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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

...

a) \(2x^2+5x-18\)

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

\(=2x\left(x-2\right)+9\left(x-2\right)\)

\(=\left(x-2\right)\left(2x+9\right)\)

b) \(4x^2-17x+15\)

\(=4x^2-12x-5x+15\)

\(=4x\left(x-3\right)-5\left(x-3\right)\)

\(=\left(x-3\right)\left(4x-5\right)\)

c) \(-8x^2+10x+7\)

\(=-8x^2-4x+14x+7\)

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

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

d) \(7x^2-30x+8\)

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

\(=7x\left(x-4\right)-2\left(x-4\right)\)

\(=\left(x-4\right)\left(7x-2\right)\)

e) \(-x^3+11x^2-30x\)

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

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

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

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

a) 2x\(^2\) + 5x - 18 = 2x\(^2\) + 9x - 4x - 18 = x(2x + 9) - 2(2x + 9) = (x-2)(2x-9)

b) 4x\(^2\) - 17x - 15 = 4x\(^2\) + 20x - 3x - 15 = 4x(x + 5 ) - 3(x + 5) = (4x - 3 )(x + 5)

c) -8x\(^2\) + 10x + 7 = -8x\(^2\) + 14x - 4x + 7 =-2x(4x - 7) - (4x - 7) = (-2x - 1)(4x - 7)

d) 7x\(^2\) - 30x + 8 = 7x\(^2\) + 2x + 28x + 8 = x(7x + 2) + 4(7x + 2) = (x + 4)(7x + 2)

e) - x\(^3\) + 11x\(^2\) - 30x = -x(x\(^2\) - 11x + 30) = -x(x\(^2\) - 5x - 6x + 30) = -x\(\left[x\left(x-5\right)-6\left(x-5\right)\right]\) = -x(x-6)(x-5)

$\dfrac{4}{-25x^2+20x-3}=\dfrac{3}{5x-1}-\dfrac{2}{5x-3}$

Điều kiện: $5x-1\ne0,\ 5x-3\ne0$

$\Leftrightarrow x\ne\dfrac{1}{5},\ x\ne\dfrac{3}{5}$

Ta có:

$-25x^2+20x-3=-(5x-1)(5x-3)$

$\Rightarrow \dfrac{-4}{(5x-1)(5x-3)}=\dfrac{3(5x-3)-2(5x-1)}{(5x-1)(5x-3)}$

$\Leftrightarrow \dfrac{-4}{(5x-1)(5x-3)}=\dfrac{5x-7}{(5x-1)(5x-3)}$

Vì $x\ne\dfrac{1}{5},\dfrac{3}{5}$ nên:

$-4=5x-7$

$\Leftrightarrow 5x=3$

$\Leftrightarrow x=\dfrac{3}{5}$

Nhưng $x=\dfrac{3}{5}$ không thỏa mãn điều kiện.

Vậy: phương trình vô nghiệm.

$\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}-\dfrac{2}{x^2-4x+3}=0$

Điều kiện:

$x^2-3x+2=(x-1)(x-2)\ne0$

$x^2-5x+6=(x-2)(x-3)\ne0$

$x^2-4x+3=(x-1)(x-3)\ne0$

$\Rightarrow x\ne1,\ x\ne2,\ x\ne3$

Ta có:

$\dfrac{1}{(x-1)(x-2)}+\dfrac{1}{(x-2)(x-3)}-\dfrac{2}{(x-1)(x-3)}=0$

Quy đồng:

$\dfrac{x-3+x-1-2(x-2)}{(x-1)(x-2)(x-3)}=0$

$\Leftrightarrow \dfrac{x-3+x-1-2x+4}{(x-1)(x-2)(x-3)}=0$

$\Leftrightarrow \dfrac{0}{(x-1)(x-2)(x-3)}=0$

Phương trình đúng với mọi $x$ thỏa mãn điều kiện.

Vậy: $x\in\mathbb{R}\setminus{1;2;3}$.

$\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}$

Điều kiện: $x\ne0,\ x\ne2$

Ta có:

$\dfrac{x-1}{2x(x-2)}-\dfrac{7}{8x}=\dfrac{5-x}{4x(x-2)}-\dfrac{1}{8(x-2)}$

Quy đồng mẫu $8x(x-2)$:

$\dfrac{4(x-1)-7(x-2)}{8x(x-2)}=\dfrac{2(5-x)-x}{8x(x-2)}$

$\Leftrightarrow \dfrac{4x-4-7x+14}{8x(x-2)}=\dfrac{10-2x-x}{8x(x-2)}$

$\Leftrightarrow \dfrac{-3x+10}{8x(x-2)}=\dfrac{-3x+10}{8x(x-2)}$

Phương trình đúng với mọi $x$ thỏa mãn điều kiện.

Vậy: $x\in\mathbb{R}\setminus{0;2}$.

