + a - x - a b + y = 0 ⇔ x - a x + a - b - y = 0 ⇔ x - a = 0 x + a - b - y = 0 ⇔ x = a x - y = b - a
+) Với x = a thay vào x + y = a + b ta có: a + y = a + b
Suy ra y = b
Do đó: x n + y n = a n + b n
+) Với x - y = b - a suy ra x = b - a + y thay vào x + y = a + b ta có:
b - a + y + y = a + b
2y = 2a
y = a
Suy ra x - a = b - a hay x = b
Do đó: x n + y n = b n + a n = a n + b n
Vậy x n + y n = a n + b n
Đáp án cần chọn là C
AA
0
TA
1
XO
8 tháng 8 2021
Ta có x + y = a + b
=> (x + y)2 = (a + b)2
=> x2 + y2 + 2xy = a2 + b2 + 2ab
=> xy = ab
Lại có x + y = a + b
=> (x + y)3 = (a + b)3
=> x3 + 3x2y + 3xy2 + y3 = a3 + 3a2b + 3ab2 + b3
=> x3 + y3 + 3xy(x + y) = a3 + b3 + 3ab(a + b)
=> x3 + y3 = a3 + b3 (vì x + y = a + b ; xy = ab)
10 tháng 7 2021
Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+b^2y^2+2axby\)
\(\Leftrightarrow a^2y^2-2axby+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay=bx\)
hay \(\dfrac{a}{x}=\dfrac{b}{y}\)
10 tháng 7 2021
Ta có : \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2\)
\(\Leftrightarrow a^2y^2-2abxy+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay-bx=0\)
\(\Leftrightarrow ay=bx\Leftrightarrow\dfrac{a}{b}=\dfrac{x}{y}\)
PT
1
CM
6 tháng 1 2018
a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).
b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2
= (a + 4 – b)(a + 4 + b).
CM
17 tháng 11 2019
a) (x - 1)(x + l)(x - 2)(x - 4). b) (x - 2)( x 2 + 4).
c) 2y(3 x 2 + y 2 ). d) 2(x + y + z) ( a - b ) 2 .
24 tháng 8 2021
a. \(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1\)
\(=\left(x-3\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left[\left(x-3\right)^2-1\right]\left(x^2-1\right)\)
\(=\left(x-3+1\right)\left(x-3-1\right)\left(x+1\right)\left(x-1\right)\)
\(=\left(x-2\right)\left(x-4\right)\left(x+1\right)\left(x-1\right)\)
b. \(x^3-2x^2+4x-8\)
\(=\left(x^3+4x\right)-\left(2x^2+8\right)\)
\(=x\left(x^2+4\right)-2\left(x^2+4\right)\)
\(=\left(x-2\right)\left(x^2+4\right)\)
c. \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3\)
\(=2y\left(3x^2+y^2\right)\)
d. \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)
\(=\left(2a^2-4ab+2b^2\right)\left(x+y+z\right)\)
\(=2\left(a^2-2ab+b^2\right)\left(x+y+z\right)\)
\(=2\left(a-b\right)^2\left(x+y+z\right)\)
1 tháng 11 2023
a: \(x^2-9-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(1-x^2\right)\)
\(=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\)
b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
d: \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
e: \(3x^2-4x-4\)
\(=3x^2-6x+2x-4\)
\(=3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x+2\right)\)
g: \(x^4+64y^4\)
\(=x^4+16x^2y^2+64y^4-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
1 tháng 11 2023
h: \(a^2+b^2+2a-2b-2ab\)
\(=a^2-2ab+b^2+2a-2b\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)
i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)
\(=\left(x+1-y+3\right)^2\)
\(=\left(x-y+4\right)^2\)
k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)