K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

1 tháng 3 2017

Theo bài ra , ta có :

\(A=2\left(x^6-y^6\right)-3\left(x^4+y^4\right)+1\)

\(\Leftrightarrow A=2[\left(x^2\right)^3-\left(y^2\right)^3]-3\left(x^4+y^4\right)+1\)

\(\Leftrightarrow A=2\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)-3\left(x^4+y^4\right)+1\)

\(\Leftrightarrow A=2x^4+2x^2y^2+2y^4-3x^4-3y^4\)(Vì x2 - y2 = 1)

\(\Leftrightarrow A=-x^4+2x^2y^2-y^4+1=-\left(x^4-2x^2y^2+y^4\right)=-\left(\left(x^2-y^2\right)^2\right)=-1+1=0\)Vậy A = 0

Chúc bạn học tốt =))ok

1 tháng 3 2017

A = 2(x^2 - y^2).(x^4 + x^2y^2 + y^4) - 3x^4 - 3y^4 +1

A = 2x^4 + 2.x^2y^2 + 2y^4 - 3x^4 - 3y^4 +1

A = -x^4 + 2.x^2y^2 -y^4 +1

A = - (x^2 - y^2) +1

A = -1 + 1 =0

7 tháng 2 2017

Ta có :

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

\(\Rightarrow\left(x^2-y^2\right)^3=1^3\)

\(\Rightarrow x^2-y^6-3x^2y^2\left(x^2-y^2\right)=1\)

\(\Rightarrow x^6-y^6=1+3x^2y^2\left(x^2-y^2\right)\)

\(\Rightarrow x^6-y^6=1+3x^2y^2\)

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

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

\(\Rightarrow x^4-2x^2y^2+y^4=1\)

\(\Rightarrow x^4+y^4=1+2x^2y^2\)

Khi đó :

\(A=2\left(x^6-y^6\right)-3\left(x^4+y^4\right)+1\)

\(=2\left(1+3x^2y^2\right)-3\left(1+2x^2y^2\right)+1\)

\(=0\)

Vậy \(A=0\).

22 tháng 6 2017

\(A=\left(x+1\right)^3-\left(x+3\right)^2\left(x+1\right)+4x^2+8\)\(=\left(x+1\right)\left[\left(x+1\right)^2-\left(x+3\right)^2\right]+4x^2+8\)

\(=\left(x+1\right)\left(x+1+x+3\right)\left(x+1-x-3\right)+4x^2+8\)\(=\left(x+1\right)\left(2x+4\right).-2+4x^2+8=-2\left(2x^2+4x+2x+4\right)+4x^2+8=-4x^2-12x-8+4x^2+8=-12x\) Với \(x=\dfrac{-1}{6}\Rightarrow A=\left(-12\right).\left(\dfrac{-1}{6}\right)=2\)

a: \(A=x^3+3x^2+3x+1-\left(x^2+6x+9\right)\left(x+1\right)+4x^2+8\)

\(=x^3+7x^2+3x+9-x^3-x^2-6x^2-6x-9x-9\)

\(=-12x\)

\(=-12\cdot\dfrac{-1}{6}=2\)

b: Sửa đề: \(B=2\left(x^6+y^6\right)-3\left(x^4+y^4\right)\)

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

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

\(=-\left(x^4+2x^2y^2+y^4\right)=-1\)

25 tháng 7 2017

Câu 1 :

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

\(A=3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)\(\Leftrightarrow3\left(6x^2-2x-6\right)-2\left(4x^2+13x-12\right)+36x-9x^2=0\)\(\Leftrightarrow18x^2-6x-18-8x^2-26x+24+36x-9x^2=0\)\(\Leftrightarrow x^2+4x+6=0\)

\(\Leftrightarrow\left(x+2\right)^2=-2\)

Ta có:

\(\left(x+2\right)^2\ge0\forall x\)

Vậy pt vô nghiệm

Vậy:ko......

Câu 3:

\(\left(5x-3\right)\left(7x+2\right)-35x\left(x-1\right)=42\)

\(\Leftrightarrow35x^2+10x-21x-6-35x^2+35x-42=0\)\(\Leftrightarrow14x=48\Leftrightarrow x=\dfrac{7}{24}\)

Câu 4:

\(\left(3x+5\right)\left(2x-1\right)+\left(5-6x\right)\left(x+2\right)=x\)

\(\Leftrightarrow6x^2-3x+10x-5+5x+10-6x^2-12x-x=0\)\(\Leftrightarrow-x=-5\Rightarrow x=5\)

câu 6,

25 tháng 7 2017

Câu 6: \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)

\(\Rightarrow10x^2+9x-\left(10x^2-2x+15x-3\right)=8\)

\(\Rightarrow10x^2+9x-10x^2+2x-15x+3=8\)

\(\Rightarrow-4x+3=8\)

\(\Rightarrow-4x=5\Rightarrow x=\dfrac{-5}{4}\)

Câu 7: \(x\left(x+1\right)\left(x+6\right)-x^3=5x\)

\(\Rightarrow\left(x^2+x\right)\left(x+6\right)-x^3=5x\)

\(\Rightarrow x^3+x^2+6x^2+6x-x^3=5x\)

\(\Rightarrow7x^2=-x\)

\(\Rightarrow7x=-1\Rightarrow x=\dfrac{-1}{7}\).