To you :)

Ta có :

\(A=-x^2+7x\)

\(=-x^2+2.x.\frac{7}{2}-\frac{49}{4}+\frac{49}{4}\)

\(=\frac{49}{4}-\left(x^2-2.x.\frac{7}{2}+\frac{49}{4}\right)\)

\(=\frac{49}{4}-\left(x-\frac{7}{2}\right)^2\)

\(\left(x-\frac{7}{2}\right)^2\ge0\)

\(\Rightarrow\frac{49}{4}-\left(x-\frac{7}{2}\right)^2\le\frac{49}{4}\)

\(\Rightarrow Min_A=\frac{49}{4}\Leftrightarrow x=\frac{7}{2}\)

Vậy ...

\(\left(x+1\right)^2-\left(x+2\right)^2=0\)

\(\Rightarrow\left[\left(x+1\right)+\left(x+2\right)\right]\left[\left(x+1\right)-\left(x+2\right)\right]=0\)

\(\Rightarrow\left[2x+3\right]\left(-1\right)=0\)

\(\Rightarrow2x+3=0\)

\(\Rightarrow2x=-3\)

\(\Rightarrow x=-\frac{3}{2}\)

Vậy ...

\(a\left(b^2+c^2\right)+b\left(a^2+c^2\right)+c\left(b^2+a^2\right)+2abc\)

\(=ab^2+ac^2+a^2b+bc^2+cb^2+a^2c+2abc\)

\(=\left(a^2b+a^2c+ab^2+abc\right)+\left(abc+ac^2+b^2c+bc^2\right)\)

\(=a\left(ab+ac+b^2+bc\right)+c\left(ab+ac+b^2+bc\right)\)

\(=\left(a+c\right)\left(ab+ac+b^2+bc\right)\)

\(=\left(a+c\right)\left[\left(ab+ac\right)+\left(b^2+bc\right)\right]\)

\(=\left(a+c\right)\left[a\left(b+c\right)+b\left(b+c\right)\right]\)

\(=\left(a+c\right)\left(b+c\right)\left(a+b\right)\)

a)lát đề

b)x³+x+2

=x³-x²+2x+x²-x+2

=x(x²-x+2)+(x²-x+2)

=(x+1)(x²-x+2)

c)x⁴+64

=(x²)²+8²+2x²*8-2x²*8 

=(x²+8)²-(4x)²

=(x²-4x+8)(x²+4x+8) . 

Vì n nguyên dương nên ta có \(n^2< n^2+n+1< n^2+2n+1\)

hay \(n^2< n^2+n+1< \left(n+1\right)^2\)

Mà n và (n+1) là hai số chính phương liên tiếp và \(n^2+n+1\)là số kẹp giữa  hai số ấy nên không thể là số chính phương.

Gọi số đó là x 

Ta có:

x²=4x³ =>x²-4x³=0

=>x²(1-4x)=0

=>x²=0 hoặc 1-4x=0

=>x=0 hoặc \(x=\frac{1}{4}\)

Vậy....

\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2\left(axby+bycz+axcz\right)\)

\(\Leftrightarrow\left(a^2y^2-2axby+b^2x^2\right)+\left(a^2z^2-2axcz+x^2c^2\right)+\left(c^2y^2-2bycz+b^2z^2\right)=0\)

\(\Leftrightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(cy-bz\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}ay=bx\\az=cx\\cy=bz\end{cases}\Leftrightarrow}\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)

\(\Leftrightarrow a^2y^2-2axby+b^2x^2=0\)

\(\Leftrightarrow\left(ay-bx\right)^2=0\Leftrightarrow ay-bx=0\Leftrightarrow ay=bx\Leftrightarrow\frac{a}{x}=\frac{b}{y}\)

Cảm ơn bạn nhiều