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\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Rightarrow36x^2-12x-36x^2+27x=30\)
\(\Rightarrow15x=30\)
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 \(x=2\)
3x(12x-4) - 9x(4x-3) = 30
<=> 36x2 - 12x - 36x2 + 27x = 30
<=> 15x = 30
<=> x = 2
\(3x\)\(\left(12x-4\right)\)\(-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2\)\(+27x=30\)
\(\Leftrightarrow15x=30\)
\(\Rightarrow x=2\)
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)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0
TH2 3x-4=0 <=>x=4/3
KL:.....
b) (x+3)(x−1)+2x(x+3)=0.
<=> (x+3)(x-1+2x)=0
TH1: x+3=0 <=> x=-3
TH2 x-1=0 <=> x=1
KL:.....
c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 \)
KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)
KL:....
a) \(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(9x^2+6x+1=0\)
\(\Leftrightarrow\left(3x+1\right)^2=0\)
\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
d) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
3x(12x - 4) - 9x(4x - 3) = 36x2 - 12x - 36x2 + 27x = 15x = 30 => x = 2
\(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=30:15\)
\(\Leftrightarrow x=2\)