Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x=81\)

\(\Leftrightarrow83x-2=81\)

\(\Leftrightarrow83x=81+2=83\)

\(\Leftrightarrow x=1\)

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

48x² – 12x – 20x + 5 + 3x – 48x² – 7 + 112x = 81.

83x – 2 = 81.

83x = 83.

x = 1.

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

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

⇔83x-2=81

⇔83x=81 +2=83

⇔x = 1

Chuc ban hoc tot❓❗⛇

Giải:

a) \(\left(3x-1\right)^2-\left(2x+3\right)^2=0\)

\(\Leftrightarrow\left(3x-1+2x+3\right)\left(3x-1-2x-3\right)=0\)

\(\Leftrightarrow\left(5x+2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=4\end{matrix}\right.\)

Vậy ...

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

\(\Leftrightarrow48x^2-20x-12x+5+3x-7-48x^2+112x=81\)

\(\Leftrightarrow83x-2=81\)

\(\Leftrightarrow83x=83\)

\(\Leftrightarrow x=1\)

Vậy ...

a/ \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

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

<=> \(83x-2=81\)

<=> \(83x=83\)

<=> \(x=1\)

b/ \(\left(2x-3\right)\left(2x+3\right)-\left(4x+1\right)x=1\)

<=> \(4x^2-9-4x^2-x=1\)

<=> \(-\left(9+x\right)=1\)

<=> \(9+x=-1\)

<=> \(x=-10\)

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

<=> \(3x^2-\left(3x^2-x+6x-2\right)=-7\)

<=> \(3x^2-3x^2+x-6x+2=-7\)

<=> \(-5x+2=-7\)

<=> \(-5x=-9\)

<=> \(x=\frac{9}{5}\)

\(\left(12x-5\right).\left(4x-1\right)+\left(3x-7\right).\left(1-16x\right)=81\)

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

\(83x-2=81\)

\(83x=83\)

=>\(x=1\)

bài dễ cậu tự làm được mÀ

len google di ban

mk chua hoc bai nay

pt tương đương:

[(4x+1)(3x+2)][(12x-1)(x+1)]=700

<=>(12x²+11x+2)(12x²+11x-1)=700

Đặt t=12x²+11x pt trở thành

(t+2)(t-1)=700

<=>t²+t-702=0

<=>t=26 hoặc t=-27

với t=26 =>x=13/12 hoặc x=-2

với t=-27 không có x