Question

[Lê Thị Hương]

1-Triễn khai lũy thừa: 1-(1-3)³ (à cho mình hỏi cái này hình như dựa vào hằng đẳng thức đúng hk,nếu có thì ghi giúp mình cái hằng đẳng đó lun nha )

2-chưng minh rằng số:70(71⁹+71⁸+71⁷+...+71²+72)+1 là 1 số chính phương

3-tìm GTLN của: A=4x-x²+3 B=x-x² C=2x-2x²-5

4-tìm GTNN của: M=x²+y²-x+6y+10

giúp mình nha

Answer 1

[Đức Hiếu]

Bài 1:

Áp dụng hằng đẳng thức số 5 ta có:

\(1-\left(1-3\right)^3=1-\left(1-3.1.3+3.1.3^2-3^2\right)\)

\(=1-\left(1-9+27-9\right)=1-1+9-27+9=-9\)

Chúc bạn học tốt!!!

Answer 2

[Nguyễn Huy Tú]

Bài 1:

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

hđt: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

Bài 3:

a, \(A=4x-x^2=-x^2+4x\)

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

\(=-\left[\left(x-2\right)^2-4\right]\)

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

Ta có: \(-\left(x-2\right)^2\le0\)

\(\Leftrightarrow A=-\left(x-2\right)^2+4\le4\)

Dấu " = " xảy ra khi \(-\left(x-2\right)^2=0\Leftrightarrow x=2\)

Vậy \(MAX_A=4\) khi x = 2

b, \(B=x-x^2=-x^2+x\)

\(=-\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}\right)\)

\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\right]\)

\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

Dấu " = " khi \(-\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(MAX_B=\dfrac{1}{4}\) khi \(x=\dfrac{1}{2}\)

c, \(C=2x-2x^2-5\)

\(=-2\left(x^2-x+\dfrac{5}{2}\right)\)

\(=-2\left(x^2-2.x\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{9}{4}\right)\)

\(=-2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\right]\)

\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le\dfrac{-9}{2}\)

Dấu " = " khi \(-2\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(MAX_C=\dfrac{-9}{2}\) khi \(x=\dfrac{1}{2}\)

Bài 4:

\(M=x^2+y^2-x+6y+10\)

\(=\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\)

Ta có: \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\)

\(\Leftrightarrow M=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

Vậy \(MIN_M=\dfrac{3}{4}\) khi \(x=\dfrac{1}{2},y=-3\)