a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

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

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)

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

\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow x=1\left(nhận\right)\)

2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)

\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)

(x+1)(x-6)(x-2)(x-3)

=(x²-5x-6)(x²-5x+6)

đặt x²-5x=a

ta có:(a-6)(a+6)

=a²-36

lớp 8?

\(A=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)+2045\)

\(=\left(x^2+6x-x-6\right)\left(x^2+3x+2x+6\right)+2045\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)+2045\)

\(=\left(x^2+5x\right)^2-6^2+2045\)

\(=\left(x^2+5x\right)^2+2009\ge2009\)

Dấu "=" xày ra khi x²+5x=0  <=> x=0 hoặc x=-5

Vậy MinA=2009 khi x=0 hoặc x=-5

\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)+2045\)

\(A=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+2045\)

\(A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)+2045\)

\(A=\left(x^2+5x\right)^2-36+2045\)

\(A=\left(x^2+5x\right)^2+2009\)

Vì \(\left(x^2+5x\right)^2\ge0\Rightarrow\left(x^2+5x\right)^2+2009\ge2009\)

\(\Rightarrow A\ge2009\)

=> GTNN của A bằng 2009 

Dấu '=' xảy ra khi \(\left(x^2+5x\right)^2=0\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)

=> x = 0 hoặc x + 5 = 0 <=>  x = -5

Vậy GTNN của A bằng 2009

Tham khảo nha ^^

a) Ta có \(A=\left(x-3\right)^2+\left(x-11\right)^2=x^2-6x+9+x^2-22x+121=2x^2-28x+130\)

\(=2\left(x^2-14x+49\right)+32=2\left(x-7\right)^2+32\ge32\)

Vậy minA = 32 khi x = 7.

b) \(B=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)

\(=\left(x+1\right)\left(x-6\right)\left(x-2\right)\left(x-3\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)

Đặt \(x^2-5x=t\Rightarrow B=\left(t-6\right)\left(t+6\right)=t^2-36\ge-36\)

minB = -36 khi t = 0 hay \(x^2-5x=0\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)