phan h B = (4^32 - 1)(4^32 + 1) = ( 4^16 + 1)(4^32+1)(4^16 - 1)

..... => ket qua

Bạn gì ơi mình không hiểu!

Bài 2

\(a,x^3+2x^2+x\)

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

\(b,xy+y^2-x-y\)

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

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

bài 3

\(a,3x.\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2=4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=2,x=-2\end{cases}}\)

vậy x=0,x=2 hay x=-2

\(b,xy+y^2-x-y=0\)

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

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

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

vậy x=-1, y=1

kết quả bằng 0 nhé bạn

=6

chắc lak tôi có on đấy

#G2k6#

\(x^4-3x^2+9=\left(x^2\right)^2+2.x^2.3+3^2-9x^2=\left(x^2+3\right)^2-\left(3x\right)^2=\left(x^2-3x+3\right)\left(x^2+3x+3\right)\)

\(x^4-7x^2+1=\left(x^4+2x^2+1\right)-9x^2=\left(x^2+1\right)^2-\left(3x\right)^2=\left(x^2-3x+1\right)\left(x^2+3x+1\right)\)

      \(x^3+4x^2-31x-70\)

\(=x^3+2x^2+2x^2+4x-35x-70\)

\(=x^2\left(x+2\right)+2x\left(x+2\right)-35\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+2x-35\right)\)

\(=\left(x+2\right)\left[x\left(x+7\right)-5\left(x+7\right)\right]=\left(x+2\right)\left(x+7\right)\left(x-5\right)\)

Ta có:

\(x^2-2018x+1=0\)

\(\Leftrightarrow x^2+1=2018x\)

Do đó

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

\(\Leftrightarrow B=\frac{\left(2018x+x\right)\left(2018x-x\right)}{x^2}=\frac{2019x\cdot2017x}{x^2}=2019\cdot2017\)

Giải sơ qua:

1)\(B=4x^2-4xy+2y^2+1=\left(2x-y\right)^2+y^2+1\ge1\)

2) có vẻ sai đề

Đúng đề hết nhé

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

2x³ - x² + 5x + 3

= 2x³ + x² - 2x² - x + 6x + 3

= x²(2x + 1) - x(2x + 1) + 3(2x + 1)  

= (2x + 1)(x² - x + 3)

thay z = -(x+y) , y = -(z+x),... vao

=> Duoc bieu thuc trong do co 1/xy + 1/yz + 1/zx = (x+y+z)/xyz = 0

TL :

5,972E24 kg

HT

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