a) Ta có: \(x^2-x-12=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

b) Ta có: \(x^2+3x-18=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=3\end{matrix}\right.\)

Phương trình phải có vế phải bạn nhé!

a) (4x-1)(2-x)-(2x-1)²

= 8x-4x²-2+x-(4x²-4x+1) = -8x²+13x-3

b) (15x⁴y⁵-30x³y⁴+35x³y⁴):(5x³y³)

= 3xy²-6y+7y = 3xy²+y

a: \(=8x-4x^2-2+2x-4x^2+4x-1\)

\(=-8x^2+14x-3\)

\(x=\dfrac{3+\sqrt{5}}{2}\Rightarrow2x-3=\sqrt{5}\Rightarrow4x^2-12x+9=5\)

\(\Rightarrow4x^2-12x+4=0\Rightarrow x^2-3x+1=0\)

\(\Rightarrow P=\left[10\left(x^2-3x+1\right)+1\right]^2+\dfrac{\left[2\left(x^2-3x+1\right)+1\right]^{10}}{x^3\left(x^2-3x+1\right)-1}=1^2+\dfrac{1^2}{0-1}=...\)

x^2*(x-30)-31x+1

thay x=31 vao bieu thuc 

(31)^2*(31-30)-31*31+1=1

a: \(=\dfrac{5\cdot3x+4x-3-2\left(15x-1\right)}{10}=\dfrac{19x-3-30x+2}{10}=\dfrac{-11x-1}{10}\)

b: \(\Leftrightarrow x\left(x-4\right)+\left(x+1\right)^2=2x\left(x+1\right)\)

\(\Leftrightarrow x^2-4x+x^2+2x+1=2x^2+2x\)

=>-2x+1=2x

=>-4x=-1

hay x=1/4(nhận)

\(a,\dfrac{3x}{2}+\dfrac{4x-3}{10}-\dfrac{15x-1}{5}\)

\(=\dfrac{3x.5+4x-3-2\left(15x-1\right)}{10}\)

\(=\dfrac{15x+4x-3-30x+3}{10}\)

\(=\dfrac{-11x}{10}\)

Lời giải:
$3x=16y\Rightarrow \frac{x}{16}=\frac{y}{3}$
Áp dụng TCDTSBN:

$\frac{x}{16}=\frac{y}{3}=\frac{x+y}{16+3}=\frac{190}{19}=10$
$\Rightarrow x=10.16=160; y=3.10=30$

Đáp án A.

a. (3x - 1)² - (x + 3)² = 0

\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)

\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)

\(\Leftrightarrow4x+2=0\)  hoặc  \(2x-4=0\)

1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)

2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)

S=\(\left\{-\dfrac{1}{2};2\right\}\)

b. \(x^3=\dfrac{x}{49}\)

\(\Leftrightarrow49x^3=x\)

\(\Leftrightarrow49x^3-x=0\)

\(\Leftrightarrow x\left(49x^2-1\right)=0\)

\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow x=0\) hoặc  \(7x+1=0\) hoặc \(7x-1=0\)

1. x=0

2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)

3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

*Cách khác:

a) Ta có: \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(3x-1\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-x-3\\3x-1=x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{2};2\right\}\)

\(a,\frac{3y}{4}=\frac{6xy}{8x}\\\frac{3y}{4}=\frac{3y.2x}{4.2x}=\frac{6xy}{8x}\)

\(b,\frac{10}{15x}=\frac{20xy}{30x^2y}\\ \frac{10}{15x}=\frac{10.2xy}{15x.2xy}=\frac{20xy}{30x^2y}\)

\(c,\frac{3x^3y^5}{2xy^6}=\frac{3x^2}{2y}\\ \frac{3x^2}{2y}=\frac{3x^2.x.y^5}{2y.x.y^5}=\frac{3x^3y^5}{2xy^6}\)