Bài 1:
Ta có:
\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=\left(ac\right)^2+2acbd+\left(bd\right)^2+\left(ad\right)^2-2abcd+\left(bc\right)^2\)
\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=a^2.\left(c^2+d^2\right)+b^2.\left(c^2+d^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)=VT\)
\(\rightarrow\)đpcm
Chúc bạn học tốt!!!
Bài 1:
\(VT=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
\(=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2=VP\)
\(\Rightarrowđpcm\)
Bài 2
Đặt
\(A=x^3+9x^2+27x+27=\left(x+3\right)^3\)
Thay x = 97
\(\Leftrightarrow A=100^3=1000000\)
Vậy A = 1000000 khi x = 97
Bài 2:
\(x^3+9x^2+27x+27=x^3+3x^2+6x^2+18x+9x+27\)
\(=x^2.\left(x+3\right)+6x.\left(x+3\right)+9.\left(x+3\right)\)
\(=\left(x+3\right).\left(x^2+6x+9\right)=\left(x+3\right).\left(x+3\right)^2\)
\(=\left(x+3\right)^3\)(1)
Thay x=97 vào (1) ta có:
\(\left(97+3\right)^3=100^3=1000000\)
Vậy.....
Chúc bạn học tốt!!!
1: Chứng minh rằng \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
VT = \(a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
VP = \(a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)
\(=a^2c^2+a^2d^2+b^2d^2+b^2c^2=VT\)
=> đpccm
\(x^3+9x^2+27x+27\) = \(x^3+3.3.x^2+3.3^2.x+3^3\)
= \(\left(x+3\right)^3\) (hằng đẳng thức ) thay \(x=97\) vào phương trình
= \(\left(97+3\right)^3=100^3=1000000\)
vậy \(x=97\) thì \(x^3+9x^2+27x+27=1000000\)
