\(a,\left(3x+1\right)^2=x^2\)

=>  \(\left(3x+1\right)^2-x^2=0\)

=>  \(\left(3x+1-x\right)\left(3x+1+x\right)=0\)

=>  \(\left(2x+1\right)\left(4x+1\right)=0\)

=> \(\orbr{\begin{cases}2x+1=0\\4x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{1}{4}\end{cases}}}\)

\(b,x^2-8x+16=0\)

=>  \(\left(x-4\right)^2=0\)

=>  \(x-4=0\)

=>  \(x=4\)

\(a,73^2-13^2-10^2+20\cdot13=\left(73-13-10\right)^2+20\cdot13=50^2+260=2500+260=2760\)

Ta có :

\(73^2-13^2-10^2+20.13=\left(73+13\right)\left(73-13\right)-13.13+20.13\)

\(=86.60-13\left(13-20\right)\)

\(=5160-\left(-91\right)=5160+91=...\)

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

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

\(=\left(x+y\right)\left(5x-5y\right)=5\left(x+y\right)\left(x-y\right)\)

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

\(d,8x^3+1=\left(2x\right)^3+1^3=\left(2x+1\right)\left(4x^2-2x+1\right)\)

Bài 1

\(a,\left(2x+1\right)^2=4x^2+4x+1\)

\(b,x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+9\right)\)

\(c,\left(\frac{1}{2}x-1\right)^2=\frac{1}{4}x^2-x+1\)

\(d,8x^3+1=\left(2x\right)^3+1^3=\left(2x+1\right)\left(4x^2-2x+1\right)\)

\(e,x^4-1=\left(x^2-1\right)\left(x^2+1\right)=\left(x+1\right)\left(x-1\right)\left(x^2+1\right)\)

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

Mình học lớp 5 mà

em mới lớp 4 thôi nên ko bt

lớp bốn thì bài dễ hơn nhiều

(12x-5)(4x-1)+(3x-7)(1-16x)=81

<=>48x2-12x-20x+5+3x-48x2-7+112x=81

<=>-32x+115x=81+2 

<=>83x=83  

<=>x=1

(12x-5)(4x-1)+(3x-7)(1-16x)=81

\(48x^2\)\(-12x-20x\)\(+5+3x\)\(-48x^2\)-7+112x=81

<=>-32x+115x=81+2 

<=>83x=83  

<=>x=1

cho xin kb

a) Với x = 0 thì ta được

$\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)$

$=\left(0-5\right)\left(0+3\right)+\left(0+4\right)\left(0-0\right)$

$=-5.3+0$

$=-15$

b) Với x = 15 thì ta được

$\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)$

$=\left({15}^2-5\right)\left(15+3\right)+\left(15+4\right)\left(15-{15}^2\right)$

$=220.18+19.\left(-210\right)$

$=3960-3990$

$=-30$

c) Với x = -15 thì ta được

$\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)$

$=\left[{\left(-15\right)}^2-5\right]\left(-15+3\right)+\left(-15+4\right)\left[-15-{\left(-15\right)}^2\right]$

$=220.\left(-12\right)+\left(-11\right).\left(-240\right)$

$=-2640+2640$

$=0$

d) Với x = 0,15 thì ta được

$\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)$

$=\left[{\left(0,15\right)}^2-5\right]\left(0,15+3\right)+\left(0,15+4\right)\left[0,15-{\left(0,15\right)}^2\right]$

$=-4,9775.3,15+4,15.0,1275$

$=-15,679125+0,529125$

$=-15,15$

a)3y(2y-4)-2y(3y+5)=44
6y2-12y-(6y2+10y)=44
6y2-12y-6y2-10y=44
-22y=44
y=-2
Vậy y=-2
b)(12x-5)(4x-1)+(3x-7)(1-16x)=81
48x2-12x-20x+5+3x-48x2-7+112x=81
          83x-2=81
          83x    =81+2
          83x    =83
              x=1
          Vậy x=1
                                     Đấy nhé bạn,nhớ k cho mình nha.<3

ĐỀ BÀI HƠI SAI SỬA LẠI:

Tìm x, y biết a) 3y(2y-4)-2y(3y+5)=44

                     b)(12x-5)(4x-1)+(3x-7)(1-16x)=81

\(a,x+6x^2=0\)

=>  \(x\left(1+6x\right)=0\)

=>  \(\orbr{\begin{cases}x=0\\1+6x=0\Leftrightarrow6x=-1\Leftrightarrow x=-\frac{1}{6}\end{cases}}\)

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

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

=>  \(\orbr{\begin{cases}2-x=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

\(c,5x\left(x-2\right)-\left(2-x\right)=0\)

=>  \(\left(5x+2\right)\left(x-2\right)=0\)

=>  \(\orbr{\begin{cases}5x+2=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{5}\\x=2\end{cases}}}\)

\(d,\left(x+1\right)=\left(x+1\right)^2\)

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

=>   \(\left(x+1\right)\left(x+1-1\right)=0\)

=>  \(\orbr{\begin{cases}x+1=0\\x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}}\)