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a) \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Rightarrow36x^2-12x-36x^2+27x=30\)
\(\Rightarrow\left(36x^2-36x^2\right)+\left(-12+27\right)=30\)
\(\Rightarrow0+15x=30\Leftrightarrow x=30:15=2\)
b) \(x\left(5-2x\right)+2x\left(x-1\right)=15\)
\(\Rightarrow5x-2x^2+2x^2-2x=15\)
\(\Rightarrow\left(5x-2x\right)+\left(-2x^2+2x^2\right)=15\)
\(\Rightarrow3x+0=15\Leftrightarrow x=15:3=5\)
a, 3x(12x - 4) - 9x(4x-3) = 30
36x2 - 12x - 36x2 + 27x = 30
- 12x + 27x = 30
15x = 30
x = 2
b, x(5 - 2x) + 2x(x - 1) = 15
5x - 2x2 + 2x2 - 2x = 15
5x - 2x = 15
3x = 15
x = 5
3x(12x – 4) – 9x(4x – 3) = 30
3x.12x – 3x.4 – (9x.4x – 9x.3) = 30
36x2 – 12x – 36x2 + 27x = 30
(36x2 – 36x2) + (27x – 12x) = 30
15x = 30
x = 2
Vậy x = 2.
3x (12x -4 )- 9x (4x-3)= 30
<=>36x2-12x-36x3+27x=30
<=>15x=30
<=>x=2
\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(36x^2-12x-36x^2+27x=30\)
\(15x=30\)
\(x=\frac{30}{15}\)
\(x=2\)
= 36X-12X-36X-27X=30
=36X-36X-12X-27X=30
=0-12X-27X
=-12X-27X=30
=X.(-12-27)=30
hình như mình tính sai
nếu công thức đúng thì k nha
\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
3x(12x - 4 ) -9x (4x -3 ) = 30
<=> 36x² - 12x - 36x²+27x = 30
<=> 15x = 30
<=> x=2
\(3x\left(12x-4\right)-9.\left(4x-3\right)=30\)
\(=>36x^2-12x-36x^2+27x=30\)
\(=>15x=30\)
\(x=30:15=2\)
Vậy x = 2.
~ Hok tốt ~
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
\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)
\(\Leftrightarrow27x-12x=30\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=\frac{30}{15}=2\)
\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(36x^2-12x-36x^2+27x=30\)
\(15x=30\)
\(x=2\)
Giải:
\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)
\(\Leftrightarrow-12x+27x=30\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=\dfrac{30}{15}\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\).
Chúc bạn học tốt!!!