@Nguyễn Huy Tú

a) Ta có : \(\left(x+y\right)^3=1^3=1\)

\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=1\)

\(\Leftrightarrow x^3+y^3+3xy=1\) ( do x + y = 1 )

help me

1.Tính:

[(x+y)⁵-2(x+y)⁴ ] : [-5(x+y)³]

= -5(x+y)² + \(\dfrac{2}{5}\)(x+y)

2.Tìm a để đa thức 24x³ -14x² +23x+2a+4 \(⋮\) 4x+1

24x³ -14x² +23x+2a+4 \(|^{4x+1}_{6x^2-5x+7}\)

24x³ +6x²

\(\overline{-20x^2}+23x+2a+4\)

-20x² -5x

\(\overline{28x+2a+4}\)

28x +7

\(\overline{2a+11}\)

Để 24x³ -14x² +23x+2a+4 \(⋮\) 4x+1 thì 2a+11=0 \(\Leftrightarrow\) a= \(\dfrac{11}{2}\)

3. Phân tích đa thức thành NT :

a, 12x³ -12x² +3x = 3x(4x² -4x+1) = 3x (2x+1)

b, x².(x-1)+9(1-x) = x² (x-1) -9(x-1) = (x-1)(x²-9)

=(x-1)(x-3)(x+3)

c,8(x-y)-x³ (x-y) = (x-y)(8-x³)= (x-y)(2-x)(4+2x+x²)

Bài 4: 

=>x^2-x+1/4=0

=>(x-1/2)^2=0

=>x-1/2=0

=>x=1/2

Bài 5: 

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

=>\(2xy=x^2+y^2-\left(x-y\right)^2=15-5^2=-10\)

=>xy=-5

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

\(=5^3+3\cdot\left(-5\right)\cdot5=125-75=50\)

\(a,C=\left(\dfrac{2x+1}{x^3-1}-\dfrac{1}{x-1}\right):\left(1-\dfrac{x^2+3}{x^2+x+1}\right)\)

\(\Leftrightarrow\left(\dfrac{2x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1.\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{\left(x-1\right)\left(x^2+3\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right)\)

\(\Leftrightarrow\dfrac{2x+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}:\dfrac{x^3-1-x^3-3x+x^2+3}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow\dfrac{-x^2+x}{x^3-1}.\dfrac{x^3-1}{x^2-3x+2}\)

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

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

b, Để C = 3

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

\(\Leftrightarrow-x=3x-6\)

\(\Leftrightarrow-x-3x=-6\)

\(\Leftrightarrow-4x=-6\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy ..................

1: \(M=\dfrac{1}{x+1}-\dfrac{x^3-x}{x^2+1}\cdot\dfrac{1}{x^2+2x+1}-\dfrac{1}{x^2-1}\)

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

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

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

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

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

2: Để M=1 thì \(-2x^2+2x-2=\left(x^2-1\right)\left(x^2+1\right)\)

\(\Leftrightarrow x^4-1+2x^2-2x+2=0\)

\(\Leftrightarrow x^4+2x^2-2x+1=0\)

hay \(x\in\varnothing\)

a: \(x^2+y^2=\left(x+y\right)^2-2xy=4-2\cdot\left(-3\right)=10\)

\(x^4+y^4=\left(x^2+y^2\right)^2-2\left(xy\right)^2=100-2\cdot\left(-3\right)^2=100-2\cdot9=82\)

b: \(x^3+y^3+3xy\)

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

=1-3xy+3xy=1

d: \(A=\left(x+y\right)^2-4\left(x+y\right)+1=9-4\cdot3+1=10-12=-2\)

a: ĐKXĐ: x>0; x<>9

b: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{3\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{3\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+2}\)

\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+2}\)

a: ĐKXĐ: x>0; x<>9

b: \(D=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right):\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)

\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

c: Để D<-1 thì D+1<0

\(\Leftrightarrow-3\sqrt{x}+2\sqrt{x}+4< 0\)

\(\Leftrightarrow4-\sqrt{x}< 0\)

hay x>16