=(x-1)^2-y^2

=(x-1-y)(x-1+y)

\(x^2-2x-y^2+1\)

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

\(=\left(x-1\right)^2-y^2\)

\(=\left[\left(x-1\right)-y\right]\left[\left(x-1\right)+y\right]\)

\(=\left(x-1-y\right)\left(x-1+y\right)\)

Lời giải:

a. $2x^2+3(x-1)(x+1)=5x(x+1)$

$\Leftrightarrow 2x^2+3x^2-3=5x^2+5x$

$\Leftrightarrow 5x^2-3=5x^2+5x$
$\Leftrightarrow 5x=-3$

$\Leftrightarrow x=\frac{-3}{5}$

b.

PT $\Leftrightarrow (-5x^2-2x+16)+4(x^2-x-2)=4-x^2$

$\Leftrightarrow -x^2-6x+8=4-x^2$

$\Leftrightarrow -6x+8=4$
$\Leftrightarrow -6x=-4$

$\Leftrightarrow x=\frac{2}{3}$

c.

PT $\Leftrightarrow 4(x^2+4x-5)-(x^2+7x+10)=3(x^2+x-2)$

$\Leftrightarrow 4x^2+16x-20-x^2-7x-10=3x^2+3x-6$

$\Leftrightarrow 3x^2+9x-30=3x^2+3x-6$

$\Leftrightarrow 6x=24$

$\Leftrightarrow x=4$

\(\Leftrightarrow-12x+17x=20+2\\ \Leftrightarrow5x=22\\ \Leftrightarrow x=\dfrac{22}{5}\)

\(\Leftrightarrow27-27x+9x^2-x^3=0\\ \Leftrightarrow3^3-3.3^2.x+3.3.x^2-x^3=\left(3-x\right)^3=0\)

\(\Rightarrow3-x=0\Rightarrow x=3\)

a: =x^2+7x+49/4-61/4

\(=\left(x+\dfrac{7}{2}\right)^2-\dfrac{61}{4}\)

\(=\left(x+\dfrac{7-\sqrt{61}}{2}\right)\left(x+\dfrac{7+\sqrt{61}}{2}\right)\)

b: \(=9x^2+2\cdot3x\cdot\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{35}{36}\)

\(=\left(3x+\dfrac{1}{6}\right)^2+\dfrac{35}{36}\)

5:

A=x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4>0 với mọi x

=>A>0 với mọi x

B=3(x^2-2/3x+5/3)

=3(x^2-2/3x+1/9+14/9)

=3(x-1/3)^2+14/3>0

=>B>0 với mọi x

C=6x-x^2-14

=-(x^2-6x+14)

=-(x^2-6x+9+5)

=-(x-3)^2-5<0 với mọi x

a: Xét ΔADM và ΔBCM có

AD=BC

góc ADM=góc BCM

DM=CM

=>ΔADM=ΔBCM

=>MA=MB

b: ΔMAB cân tại M

mà MN là đường trung tuyến

nên MN vuông góc AB

=>x^2+6x+9+x^2+10x+25=0

=>2x^2+16x+34=0

=>x^2+8x+17=0

=>(x+4)^2+1=0(loại)

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

\(\Leftrightarrow x^2-4=2\left(x^2+3x-2x-6\right)\)

\(\Leftrightarrow x^2-4=2x^2+2x-12\)

\(\Leftrightarrow x^2-2x^2-2x=-12+4\)

\(\Leftrightarrow-x^2-2x=-8\)

\(\Leftrightarrow-x^2-2x+8=0\)

\(\Leftrightarrow-x^2+2x-4x+8=0\)

\(\Leftrightarrow-x\left(x-2\right)-4\left(x-2\right)=0\)

\(\Leftrightarrow\left(-x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x-4=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{-4;2\right\}\)

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

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=2\left(x-2\right)\left(x+3\right)\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)-2\left(x-2\right)\left(x+3\right)=0\)

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

\(\Leftrightarrow\left(x-2\right)\left(x+2-2x-6\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`(x+3)(x-4)`

`= x(x-4) + 3(x-4)`

`= x^2-4x + 3x - 12`

`= x^2 - x - 12`

=x^2-4x+3x-12

=x^2-x-12