Ta có \(x^2-y^2-z^2=0\Rightarrow z^2=x^2-y^2\)

Có \(VT=\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(5x-3y\right)^2-\left(4z\right)^2\)\(=\left(5x-3y\right)^2-16z^2=\left(5x-3y\right)^2-16\left(x^2-y^2\right)\)

\(=25x^2-30xy+9y^2-16x^2+16y^2=9x^2-30xy+25y^2\)

\(=\left(3x\right)^2-2.3x.5y+\left(5y\right)^2=\left(3x-5y\right)^2=VP\left(đpcm\right)\)

a: \(=\dfrac{15}{5}\cdot\dfrac{x^3}{x^2}\cdot\dfrac{y^5}{y^3}\cdot z=3xy^2z\)

b: \(=-\dfrac{4}{3}x^3\)

c: \(=\dfrac{30x^4y^3}{5x^2y^3}-\dfrac{25x^2y^3}{5x^2y^3}-\dfrac{3x^4y^4}{5x^2y^3}\)

\(=6x^2-5-\dfrac{3}{5}x^2y\)

d: \(=\dfrac{4x^4}{-4x^2}+\dfrac{8x^2y^2}{4x^2}-\dfrac{12x^5y}{4x^2}\)

\(=-x^2+2y^2-3x^3y\)

a, 15x³y⁵z : 5x²y³ = 3xy²z.

b, 12x⁴y² : ( - 9xy² ) = \(\frac{3}{4}x^3\).

c, ( 30x⁴y³ - 25x²y³ - 3x⁴y⁴ ) : 5x²y³ = \(6x^2-5-\frac{3}{5}x^2y.\)

d, ( 4x⁴ - 8x²y² + 12x⁵y ) : ( - 4x² ) = -x² + 2y² - 3x³y.

\(VT=\left[\left(x-2\right)^2+4\left(x+y+1\right)\right]\left[\left(y-2\right)^2+4\left(x+y+1\right)\right]\)

\(VT=\left(x-2\right)^2\left(y-2\right)^2+4\left(x+y+1\right)\left[\left(x-2\right)^2+\left(y-2\right)^2\right]+16\left(x+y+1\right)^2\)

\(VP=\left[4\left(x+y+1\right)-\left(x-y\right)\right]\left[4\left(x+y+1\right)+\left(x-y\right)\right]\)

\(VP=16\left(x+y+1\right)^2-\left(x-y\right)^2\)

Ta có \(VT=VP\)

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

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

Nhận xét:

\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(x-2\right)^2\left(y-2\right)^2\ge0\\x;y\ge0\Rightarrow4\left(x+y+1\right)>0\Rightarrow4\left(x+y+1\right)\left[\left(x-2\right)^2+\left(y-2\right)^2\right]\ge0\end{matrix}\right.\)

Vậy (1) xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-2\right)^2\left(y-2\right)^2=0\\\left(x-2\right)^2+\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow x=y=2\)

Vậy phương trình đã cho có nghiệm duy nhất \(x=y=2\)

a) 6x².(3x² - 4x + 5) = 18x⁴ - 24x³ + 30x²

b) (x - 2y)(3xy + 6y² + x) = 3x²y + 6xy² + x² - 6xy² - 12y³ - 2xy = -12y³ + 3x²y - 2xy + x²

c) (18x⁴y³ - 24x³y⁴ + 12x³y³) : (-6x²y³) = -6x²y³(-3x² + 4xy - 2x) : (-6x²y³) = 4xy - 3x² - 2x

Ta có : \(\left(5x+5y+5z\right)^2-\left(25xy+25yz+25zx\right)\)

\(=25\left(\left(x+y+z\right)^2-\left(xy+yz+zx\right)\right)\)

Xét : \(\left(x+y+z\right)^2-\left(xy+yz+zx\right)=0\)

\(=>x^2+y^2+z^2+2xy+2yz+2zx-xy-yz-zx=0\)

\(=>x^2+y^2+z^2+xy+yz+zx=0\)

Nhân biểu thức với 2 ta được:

\(2x^2+2y^2+2z^2+2xy+2yz+2zx=0\)

\(=>\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2=0\)

\(=>x+y=y+z=z+x=0\)

Vạy để phân thức A xác định thì x,y,z không đồng thời bằng 0;

CHÚC BẠN HỌC TỐT...

sao 

bn ko 

tách 

ra 

từng cái 1 cho dễ

Ai bt thì làm giúp mình câu 2 và câu 3 nhé. Câu 1 mình tự làm đc r

Câu 20 thôi nha