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 )

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

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

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

...

\(=x^3-5x^2-6x^2+30x\)

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

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

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

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

x³ - 11x² + 30x = 0

x(x² - 11x + 30) = 0

x(x² - 5x - 6x + 30) = 0

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

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

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

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

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: \(x^2+3x-18=0\)

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

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

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

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

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

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

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 

\(1,x^2+4x-2=\left(x+2\right)^2-6\ge6\)

Dấu \("="\Leftrightarrow x=-2\)

\(2.x^2+7x+1=\left(x+\dfrac{7}{2}\right)^2-\dfrac{45}{4}\ge-\dfrac{45}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{7}{2}\)

\(3,25x^2+30x+11=\left(5x+3\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{5}\)

a) \(8x^2+30x+7=0\)
\(\Rightarrow8x^2+2x+28x+7=0\)
\(\Rightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\)
\(\Rightarrow\left(2x+7\right)\left(4x+1\right)=0\)
\(\Rightarrow\)\(2x+7=0\)  hoặc  \(4x+1=0\)
\(\Rightarrow\)\(2x=-7\)          ;     \(4x=-1\)
\(\Rightarrow\)\(x=\frac{-7}{2}\)             ;     \(x=\frac{-1}{4}\)
Vậy \(x\in\left\{\frac{-7}{2};\frac{-1}{4}\right\}\)

b) \(x^3-11x^2+30x=0\)
\(\Rightarrow x\left(x^2-11x+30\right)=0\)
\(\Rightarrow x\left(x^2-6x-5x+30\right)=0\)
\(\Rightarrow x\left[x\left(x-6\right)-5\left(x-6\right)\right]=0\)
\(\Rightarrow x\left(x-5\right)\left(x-6\right)=0\)
\(\Rightarrow\)\(x=0\)  hoặc  \(x-5=0\)  hoặc  \(x-6=0\)
\(\Rightarrow\)\(x=0\)     ;      \(x=5\)               ;     \(x=6\)
Vậy \(x\in\left\{0;5;6\right\}\)

a)\(8x^2+30x+7=0\Leftrightarrow8x^2+2x+28x+7=0\Leftrightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\)

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

b)\(x^3-11x^2+30x=0\Leftrightarrow x\left(x^2-11x+30\right)=0\Leftrightarrow x\left(x^2-5x-6x+30\right)=0\)

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

<=>x=0 hoặc x-6=0 hoặc x-5=0 <=> x=0 hoặc x=6 hoặc x=5

`a)A=-x^2+x+1`

`=-(x^2-x)+1`

`=-(x^2-2.x. 1/2+1/4-1/4)+1`

`=-(x-1/2)^2+5/4<=5/4`

Dấu "=" xảy ra khi `x-1/2=0<=>x=1/2`

`b)B=x^2+3x+4`

`=x^2+2.x. 3/2+9/4+7/4`

`=(x-3/2)^2+7/4>=7/4`

Dấu "=" xảy ra khi `x-3/2=0<=>x=3/2`

`c)=x^2-11x+30`

`=x^2-2.x. 11/2+121/4-1/4`

`=(x-11/2)^2-1/4>=-1/4`

Dấu "=" xảy ra khi `x+1/4=0<=>x=-1/4`