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3x(12x - 4) - 9x(4x - 3) = 36x2 - 12x - 36x2 + 27x = 15x = 30 => x = 2
\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=30:15\)
\(\Leftrightarrow x=2\)
\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Rightarrow36x^2-12x-36x^2+27x=30\)
\(\Rightarrow15x=30\)
Vậy \(x=2\)
\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Rightarrow x\left(36x-12\right)-x\left(36x-27\right)=30\)
\(\Rightarrow x.\left[\left(36x-12\right)-\left(36x-27\right)\right]=30\)
\(\Rightarrow x.\left(36x-12-36x+27\right)=30\)
\(\Rightarrow x.\left(-12+27\right)=30\)
\(\Rightarrow15x=30\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
Bài giải:
a) 3x (12x - 4) - 9x (4x - 3) = 30
36x2 – 12x – 36x2 + 27x = 30
15x = 30
Vậy x = 2.
b) x (5 - 2x) + 2x (x - 1) = 15
5x – 2x2 + 2x2 – 2x = 15
3x = 15
x =5
a) 3x (12x - 4) - 9x (4x - 3) = 30
36x2 – 12x – 36x2 + 27x = 30
15x = 30
Vậy x = 2.
b) x (5 - 2x) + 2x (x - 1) = 15
5x – 2x2 + 2x2 – 2x = 15
3x = 15
x =5
a) ta có : \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x=30\Leftrightarrow15x=30\Leftrightarrow x=2\)
b) điều kiện : \(x\ne\dfrac{1}{5};x\ne1;x\ne\dfrac{3}{5}\)
ta có : \(\dfrac{3}{5x-1}+\dfrac{2}{3-3x}=\dfrac{4}{\left(1-5x\right)\left(5x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-3x\right)+2\left(5x-1\right)}{\left(5x-1\right)\left(3-3x\right)}=\dfrac{4}{\left(1-5x\right)\left(5x-3\right)}\)
\(\Leftrightarrow\dfrac{x+7}{3-3x}=\dfrac{4}{3-5x}\Leftrightarrow\left(x+7\right)\left(3-5x\right)=4\left(3-3x\right)\)
\(\Leftrightarrow-5x^2-20+9=0\)
ta có : \(\Delta'=\left(10\right)^2+5\left(9\right)=145>0\) \(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x=\dfrac{10+\sqrt{145}}{-5};x=\dfrac{10-\sqrt{145}}{-5}\)
a) \(36x^2-12x-36x^2+27x=30\)
\(15x=30\)
\(x=2\)
b) \(5x-2x^2+2x^2-2x=15\)
\(3x=15\)
\(x=5\)
a: \(=\dfrac{2x\left(3x^2+2\right)+3x^2+2}{3x^2+2}=2x+1\)
b: \(=\dfrac{2x^3-10x^2-17x^2+85x+30x-150}{x-5}=2x^2-17x+30\)
c: \(=\dfrac{12x^4-8x^3+12x^3-8x^2+8x^2-\dfrac{16}{3}x+\dfrac{43}{3}x-\dfrac{86}{9}+\dfrac{113}{9}}{3x-2}\)
\(=4x^3+4x^2+\dfrac{8}{3}x+\dfrac{43}{9}x+\dfrac{\dfrac{113}{9}}{3x-2}\)
a: \(=\dfrac{2x\left(3x^2+2\right)+3x^2+2}{3x^2+2}=2x+1\)
b: \(=\dfrac{2x^3-10x^2-17x^2+85x+30x-150}{x-5}=2x^2-17x+30\)
c: \(=\dfrac{12x^4-8x^3+12x^3-8x^2+8x^2-\dfrac{16}{3}x+\dfrac{43}{3}x-\dfrac{86}{9}+\dfrac{113}{9}}{3x-2}\)
\(=4x^3+4x^2+\dfrac{8}{3}x+\dfrac{43}{9}x+\dfrac{\dfrac{113}{9}}{3x-2}\)
a) -4x(x - 7) + 4x(x2 - 5) = 28x2 - 13
=> -4x2 + 28x + 4x2 - 20x = 28x2 - 13
=> (-4x2 + 4x2) + (28x - 20x) = 28x2 - 13
=> 8x = 28x2 - 13
=> 8x - 28x2 + 13 = 0
=> phương trình vô nghiệm
b) (4x2 - 5x)(3x + 2) - 7x(x + 5) = (-4 + x)(-2x - 3) + 12x2 + 2x2
=> 4x2(3x + 2) - 5x(3x + 2) - 7x2 - 35x = -4(-2x - 3) + x(-2x - 3) + 14x2
=> 12x3 + 8x2 - 15x2 - 10x - 7x2 - 35x = 8x + 12 - 2x2 - 3x + 14x2
=> 12x3 + (8x2 - 15x2 - 7x2) + (-10x - 35x) = (8x - 3x) + 12 + (-2x2 + 14x2)
=> 12x3 - 14x2 - 45x = 5x + 12 + 12x2
=> 12x3 - 14x2 - 45x - 5x - 12 - 12x2 = 0
=> 12x3 + (-14x2 - 12x2) + (-45x - 5x) - 12 = 0
=> 12x3 - 26x2 - 50x - 12 = 0
Làm nốt
Cái câu b sửa cái đề lại nhé dấu " = " ở chỗ (-2x = 3) là gì vậy?
3x(12x-4) - 9x(4x-3) = 30
<=> 36x2 - 12x - 36x2 + 27x = 30
<=> 15x = 30
<=> x = 2
\(3x\)\(\left(12x-4\right)\)\(-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2\)\(+27x=30\)
\(\Leftrightarrow15x=30\)
\(\Rightarrow x=2\)