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

c: \(x^4+x^3-4x^2+x+1\)

\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)

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

\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\right]\)

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

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) \(x^2-x+x=4\)

\(x^2=4\)

\(x=\pm2\)

b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)

\(\left(x-5\right)\left(3x-2\right)=0\)

\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c) Ta có: \(a+b+c=5-3-2=0\)

\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)

d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :

\(t^2-11t+18=0\)

\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)

\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)

\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)

a:Ta có: \(x\left(x-1\right)+x=4\)

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

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

b: Ta có: \(3x\left(x-5\right)-2x+10=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c: Ta có: \(5x^2-3x-2=0\)

\(\Leftrightarrow5x^2-5x+2x-2=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d: Ta có: \(x^4-11x^2+18=0\)

\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)

\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

a) x(x-1)+x=4

⇔x²=4⇔\(x=\pm2\)

b)3x(x-5)-2x+10=0

⇔3x(x-5)-2(x-5)=0

⇔(x-5)(3x-1)=0

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

c)5x²-3x-2=0

⇔ 5x(x-1)+2(x-1)=0

⇔ (x-1)(5x+2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d)x⁴-11x²+18=0

⇔ x²(x²-2)-9(x²-2)=0

⇔ (x²-2)(x²-9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)

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 )

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

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

\(30x-15x^2-0\)

\(\Leftrightarrow15x\left(2-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}15x=0\\2-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

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

...

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

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

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

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

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