a) A=(4-5x)²-(3+5x)²=(4-5x-3-5x)(4-5x+3+5x)=(-25x+1)1=-25x+1

B=(3x-1)(1+3x)-(3x+1)²=9x²-1-(3x+1)²=9x²-1-(9x²+6x+1)=9x²-1-9x²-6x-1=-6x-2=-2(3x+1)

êffeeffe

a/ \(4x^2-49=\left(2x\right)^2-7^2=\left(2x-7\right)\left(2x+7\right)\)

b/ \(a^2-2a-b^2-2b=\left(a^2-2a+1\right)-\left(b^2+2b+1\right)=\left(a-1\right)^2-\left(b+1\right)^2\)

\(=\left(a-1-b-1\right)\left(a-1+b+1\right)=\left(a-b-2\right)\left(a+b\right)\)

1:

Để đây là 1 cấp số nhân thì

\(\left[{}\begin{matrix}\left(2a-1\right)^2=a\left(2a+1\right)\\a^2=\left(2a-1\right)\left(2a+1\right)\\\left(2a+1\right)^2=a\left(2a-1\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}4a^2-4a+1-2a^2-a=0\\4a^2-1-a^2=0\\4a^2+4a+1-2a^2+a=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2a^2-5a+1=0\\3a^2-1=0\\2a^2+5a+1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}a=\dfrac{5\pm\sqrt{17}}{4}\\a=\pm\dfrac{\sqrt{3}}{3}\\a=\dfrac{-5\pm\sqrt{17}}{4}\end{matrix}\right.\)

2:

Để đây là 1 cấp số nhân thì

\(\left[{}\begin{matrix}\left(2b+3\right)^2=7\cdot49\\7^2=49\left(2b+3\right)\\49^2=7\left(2b+3\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\left(2b+3\right)^2=343\\2b+3=1\\2b+3=343\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}b=-1\\b=170\\2b+3=\pm7\sqrt{7}\end{matrix}\right.\)

=>\(b\in\left\{-1;170;\dfrac{7\sqrt{7}-3}{2};\dfrac{-7\sqrt{7}-3}{2}\right\}\)

@tth

@Neet

@ngonhuminh

bạn vừa đăng câu này rồi mà

`4b^2+a^2+4ab`

`=(2b)^2+2.2b.a+a^2`

`=(a+2b)^2`

`-49-2a^4+14sqrt2a^2`

`=-(2a^4-14sqrt2a^2+49)`

`=-((sqrt2a^2)^2-2.sqrt2a^2.7+7^2)`

`=-(sqrt2a^2-7)^2`

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

\(-49-2a^4+14\sqrt{2a^2}=-\left(\sqrt{2a^2}-7\right)^2\)

a: =400+50-450=0