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Tử số có : \(\left(xy+z\right)\left(yz+x\right)\left(zx+y\right)\)\(=\left[\left(1-y-z\right)y+z\right]\left[\left(1-x-z\right)z+x\right]\left[\left(1-x-y\right)x+y\right]\)

\(=\left[y-y^2-yz+z\right]\left[z-xz-z^2+x\right]\left[x-x^2-xy+y\right]\)

\(=\left[y\left(1-y\right)+z\left(1-y\right)\right]\left[z\left(1-z\right)+x\left(1-z\right)\right]\)\(\left[x\left(1-x\right)+y\left(1-x\right)\right]\)

\(=\left(1-y\right)\left(y+z\right)\left(1-z\right)\left(z+x\right)\left(1-x\right)\left(x+y\right)\)

Rút gọn có \(S=\frac{\left(y+z\right)\left(z+x\right)\left(x+y\right)}{\left(1-x\right)\left(1-z\right)\left(1-y\right)}\)

Mà \(\left(1-x\right)\left(1-z\right)\left(1-y\right)=\left(x+y+z-x\right)\)\(\left(x+y+z-z\right)\left(x+y+z-y\right)\)\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(\Rightarrow S=1\)

a/  (5x^2 -10xy+5y^2)-20z^2=5[(x-y)^2-(2z)^2]=5(x-y-2x)(x-y+2z)

b/16x-5x^2-3=-[5x^2-16x+3]=-[(5x^2-x)-(15x+3)]=-[x(5x-1)-3(5x-1)]=(3-x)(5x-1)

c/x^2+4x+3=(x^2+x)+(3x+3)=x(x+1)+3(x+1)=(x+1)(x+3)

2a/ 5x(X^2-9)=0=>x=0 hoặc x^2=9=>x=0 hoặc x=+-3

b/x^2-7x+10=0=>(x^2-2x)-(5x-10)=0=>x(x-2)-5(x-2)=0=>x-2=0 hoặc x-5 =0 => tự tính nhé!

Answer:

Bài 1:

\(5x² - 10xy + 5y² - 20z²\)

\(= 5( x² - 2xy + y² - 4z²)\)

\(= 5 [(x² - 2xy + y²) - (2z)²]\)

\(= 5 [(x - y)² - (2z)²]\)

\(= 5 (x - y - 2z) ( x - y + 2z)\)

\(16x - 5x² - 3 \)

\(= -( 5x² - 16x + 3)\)

\(= -( 5x² - 15x - 1x + 3)\)

\(= - [ (5x² -x) - (15x -3)]\)

\(= - [ x(5x -1) -3(5x -1)]\)

\(= - (5x-1)(x-3)\)

\(x² + 4x + 3\)

\(= x² + x + 3x + 3\)

\(= (x² + x) + (3x + 3)\)

\(= x( x + 1) +3 (x+1)\)

\(= (x+1) (x+3)\)

Bài 2:

\(5x\left(x^2-9\right)=0\)

\(\Rightarrow5x\left(x-3\right)\left(x+3\right)=0\)

Trường hợp 1: \(5x=0\Leftrightarrow x=0\)

Trường hợp 2: \(x-3=0\Leftrightarrow x=3\)

Trường hợp 3: \(x+3=0\Leftrightarrow x=-3\)

\(x^2-7x+10=0\)

\(\Rightarrow x^2-5x-2x+10=0\)

\(\Rightarrow x\left(x-5\right)-2\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x-2\right)=0\)

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

Answer:

\(2\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)+6=0\) (1)

ĐK: \(x\ne0\)

Ta đặt: \(x+\frac{1}{x}=t\)

\(\Rightarrow t^2=x^2+\frac{1}{x^2}+2\)

\(\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)

Ta có:

(1) \(\Rightarrow2\left(t^2-2\right)-5t+6=0\)

\(\Rightarrow2t^2-4-5t+6=0\)

\(\Rightarrow2t^2-5t+2=0\)

\(\Rightarrow2t^2-4t-t+2=0\)

\(\Rightarrow2t\left(t-2\right)-\left(t-2\right)=0\)

\(\Rightarrow\left(t-2\right)\left(2t-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}t-2=0\\2t-1=0\end{cases}\Rightarrow\orbr{\begin{cases}t=2\\t=\frac{1}{2}\end{cases}}}\)

Trường hợp 1: \(t=2\)

\(\Rightarrow x+\frac{1}{x}=2\)

\(\Rightarrow x+\frac{1}{x}-2=0\)

\(\Rightarrow\frac{x^2-2x+1}{x}=0\)

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

\(\Rightarrow\left(x-1\right)^2=0\)

\(\Rightarrow x=1\)

Trường hợp 2: \(t=\frac{1}{2}\)

\(\Rightarrow x+\frac{1}{x}=\frac{1}{2}\)

\(\Rightarrow x+\frac{1}{x}-\frac{1}{2}=0\)

\(\Rightarrow\frac{2x^2+2-x}{2x}=0\)

\(\Rightarrow2x^2-x+2=0\)

\(\Rightarrow\Delta=\left(-1\right)^2-2.2.4=-15< 0\) (Loại)

Vậy phương trình có nghiệm là \(x=1\)

Answer:

\(\frac{4}{x-3}-\frac{x}{x+3}-\frac{x^2+9}{x^2-9}\left(x\ne\pm3\right)\)

\(=\frac{4}{x-3}-\frac{x}{x+3}-\frac{x^2+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{4\left(x+3\right)-x\left(x-3\right)-\left(x^2+9\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{4x+12-x^2+3x-x^2-9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{-2x^2+7x+3}{x^2-9}\)