Question

Answer 1

a)

\(P=(x-10)^2-x(x+80)\\=x^2-2\cdot x\cdot10+10^2-x^2-80x\\=x^2-20x+100-x^2-80x\\=-100x+100\\=-100(x-1)\)

Thay \(x=0,87\) vào \(P\), ta được:

\(P=-100(0,87-1)=-100\cdot(-0,13)=13\)

\(---\)

b)

\(Q=4a^2+8ab+4b^2\\=4(a^2+2ab+b^2)\\=4(a+b)^2\)

Thay \(a=65;b=35\) vào \(Q\), ta được:

\(Q=4(65+35)^2\\=4\cdot 100^2\\=4\cdot 10000\\=40000\)

\(---\)

c)

\(R=x^3-3x^2+3x-1\\=x^3-3\cdot x^2\cdot1+3\cdot x\cdot 1^2-1^3\\=(x-1)^3\)

Thay \(x=101\) vào \(R\), ta được:

\(R=(101-1)^3\\=100^3\\=1000000\)

\(Toru\)

Answer 2

a: \(P=\left(x-10\right)^2-x\left(x+80\right)\)

\(=x^2-20x+100-x^2-80x=-100x+100\)

Khi x=0,87 thì \(P=100-100\cdot0.87=100-87=13\)

c: \(R=x^3-3x^2+3x-1\)

\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=\left(x-1\right)^3\)

Khi x=101 thì \(R=\left(101-1\right)^3=100^3=1000000\)

b: \(Q=4a^2+8ab+4b^2\)

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

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

Khi a=65 và b=35 thì \(Q=4\left(35+65\right)^2=4\cdot100^2=40000\)