\(x^2-2xy+y^2-z^2+2zt-t^2\)

\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)

\(=\left(x-y\right)^2-\left(z-t\right)^2\)

\(=\left(x-y+z-t\right)\left(x-y-z+t\right)\)

hok tốt!

\(x^2-2xy+y^2-z^2+2zt-t^2\)

\(=\left(x-y\right)^2-\left(z-t\right)^2\)

\(=\left(x-y+z-t\right)\left(x-y-z+t\right)\)

x³+3x-3=0

x³+3x=0+3

x³+3x=3

x(x²+3)=3

3:x=x²:3

=>ko co gia tri nao thoa man

a) \(25x^2-9=0\)

\(\Leftrightarrow\left(5x\right)^2-3^2=0\)

\(\Leftrightarrow\left(5x+3\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)

Vậy \(S=\left\{\frac{3}{5};\frac{-3}{5}\right\}\)

b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x+17=16\)

\(\Leftrightarrow8x=-1\)

\(\Leftrightarrow x=-\frac{1}{8}\)

Vậy.........

c)\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(x^2+6x+9\right)-5\left(x^2-49\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow2x=-255\)

\(\Leftrightarrow x=-127,5\)

Vậy.............

có j sai xót mong m.n bỏ qua☺

a) \(25x^2-9=0\)                      

<=> \(\left(5x\right)^2=9\)

<=> \(\left(5x\right)^2=3^2\)

<=> \(5x=3\)

<=> \(x=\frac{3}{5}\)

b) \(\left(x+4\right)^2-\left(x-1\right)\left(x+1\right)=16\)

<=> \(x^2+2.x.4+4^2-\left(x^2-1^2\right)=16\)

<=> \(x^2+8x+16-x^2+1=16\)

<=> \(\left(x^2-x^2\right)+8x+\left(16+1\right)=16\)

<=> \(8x+17=16\)

<=> \(8x=-1\)

<=> \(x=\frac{-1}{8}\)

c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

<=> \(\left(2x\right)^2-2.2x.1+1^2+x^2+2.x.3+3^2-5\left(x^2-7^2\right)=0\)

<=> \(4x^2-4x+1+x^2+6x+9-5x^2+5.7^2=0\)

<=> \(\left(4x^2+x^2-5x^2\right)-\left(4x-6x\right)+\left(1+9+5.7^2\right)=0\)

<=> \(2x+245=0\)

<=> \(2x=-245\)

<=> \(x=\frac{-245}{2}\)

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

2 )   \(3x^2\left(x-1\right)+5x\left(1-x\right)=3x^2\left(x-1\right)-5x\left(x-1\right)=\left(3x^2-5x\right)\left(x-1\right)\)

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

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

5 )    \(4x^2-36=4\left(x^2-9\right)=4\left(x+3\right)\left(x-3\right)\)

4x² - 36

= 4.x² - 4.9

= 4(x² - 9)

x² + xy + x 

= x.x + xy + x

= x(x + y + 1)

chỉnh lại đề là tìm x nhé:

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

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...

Gọi N là trung điểm của HD .

Ta có : MN là đường trung bình của tam giác HDC 

\(\Rightarrow MN//DC\)

\(MN=\frac{1}{2}DC\) (T/c đường TB )

Ta lại có : 

\(AB//DC\)và  \(AB=MN\)

=> ABMN là hình bình hành .

\(\Rightarrow AN//BM\)(1)

Xét tam giác ADM có :

\(\hept{\begin{cases}DH\perp AM\\MN\perp AD\end{cases}}\)

\(\Rightarrow AN\perp DM\)(2)

Từ (1) và (2)

\(\Rightarrow\widehat{BMD}=90^o\)(đpcm)

a) = x^2 + 2x + 1 - x^2 +2x - 1 -3x^2 +x - x - 1

= - 3x^2 +4x -1

b) =5x^2 + 10x - 10x - 20 - 1/2 .(36 - 96x + 64x^2 ) +17

= 5x^2 - 20 - 18 - 48 x - 32x^2 +17

= -27x^2 - 48x - 3

Chúc bn hok tốt a !