Question

Tính giá trị biểu thức

\(9x^2\)+  42x+49 tại x=1

\(\frac{1}{4}a^2+2ab^2+4b^4\)tại a=2 , b=-1

Answer 1

\(9x^2+42x+49=\left(3x+7\right)^2\)

Thay x=1 ta có 

\(\left(3.1+7\right)^2=10^2=100\)

\(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a+2b^2\right)^2\)

Thay a=2;b=-1 ta có 

\(\left(\frac{1}{2}.2+2\left(-1\right)^2\right)^2=\left(1+2\right)^2=3^2=9\)

Answer 2

\(\(9x^2+42x+49\)\)tại x = 1

Ta có:\(\(9x^2+42x+49=\left(3x\right)^2+2.3x.7+7^2=\left(3x+7\right)^2\)\)

Thay x = 1 vào \(\(\left(3x+7\right)^2\)\)ta được:

\(\(\left(3.1+7\right)^2=10^2=100\)\)

\(\(\frac{1}{4}a^2+2ab^2+4b^4\)\)tại a = 2 ; b = -1

Ta có: \(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b^2+\left(2b\right)^2=\left(\frac{1}{2}a+2b^2\right)^2\)

Thay a = 2 ; b = -1 vào\(\left(\frac{1}{2}a+2b^2\right)^2\)ta được:

\(\(\left(\frac{1}{2}.2+2.\left(-1\right)^2\right)^2=\left(3\right)^2=9\)\)