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 )

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

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

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: \(x^2+3x-18=0\)

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

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

a)

ta có \(x^2+8x+7\)

\(=x^2+7x+x+7\)

\(=x\left(x+7\right)+\left(x+7\right)\)

\(=\left(x+7\right)\left(x+1\right)\)

b)

Ta có \(8x^2+30x+7\)

\(=8x^2+2x+28x+7\)

\(=2x\left(4x+1\right)+7\left(4x+1\right)\)

\(\left(4x+1\right)\left(2x+7\right)\)

A)(x+1)(x+7)

B)(x+1/4)(x+7/2)

giải rõ hơn dc không bạn ??

\(x^2+3x-18=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}}\)

\(8x^2+30x+7=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=0\\x+\frac{7}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=-\frac{7}{2}\end{cases}}}\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}}\)hoặc \(x=0\)

\(x^2+3x-18=x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x-2\right)\left(x+6\right)\)

8x²+30x+7

=8x²+2x+28x+7

=(8x²+2x)+(28x+7)

=2x(4x+1)+7(4x+1)

=(2x+7)(4x+1)

b) 8x² + 30x + 7  = 0

8x² + 16x + 14x + 7  = 0

8x.(x+2) + 7.(x+2) = 0

(x+2).(8x+7) = 0

..

bn tự làm tiếp nhé! ^-^

c) x³ - 11x² + 30x = 0

x.(x² - 11x +30) = 0

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

x.[ x.(x-5) - 6.(x-5) ] = 0

x.(x-5).(x-6) = 0

...

Đề bài là gì bạn? Lần sau bạn đăng câu hỏi thì nhớ thêm cái đề vào nhé!

2x\(^2\)+8x+6= 2x\(^2\)+2x+6x+6= 2x(x+1)+6(x+1)=(2x+6)(x+1)

3x\(^2\)-8x+5= 3x\(^2\)-3x-5x+5=3x(x-1)-5(x-1)= (3x-5)(x+1)

2x\(^2\)-30x+28= 2x\(^2\)-2x-28x+28= 2x(x-1)-28(x-1)= 2(x-19)(x-1)

4x\(^2\)+8x+3= 4x\(^2\)+2x+6x+3= 2x(2x+1)+3(2x+1)= (2x+3)(2x+1)