(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

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

=>  \(x\left(1+6x\right)=0\)

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

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

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

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

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

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

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

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

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

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

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

\(a,75\cdot20,9+5^2\cdot20,9=20,9\left(75+25\right)=20,9\cdot100=2090\)

\(b,85\cdot12,7+5\cdot3\cdot12,7=12,7\left(85+15\right)=12,7\cdot100=1270\)

• a. 2090 b.1270

\(a,\frac{3}{7}x^2+6x+x^2y=x\left(\frac{3}{7}x+6+xy\right)\)

\(b,3x\left(x+3y\right)-6xy\left(x+3y\right)=3x\left(x+3y\right)\left(1-2y\right)\)

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

\(d,3\left(x-y\right)-5x\left(y-x\right)=\left(3+5x\right)\left(x-y\right)\)

a.Ta có $\Delta ABC$ cân tại $A\to AB=AC,\widehat{ABC}=\widehat{ACB}\to \widehat{EBC}=\widehat{DCB}$

Xét $\Delta ABD,\Delta ACE$ có

$\widehat{ADB}=\widehat{AEC}={90}^o$ 

$AB=AC$

$\widehat{BAD}=\widehat{EAC}$

$\to \Delta ABD=\Delta ACE(c.g.c)$

$\to AE=AD$

b.Từ câu a$\to \frac{AE}{AB}=\frac{AD}{AC}$

$\to DE//BC$

Mà $\widehat{EBC}=\widehat{DCB}$

$\to BCDE$ là hình thang cân