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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
3x(12x-4) - 9x(4x-3) = 30
<=> 36x2 - 12x - 36x2 + 27x = 30
<=> 15x = 30
<=> x = 2
\(3x\)\(\left(12x-4\right)\)\(-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2\)\(+27x=30\)
\(\Leftrightarrow15x=30\)
\(\Rightarrow x=2\)
a) 3x(12x - 4) - 9x(4x - 3) = 30
36x2 - 12x - 36x2 + 27x = 30
15x = 30
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
=> x = 30 : 15 = 2
b) x (5 - 2x) + 2x (x - 1) = 15
5x - 2x2 + 2x2 - 2x = 15
3x = 15
=> x = 15 : 3 = 5
Bài 1:
a)\(-x^2\left(3x^3-2x+\frac{1}{2}\right)=-3x^5+2x^3-\frac{x^2}{2}\)
b) Sửa đề: \(\left(2xy-y^2+x\right)\frac{2}{3}x^3y=\frac{4}{3}x^4y^2-\frac{2}{3}x^3y^3+\frac{2}{3}x^4y\)
Không sửa đề: \(\left(2xy-y2+x\right)\frac{2}{3}x^3y=\frac{4}{3}x^4y^2-\frac{4}{3}x^3y^2+\frac{2}{3}x^4y\)
Bài 2:
a)\(3x\left(12x-5\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-15x-36x^2+27x=30\)
\(\Leftrightarrow12x=30\Leftrightarrow x=\frac{5}{2}\)
b) \(x\left(7-2x\right)+2x\left(x-4\right)=15\)
\(\Leftrightarrow7x-2x^2+2x^2-8x=15\)
\(\Leftrightarrow-x=15\Leftrightarrow x=-15\)
Hok tốt
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, Điều kiện xác định: x<>0
b, Điều kiện xác định: x <> -1/3
c, Điều kiện xác định: x<>2
d, Điều kiện xác định: a<>0 và b<>0; b<>2a
A : không rút gọn được
\(B=\frac{4x^2\left(x-2\right)+3\left(x-2\right)}{3x\left(4x^2+3\right)+4x^2+3}=\frac{\left(4x^2+3\right)\left(x-2\right)}{\left(4x^2+3\right)\left(3x+1\right)}=\frac{x-2}{3x+1}\)
\(C=\frac{x^4-1}{x^3+2x^2-x-2}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x+2\right)\left(x^2-1\right)}=\frac{x^2+1}{x+2}\)
\(D=\frac{a^3+b^3}{a^3+\left(a-b\right)^3}=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a+a-b\right)\left(a^2-a^2+ab+a^2-2ab+b^2\right)}\)\(=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(2a-b\right)\left(a^2-ab+b^2\right)}=\frac{a+b}{2a-b}\)
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.