a, (y-x^2)^2:(y-x^2) =y-x^2

b, (x-y^2)^2:(y-x^2)=x-y^2

học tốt

Bài làm:

a) \(\left(x^4-2x^2y+y^2\right)\div\left(y-x^2\right)\)

\(=\left(x^2-y\right)^2\div\left(y-x^2\right)\)

\(=\left(y-x^2\right)^2\div\left(y-x^2\right)\)

\(=y-x^2\)

b) \(\left(x^2-2xy^2+y^4\right)\div\left(x-y^2\right)\)

\(=\left(x-y^2\right)^2\div\left(x-y^2\right)\)

\(=x-y^2\)

a) ( x⁴ - 2x²y + y² ) : ( y - x² )

= ( x² - y )² : ( y - x² )

= ( x² - y )² : ( x² - y )

= x² - y 

b) ( x² - 2xy² + y⁴ ) : ( x - y² )

= ( x - y² )² : ( x - y² )

= x - y²

Lời giải:

$2x^2+y^2+2xy-8x-6y+30$

$=(x^2+y^2+2xy)+x^2-8x-6y+30$
$=(x+y)^2-6(x+y)+(x^2-2x)+30$
$=(x+y)^2-6(x+y)+9+(x^2-2x+1)+20$

$=(x+y-3)^2+(x-1)^2+20\geq 20$
Vậy GTNN của biểu thức là $20$ khi $x+y-3=x-1=0$

$\Leftrightarrow x=1; y=2$

(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁a^{n − 1}b + ⁿC₂a^{n − 2}b² + ⁿC₃a^{n − 3}b³ + ... + ⁿC_nbn
Đã nghĩ ra 
Nhờ công thức tổ hợp và chỉnh hợp lớp 11
 

( x + y )⁵ = x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + b⁵

nhầm 1 chút cái cuối cùng là công với x⁵  không pải cộng với b​⁵  sr bn nha

a) \(-\left(x+2\right)\cdot\left(x^2-1x+4\right)\)

\(=-\left(x+2\right)\cdot\left(x^2-x+4\right)\)

\(=-\left(x^3-x^2+4x+2x^2-2x+8\right)\)

\(=-\left(x^3+x^2+2x+8\right)\)

\(=-x^3-x^2-2x-8\)

b) \(-\left(x+2y\right)\cdot\left(x^2-2xy+y^2\right)\)

\(=-\left(x^3-2x^2y+xy^2+2x^2y-4xy^2+2y^3\right)\)

\(=-\left(x^3-3xy^2+2y^3\right)\)

\(=-x^3+3xy^2-2y^3\)

c) \(-\left(5-a\right)\cdot\left(25+5a+a^2\right)\)

\(=-\left(125-a^3\right)\)

\(=-125+a^3\)

d) \(-\left(x-2y\right)\cdot\left(x^2+2xy+4y^2\right)\)

\(=-\left(x^3-8y^3\right)\)

\(=-x^3+8y^3\)

a, sai đề

b, \(-\left(x+2y\right)\left(x^2-2xy+y^2\right)=-\left(x^3+y^3\right)\)

\(=-x^3-y^3\)

c, \(-\left(5-a\right)\left(25+5a+a^2\right)=-\left(125-a^3\right)\)

\(=a^3-125\)

d, \(-\left(x-2y\right)\left(x^2+2xy+4y^2\right)=-\left(x^3-8y^3\right)\)

\(=8y^3-x^3\)

Giải:

5) \(-x^2+x-\dfrac{1}{2}\)

\(=-x^2+x-\dfrac{1}{4}+\dfrac{3}{4}\)

\(=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)

\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\le\dfrac{3}{4}\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy ...

6) \(-\dfrac{1}{4}x^2+x-2\)

\(=-\dfrac{1}{4}x^2+x-1-1\)

\(=-\left(\dfrac{1}{4}x^2-x+1\right)-1\)

\(=-\left(\dfrac{1}{2}x-1\right)^2-1\le-1\)

\(\Leftrightarrow\dfrac{1}{2}x-1=0\Leftrightarrow x=2\)

Vậy ...

