\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\) (sửa đề)

\(=\left[\left(x+1\right)\left(x+4\right)\right].\left[\left(x+2\right).\left(x+3\right)\right]-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt \(y=x^2+5x+4\), thay vào đa thức, ta được:

\(y\left(y+2\right)-24\)

\(=y^2+2y-24\)

\(=\left(y^2+2y+1\right)-25\)

\(=\left(y+1\right)^2-5^2\)

\(=\left(y+1-5\right)\left(y+1+5\right)\)

\(=\left(y-4\right)\left(y+6\right)\)

\(=\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)

\(x^4-4\left(x^2+25\right)-25\) 

\(=x^4-4x^2-125\) 

\(=\left(x^4-4x^2+4\right)-129\) 

\(=\left(x^2-2\right)^2-\left(\sqrt{129}\right)^2\) 

\(=\left(x^2-2+\sqrt{129}\right)\left(x^2-2-\sqrt{129}\right)\)

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

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

\(=\left(x^2+5\right)\left(x^2-9\right)\)

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

EZ :))

MAI CÔ KIỂM TRA HẢ? CHÚC BẠN MAY MẮN LẦN SAU

\(4x^4+4x^2+1=\left(2x^2+1\right)^2\)

\(9x^4-6x^2+1=\left(3x^2-1\right)^2\)

\(\dfrac{x^2}{9}-\dfrac{2}{3}x+1=\left(\dfrac{x}{3}+1\right)^2\)

\(x^2-25=\left(x-5\right)\left(x+5\right)\)

\(x^4+2x^3+10x-25\)

\(=x^4+5x^2+2x^3+10x-5x^2-25\)

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

\(x^4-6x^2+25\)

\(=x^4+4x^3+5x^2-4x^3-16x^2-20x+5x^2+20x+25\)

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

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

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)

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