đề là ptđt thành nhân tử hả bạn ? 

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+10=t\)

\(t\left(t+2\right)-24=t^2+2t-24=\left(t-4\right)\left(t+6\right)\)

Theo cách đặt : \(\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+11=t\)

\(\Rightarrow\left(t+1\right)\left(t-1\right)-24\)

\(=t^2-1-24\)

\(=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)\)

\(\Rightarrow\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

Ta có : \(VT=3+\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge3+\frac{\left(1+1+2\right)^2}{a+b+c}=3+16=19\)

Dấu "=" tự tìm nha b yeuuuuu

\(\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+4}{c}=1+\frac{1}{a}+1+\frac{1}{b}+1+\frac{4}{c}\)

Theo BĐT Cauchy Schwarz dạng Engel ta có : 

\(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\ge\frac{\left(1+1+2\right)^2}{a+b+c}=16\)

\(\Rightarrow\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+4}{c}\ge16+3=19\)

Dấu ''='' xảy ra khi \(a=b=c=\frac{1}{3}\)

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

\(< =>x^2+6x+9+x^2-4x+4=2x^2\)

\(< =>2x+13=0< =>x=-\frac{13}{2}\)

b,\(5x\left(x-2\right)=x-2< =>\left(x-2\right)\left(5x-1\right)=0< =>\hept{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)

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

\(\Leftrightarrow x^2+6x+9+x^2-4x+4-2x^2=0\)

\(\Leftrightarrow2x+13=0\)

\(\Leftrightarrow2x=-13\)

\(\Leftrightarrow x=-\frac{13}{2}\)
Vậy \(S=\left\{-\frac{13}{2}\right\}\)

b) \(5x\left(x-2\right)=x-2\)

\(\Leftrightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\5x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)

Vậy \(S=\left\{2;\frac{1}{5}\right\}\)

a, \(5y^2-5x^2+6x+6y=5\left(y-x\right)\left(x+y\right)+6\left(x+y\right)\)

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

b, \(12x^2+19x+7=12x^2+12x+7x+7\)

\(=12x\left(x+1\right)+7\left(x+1\right)=\left(12x+7\right)\left(x+1\right)\)

5y² - 5x² + 6x + 6y 

= 5(y² - x²) + 6(x + y) 

= 5(y - x)(x + y) + 6(x + y) 

= (x + y)(5y - 5x + 6) 

b) 12x² + 19x + 7 

= 12x² + 12x + 7x + 7 

= 12x(x + 1) + 7(x + 1) 

= (x + 1)(12x + 7) 

mik ko nhìn thấy đa thức đâu bạn ơiiiiiiiiiiiiiiiiiiiiiiii

\(\left(x-2\right)^2-x\left(x-1\right)\left(x+1\right)+x\left(7x-6\right)=0\)

\(\Leftrightarrow x^2-4x+4-x\left(x^2-1\right)+7x^2-6x=0\)

\(\Leftrightarrow8x^2-10x+4-x^3+x=0\Leftrightarrow-x^3+8x^2-9x+4=0\Leftrightarrow x=6,7...\)

\(\left(x-2\right)^2-x\left(x-1\right)\left(x+1\right)+x\left(7x-6\right)=0\)

\(\Leftrightarrow x^2-4x+4-x^3-x^2+x^2+x+7x^2-6x=0\)

\(\Leftrightarrow-x^3-8x^2-9x+4=0\)

Làm tiếp nhé =))

mik cần gấp , ai làm nhanh mik k cho .

\(a,\)\(x^2+2x-15=0\Rightarrow\left(x+1\right)^2-16=0\Rightarrow\left(x+1\right)^2=16\)

\(\hept{\begin{cases}x+1=4\\x+1=-4\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\x=-5\end{cases}}\)\(\)vậy \(S=\left\{3;-5\right\}\)

\(b,\)\(2x^2-7x+6=0\Rightarrow2x\left(x-1\right)-5\left(x-1\right)+1=0\)

\(\Rightarrow\left(x-1\right)\left(2x-5\right)=-1=1.-1=-1.1\)

\(x-1=1;2x-5=-1\)và \(x-1=-1;2x-5=1\)

\(\Rightarrow x=2\)và\(x=0;x=3\)vậy \(S=\left\{2;0;3\right\}\)

Mình làm nốt nhé

c) \(x^3-4x^2+5x=x\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow x\left(x^2-4x+4+1\right)=0=x\left[\left(x-2\right)^2+1\right]=0\)

Vì \(\left(x-2\right)^2+1>0\forall x\Leftrightarrow x=0\)

d) \(x^3+5x^2+3x-9=0\Leftrightarrow x^3+3x^2+2x^2+6x-3x-9=0\)

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

\(\Leftrightarrow\left(x+3\right)\left[\left(x+1\right)^2-4\right]=\left(x+3\right)\left(x+1-2\right)\left(x+1+2\right)=0\)

\(\Leftrightarrow\left(x+3\right)^2\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}}\)

Tự KL

x³ + 9x = 0 <=> x( x² + 9 ) = 0 <=> x = 0 ( vì x² + 9 > 0 )

Vậy pt có nghiệm x = 0

x³ + 9x = 0

<=> x(x²+9) = 0

=> x = 0 ( vì x²+9>0 với mọi x )

Vậy x = 0

a) 5x² -20

=  5(x² -4)

=5 (x² -2²)

= 5(x-2)(x+2)

b) 16 - (x+y)²

=4² -(x+y)²

= (4-x-y)(4+x+y) 

a, \(5\left(x^2-4\right)=5\left(x-2\right)\left(x+2\right)\)

b, \(16-\left(x+y\right)^2=\left(4-x-y\right)\left(4+x+y\right)\)

mấy bài này áp dụng hđt là được nhé