a: \(\Leftrightarrow8x^2+16x+14x+7=0\)

=>(2x+1)(8x+7)=0

=>x=-1/2 hoặc x=-7/8

b: \(=x^3-x-6x-6\)

\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-6\right)=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)

\(a,\Rightarrow8x^2+2x+28x+7=0\\ \Rightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\\ \Rightarrow\left(2x+7\right)\left(4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\\ b,Sửa:x^3-7x-6=0\\ \Rightarrow x^3-x-6x-6=0\\ \Rightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)

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) 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: \(x^2+3x-18=0\)

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

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

i) x³- 11x² + 30x

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

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

j) 4x⁴- 21x²y² + y⁴

=4x^4+4x^2y^2+y^4-25x^2y^2

=(2x^2+y^2)^2-(5xy)^2

=(2x^2+y^2-5xy)(2x^2+y^2+5xy)

              k,x³ + 4x²- 7x - 10

\(a,\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ b,\Leftrightarrow\left(3x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{3}\)

a) pt <=> x^2(x - 4)(x + 4) = 0

<=> x = 0 hoặc x = 4 hoặc x = -4

b) pt <=> (3x -5)^2=0

<=> x = 5/3

a: \(x^4-16x^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

b: \(9x^2-30x+25=0\)

\(\Leftrightarrow3x-5=0\)

hay \(x=\dfrac{5}{3}\)

a: \(9x^2-30x+25=0\)

\(\Leftrightarrow3x-5=0\)

hay \(x=\dfrac{5}{3}\)

c: \(9x^2-25=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

a) \(9x^2-30x+25=0\Rightarrow\left(3x-5\right)^2=0\Rightarrow x=\dfrac{5}{3}\)

b) \(25x^2-5x+\dfrac{1}{4}=0\Rightarrow\left(10x-1\right)^2=0\Rightarrow x=\dfrac{1}{10}\)

c) \(9x^2-25=0\Rightarrow\left(3x-5\right)\left(3x+5\right)=0\)

    \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

d) \(\left(2x-1\right)^2-\left(3x+2\right)^2=0\)

   \(\Rightarrow\left(2x-1+3x+2\right)\left(2x-1-3x-2\right)=0\)

  \(\Rightarrow-\left(5x+1\right)\left(5x+3\right)=0\)

 \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

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

...

b: Δ=(-12)^2-4*2*(9+4căn 2)

=144-72-32căn 2=72-32căn 2

=(8-2căn 2)^2

=>PT có hai nghiệm pb là:

\(\left\{{}\begin{matrix}x=\dfrac{12-8+2\sqrt{2}}{4}=\dfrac{2+\sqrt{2}}{2}\\x_2=\dfrac{2-\sqrt{2}}{2}\end{matrix}\right.\)

c: Δ=(-30)^2-4*3*(-26+8căn 3)

=900+312-96căn 3

=1212-2*căn 3072

=>Phương trình có hai nghiệm pb là:

\(\left\{{}\begin{matrix}x=\dfrac{30-2\sqrt{1212-2\sqrt{3072}}}{6}\\x=\dfrac{30+2\sqrt{1212-2\sqrt{3072}}}{6}\end{matrix}\right.\)

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

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

\(=>\orbr{\begin{cases}x-6=0\\x-5=0,x=0\end{cases}}\)

\(=>\orbr{\begin{cases}x=6\\x=5,x=0\end{cases}}\)

P/S: mk mới lớp 7 sai sót mong bỏ qua 

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

\(8x^2+28x+2x+7=0\)

\(2x.\left(4x+1\right)+7.\left(4x+1\right)=0\)

\(\left(2x+7\right).\left(4x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=-7\\4x=-1\end{cases}}\)

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

vậy ....

P/S sorry mk làm hơi lâu :)__chờ tí làm câu a cho

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

\(x^2-3x+6x-18=0\)

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

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

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

P/S:nếu cách làm sai sót hay dài thì bn thồn cảm nha mk kiểm tra kết quả thấy ko sai