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: \(x^2+3x-18=0\)

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

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

\(15x^4+30x^3+13x^2-2x-1=0\)

<=> \(15x^4+15x^3+15x^3+15x^2-2x^2-2x-1=0\)

<=> \(15x^2\left(x^2+x\right)+15x\left(x^2+x\right)-2\left(x^2+x\right)-1\)

<=> \(15\left(x^2+x\right)^2-2\left(x^2+x\right)-1=0\)

<=> \(\orbr{\begin{cases}x^2+x=\frac{1}{3}\\x^2+x=\frac{1}{5}\end{cases}}\)

Em tự giải tiếp nhé!

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.

Chọn D

D

Vì PTT 1 là âm còn PTT 2 là \(\dfrac{-A}{-B}\)    ==>    \(\dfrac{A}{B}\)   vậy PTT 2 là dương   (Âm<Dương)

a) \(8x^2+30x+7=0\)

\(\Leftrightarrow8\left(x^2+\frac{15}{4}x+7\right)=0\)

\(\Leftrightarrow x^2+\frac{1}{4}x+\frac{7}{2}x+\frac{7}{8}=0\) 

\(\Leftrightarrow x\left(x+\frac{1}{4}\right)+\frac{7}{2}\left(x+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(x+\frac{1}{4}\right)\left(x+\frac{7}{2}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+\frac{1}{4}=0\\x+\frac{7}{2}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=-\frac{7}{2}\end{array}\right.\)

b)\(x^3-11x^2+30x=0\)

\(\Leftrightarrow x\left(x^2-11x+30\right)=0\)

\(\Leftrightarrow x\left(x^2-5x-6x+30\right)=0\)

\(\Leftrightarrow x\left[x\left(x-5\right)-6\left(x-5\right)\right]=0\)

\(\Leftrightarrow x\left(x-5\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-5=0\\x-6=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\\x=6\end{array}\right.\)

a) \(\Delta=\left(30\right)^2-4.8.7=676>0\) ( PTC2NPB )

\(X_1=\frac{-30+\sqrt{676}}{16}\)

\(X_2=\frac{-30-\sqrt{676}}{16}\)