7) \(-\dfrac{1}{9}x^2-\dfrac{1}{3}x+1\)

\(=-\dfrac{1}{9}x^2-\dfrac{1}{3}x-\dfrac{1}{4}+\dfrac{5}{4}\)

\(=-\left(\dfrac{1}{9}x^2+\dfrac{1}{3}x+\dfrac{1}{4}\right)+\dfrac{5}{4}\)

\(=-\left(\dfrac{1}{3}x+\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy ...

8) \(-2x^2+2xy-2y^2+2x+2y-8\)

\(=-x^2+2xy-y^2+2x-x^2+2y-y^2-1-1-6\)

\(=-\left(x^2-2xy+y^2\right)-\left(x^2-2x+1\right)-\left(y^2-2y+1\right)-6\)

\(=-\left(x-y\right)^2-\left(x-1\right)^2-\left(y-1\right)^2-6\le-6\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y-1=0\end{matrix}\right.\Leftrightarrow x=y=1\)

Vậy ...

a) Ta có  (1 + 2y)² + (1 - 2y)² + 2(1 + 2y)(1 - 2y) 

= (1 + 2y + 1 - 2y)² = 2² = 4

b) Ta có (7x + 2y)² + (7x - 2y)² - 2(49x² - 4y²) 

= (7x + 2y)² + (7x - 2y)² - 2[(7x)² - (2y)²]

= (7x + 2y)² + (7x - 2y)² - 2(7x - 2y)(7x + 2y)

= (7x + 2y - 7x + 2y)² 

= (4y)² = 16y²

\(\text{1) }\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\\ =\dfrac{\left(x^7+x^6\right)+\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^6+x^4+x^2+1}{x-1}\)

\(\text{3) }\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\\ =\dfrac{\left(x^2-2xy+y^2\right)+\left(2xz-2yz\right)+z^2}{\left(x^2-2xy+y^2\right)-z^2}\\ =\dfrac{\left(x-y\right)^2+2\left(x-y\right)z+z^2}{\left(x-y\right)^2-z^2}\\ =\dfrac{\left(x-y+z\right)^2}{\left(x-y+z\right)\left(x-y-z\right)}\\ =\dfrac{x-y+z}{x-y-z}\)

Câu 1 :

Ta có :

\(\dfrac{x^7+x^6+x^5+x^4+x^3+x^2+x+1}{x^2-1}\)

\(=\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x+1\right)\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)}\)

\(=\dfrac{x^4\left(x^2+1\right)+\left(x^2+1\right)}{\left(x-1\right)}\)

\(=\dfrac{\left(x^2+1\right)\left(x^4+1\right)}{x-1}\)

Câu 2 : Ta có VT :

\(\dfrac{2x^2+3xy+y^2}{2x^3+x^2y-2xy^2-y^3}\)

\(=\dfrac{2x^2+2xy+xy+y^2}{x^2\left(2x+y\right)-y^2\left(2x+y\right)}\)

\(=\dfrac{2x\left(x+y\right)+y\left(x+y\right)}{\left(2x+y\right)\left(x^2-y^2\right)}\)

\(=\dfrac{\left(x+y\right)\left(2x+y\right)}{\left(x+y\right)\left(x-y\right)\left(2x+y\right)}\)

\(=\dfrac{1}{x-y}=VP\left(đpcm\right)\)

Câu 3 :

\(\)Ta có :

\(\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)

\(=\dfrac{\left(x-y+z\right)^2}{\left(x-y\right)^2-z^2}=\dfrac{\left(x-y+z\right)\left(x-y+z\right)}{\left(x-y-z\right)\left(x-y+z\right)}\)

\(=\dfrac{\left(x-y+z\right)}{\left(x-y-z\right)}\)

\(\)