ta có : (10^50)^3<10^150+5*10^50+1<10^150+3*(10^50)^2+3*10^50+1= (10^50+1)^3

vay10^150+5*10^50+1 khong la lap phuong cua 2 so tu nhien

Tham khảo .

Ta có :

\(\left(10^{53}\right)^3< 10^{150}+5.10^{50}+1< 10^{150}+3.\left(10^{50}\right)^2+1\)

\(=\left(10^{50}+1\right)^3\)

Vậy \(10^{150}+5.10^{50}+1\)không là lập phương của 1 số tự nhiên 

đpcm

= (-2a ^2 +4a)(1-x)

= -2 (a^2 -2a)(1-x)

= -2a(a-2)(1-x)

= 2a(a-2)(x-1)

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

\(=-a\cdot2a\left(x-1\right)-2\cdot2a\left(x-1\right)\)

\(=2a\left(x-1\right)\left(-a-2\right)\)

a)  \(x^5+x+1\)

\(=\left(x^5-x^4+x^2\right)+\left(x^4-x^3+x\right)+\left(x^3-x^2+1\right)\)

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

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

b)  \(x^5+x^4+1=\left(x^2+x+1\right)\left(x^3-x+1\right)\)

c)  \(x^8+x+1=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

d)  \(x^8+x^7+1=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

\(x^8+x^7+1\)

\(=\left(x^8-x^6+x^5-x^3+x^2\right)+\left(x^7-x^5+x^4-x^2+x\right)+\left(x^6-x^4+x^3-x+1\right)\)

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

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

a) \(x^2-2x+x\)

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

b) \(x^2+y^2-2xy-9\)

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

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

c) \(x^2-y^2+6x+9\)

\(=\left(x^2+6x+9\right)-y^2\)

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

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

\(x^2-2x+x\)

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

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

\(=\left(x-1\right)\left(x-1+1\right)\)

\(=x.\left(x-1\right)\)

\(x^2-4xy+y^2\cdot4=0\)

\(\left(x-2y\right)^2=0\)

\(x-2y=0\)

\(x=2y\)

\(x^2+2x+1-y^2=0\)

\(\left(x+1\right)^2-y^2=0\)

\(\left(x+1-y\right)\left(x+1+y\right)=0\)( Bạn làm tiếp nếu có thể )

\(\left(x+y\right)^2-9y^2=0\)

\(\left(x+y-3y\right)\left(x+y+3y\right)=0\)

\(\left(x-2y\right)\left(x+4y\right)=0\)( Tương tự trên )

\(x^2-4xy+y^2.4=0\)

\(x^2-2x.2y+\left(2y\right)^2=0\)

\(\left(x-2y\right)^2=0\)

\(\Rightarrow x-2y=0\)

\(\Rightarrow x=2y\)

Vậy \(x=2y\)

\(x^2+2x+1-y^2=0\)

\(\left(x^2+2x.1+1^2\right)-y^2=0\)

\(\left(x+1\right)^2-y^2=0\)

\(\left(x+1-y\right)\left(x+1+y\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1-y=0\\x+1+y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=y-1\\x=-y-1\end{cases}}}\)

Vậy \(x=y-1\)hoặc \(x=-y-1\)

\(\left(x+y\right)^2-9y^2=0\)

\(\left(x+y\right)^2-\left(3y\right)^2=0\)

\(\Rightarrow\left(x+y-3y\right)\left(x+y+3y\right)=0\)

\(\Rightarrow\left(x-2y\right)\left(x+4y\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2y=0\\x+4y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2y\\x=-4y\end{cases}}}\)

Vậy \(x=2y\)hoặc \(x=-4y\)

Bài này không có y cho nên mình không tìm x cụ thể được, bạn thông cảm

ta co

A=x-5/x+7=x+7-12/x+7=1-12/x-7<1

suy ra x<1 

a) Ta có :

\(889=900-1\)

\(=30^2-1\)

\(=\left(30+1\right)\left(30-1\right)\)

\(=31.29\)

b) \(9991=10000-9\)

\(=100^2-3^2\)

\(=\left(100+3\right)\left(100-3\right)\)

\(=103.97\)

899 mới đúng nhé bạn ơi

Bỏ đi phần a=b=c =0 mới giải được nha .

Ta có :

Bình phương 2 vế của a+b+c =0   ta được :

\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)(1)

Bình phương 2 vế của (1) ta được :

\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)

\(=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

Ta có: (a+b+c)^2 + a^2 + b^2 + c^2

= a^2 +b^2 +c^2 + 2ab + 2ac + 2bc + a^2 + b^2 + c^2

= (a^2 +2ab+ b^2) + (b^2 +2bc+ c^2) +(c^2 +2ac+ a^2 )

= (a+b)^2 +(b+c)^2 +(c+a)^2

\(\left(a^2+b^2+c^2\right)+a^2+b^2+c^2\)

\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)

\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)