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\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\text{⇔}48x^2-32x+5-48x^2-7+115x=81\)
\(\text{⇔}83x-2=81\)
\(\text{⇔}83x=83\)
\(\text{⇔}x=1\)
Vậy: x=1
Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x=81\)
\(\Leftrightarrow83x=83\)
hay x=1
\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-32x+5+48x^2+115x-7=81\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x=81\\ \Leftrightarrow83x=83\Leftrightarrow x=1\)
Rút gọn vế trái:
VT = (12x – 5)(4x – 1) + (3x – 7)(1 – 16x)
= 12x.(4x – 1) + (–5).(4x – 1) + 3x.(1 – 16x) + (–7).(1 – 16x)
= 12x.4x+ 12x.(–1) + (–5).4x + (–5).(–1) + 3x.1 + 3x.(–16x) + (–7).1 + (–7).(–16x)
= 48x2 – 12x – 20x + 5 + 3x – 48x2 – 7 + 112x
= (48x2 – 48x2) + (– 12x – 20x + 3x + 112x) + (5 – 7)
= 83x – 2
Vậy ta có:
83x – 2 = 81
83x = 81 + 2
83x = 83
x = 83 : 83
x = 1.
\(\Rightarrow48x^2-32x+5-48x^2+115x-7=81\)
\(\Rightarrow83x=83\Rightarrow x=1\)
1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)
\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)
\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)
\(\Leftrightarrow5x-6=0\)
\(\Leftrightarrow5x=6\)
\(\Leftrightarrow x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)
\(\Leftrightarrow12x+24=0\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy: x=-2
3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)
\(\Leftrightarrow15x-30=0\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
Vậy: x=2
4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)
\(\Leftrightarrow83x-83=0\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
Vậy: x=1
Ta có : ( 12x - 5 ) ( 4x - 1 ) + ( 3x - 7 ) ( 1 -16x ) = 81
=> 48x2 - 20x - 12x + 5 + 3x - 7 - 48x2 + 112x = 81
=> 80x - 2 = 81
=> 80x = 83
=> x = 83/80
<=> 48x^2 - 12x - 20x + 5 + 3x - 48x^2 - 7 + 112x = 81
<=> -32x + 115x = 81 + 2
<=> 83x = 83
<=> x = 1
Có viết sai gì ko đấy
(12x-5).(4x-1)-(3x-7).16x=81
48x²-12x-20x+5-(48x²-112x)=81
48x²-12x-20x+5-48x²+112x=81
(48x²-48x²)+(112x-12x-20x)+5=81
0+100x-20x+5=81
80x+5=81
80x=81-5
80x=76
x=76:80
x=19/20
Vậy x=19/20
Hok tốt