$\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}$

Điều kiện:

$x^2+9x+20=(x+4)(x+5)\ne0$

$x^2+11x+30=(x+5)(x+6)\ne0$

$x^2+13x+42=(x+6)(x+7)\ne0$

$\Rightarrow x\ne-4,-5,-6,-7$

Ta có:

$\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$

Quy đồng:

$\dfrac{(x+6)(x+7)+(x+4)(x+7)+(x+4)(x+5)}{(x+4)(x+5)(x+6)(x+7)}=\dfrac{1}{18}$

$\Leftrightarrow 18[(x+6)(x+7)+(x+4)(x+7)+(x+4)(x+5)]=(x+4)(x+5)(x+6)(x+7)$

Khai triển và rút gọn:

$\Leftrightarrow x^2+11x-26=0$

$\Leftrightarrow (x-2)(x+13)=0$

$\Leftrightarrow x=2$ hoặc $x=-13$

Cả hai giá trị đều thỏa mãn điều kiện.

Vậy: $x\in{-13;2}$.

hic, mk chx học

a)

\(x^3-7x-6=x^3-x-6x-6\)

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

\(=x(x-1)(x+1)-6(x+1)=(x+1)[x(x-1)-6]\)

\(=(x+1)(x^2-x-6)=(x+1)[x^2-3x+2x-6]\)

\(=(x+1)[x(x-3)+2(x-3)]=(x+1)(x+2)(x-3)\)

b) \(x^3-6x^2+8x\)

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

\(=x(x^2-4x-2x+8)\)

\(=x[x(x-4)-2(x-4)]=x(x-2)(x-4)\)

c) \(x^4+2x^3-16x^2-2x+15\)

\(=(x^4+2x^3-x^2-2x)-15x^2+15\)

\(=[(x^4-x^2)+(2x^3-2x)]-15(x^2-1)\)

\(=[x^2(x^2-1)+2x(x^2-1)]-15(x^2-1)\)

\(=(x^2-1)(x^2+2x)-15(x^2-1)=(x^2-1)(x^2+2x-15)\)

\(=(x^2-1)(x^2-3x+5x-15)=(x^2-1)[x(x-3)+5(x-3)]\)

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

d)

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

\(=x(x^2-5x-6x+30)\)

\(=x[x(x-5)-6(x-5)]=x(x-6)(x-5)\)

\(a,\)\(x^4-4x^3+4x^2=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b,\)\(x^2+5x+4=0\)

\(\Leftrightarrow x^2+x+4x+4=0\)

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

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

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

\(c,\)\(9x-6x^2-3=0\)

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

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow2x^2-2x-x+1=0\)

\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)

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

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

\(d,\)\(2x^2+5x+2=0\)

\(\Leftrightarrow2x^2+4x+x+2=0\)

\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)

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

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

a)\(x^2-13x+36=x^2-4x-9x+36=x\left(x-4\right)-9\left(x-4\right)=\left(x-9\right)\left(x-4\right)\)

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

c)\(x^2-5x-24=x^2+3x-8x-24=x\left(x+3\right)-8\left(x+3\right)=\left(x-8\right)\left(x+3\right)\)

d)\(3x^2-16x+5=3x^2-x-15x+5=x\left(3x-1\right)-5\left(3x-1\right)=\left(x-5\right)\left(3x-1\right)\)

e)\(8x^2+30x+7=8x^2+28x+2x+7=4x\left(2x+7\right)+\left(2x+7\right)=\left(4x+1\right)\left(2x+7\right)\)

g)\(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(2x-1\right)\left(x-3\right)\)

h)\(6x^2-7x+3=6x^2-9x-2x+3=3x\left(2x-3\right)-\left(2x-3\right)=\left(3x-1\right)\left(2x-3\right)\)

i)\(3x^2-14x+11=3x^2-3x-11x+11=3x\left(x-1\right)-11\left(x-1\right)=\left(3x-11\right)\left(x-1\right)\)

k)\(5x^2+8x-13=5x^2-5x+13x-13=5x\left(x-1\right)+13\left(x-1\right)=\left(5x+13\right)\left(x-1\right)\)

a ) \(x^2-13x+36=x^2-4x-9x+36=x\left(x-4\right)-9\left(x-4\right)=\left(x-9\right)\left(x-4\right)\)

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

c ) \(x^2-5x-24=x^2-3x+8x-24=x\left(x-3\right)+8\left(x-3\right)=\left(x+8\right)\left(x-3\right)\)

d ) \(3x^2-16x+5=3x^2-15x-x+5=3x\left(x-5\right)-\left(x-5\right)=\left(3x-1\right)\left(x-5\right)\)

e ) \(8x^2+30x+7=8x^2+2x+28x+7=2x\left(4x+1\right)+7\left(4x+1\right)=\left(2x+7\right)\left(4x+1\right)\)

g ) \(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(2x-1\right)\left(x-3\right)\)

h ) \(6x^2-7x-20=6x^2-15x+8x-20=3x\left(2x-5\right)+4\left(2x-5\right)=\left(3x+4\right)\left(2x-5\right)\)

i ) \(3x^2-14x+11=3x^2-3x-11x+11=3x\left(x-1\right)-11\left(x-1\right)=\left(3x-11\right)\left(x-1\right)\)

k ) \(5x^2+8x-13=5x^2-5x+13x-13=5x\left(x-1\right)+13\left(x-1\right)=\left(5x+13\right)\left(x-1\right)\)