Phân thức đại số

dù là số âm hay số dương nhưng khi có mũ 2 (²) thì nó luôn luôn dương

A=4x²+9

A=(2x)²+3²

B=25x²+10x+4

B=(5x)²+10x+2²

B=(5x+2)²

C=4x²+6x+8

C=2(2x²+3x+4)

C=2(2x+2)²

Từ \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\) + \(\dfrac{1}{z}\) = 0

\(=>yz+xz+xy=0\)

\(=>yz=-xz-xy\)

Ta có : \(x^2+2yz=x^2+yz+yz=x^2+yz-yx-xz=\left(y-x\right)\left(z-x\right)=-\left(x-y\right)\left(z-x\right)\)

Tương tư :

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

\(z^2+2xy=z^2+xy+xy=z^2+xy-yz-xz=\left(z-y\right)\left(z-x\right)=-\left(y-z\right)\left(z-x\right)\)Nên A = \(\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}\)

=\(\dfrac{-yz}{\left(x-y\right)\left(z-x\right)}+\dfrac{-xz}{\left(x-y\right)\left(y-z\right)}+\dfrac{-xy}{\left(y-z\right)\left(z-x\right)}\)

=\(\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(\dfrac{(-y^2z+yz^2-z^2x+x^2z-x^2y+xy^2)+(xyz-xyz)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(\dfrac{\left(xyz-y^2z-z^2x+yz^2\right)+\left(-x^2y+xy^2+x^2z-xyz\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(\dfrac{z\left(xy-y^2-xz+zy\right)-x\left(xy-y^2-xz+zy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(\dfrac{\left(z-x\right)\left(xy-y^2-xz+zy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(\dfrac{\left(z-x\right)\left(x-y\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=1

= =" ... bài này làm dài ..bấm máy mỏi tay lắm...

nhanh gọn lẹ.... A = 0

Mọi người giúp mình với, mình đang cần gấp

a) x² + 6x + 9 = x² + 2.3.x + 3² = (x + 3)²

b) 10x - 25 - x² = - (x² - 2.5.x + 5²) = - (x - 5)²

c) \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

d) 25x² - 64y² = (5x)² - (8y)² = (5x - 8y)(5x + 8y)

\(a,x^2+6x+9=\left(x+3\right)^2\)

\(b,10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

\(a,ĐKXĐ;x\ne\pm1\)

\(a,\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right)-\dfrac{x+2003}{x}\)

\(=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)-\dfrac{x+2003}{x}\)

\(=\dfrac{\left(x+1\right)^2-\left(x-1\right)^2+x^2-4x-1}{\left(x-1\right).\left(x+1\right)}-\dfrac{x+2003}{x}\)

\(=\dfrac{2.2x+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2003}{x}\)

\(=\dfrac{4x+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2003}{x}\)

\(=\dfrac{\left(x-1\right).\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2003}{x}\)

\(=1-\dfrac{x+2003}{x}\)

\(=\dfrac{x-\left(x+2003\right)}{x}\)

\(=\dfrac{x-x-2003}{x}\)

\(=-\dfrac{2003}{x}\)

c/ để B nguyên ⇔ \(-\dfrac{2003}{x}\in Z\)

\(\Leftrightarrow x\inƯ\left(2003\right)\)

\(\Leftrightarrow x=\left\{\pm1;\pm2003\right\}\)

kết vợp với đkxđ

=> \(x=\left\{\pm2003\right\}\)

vậy.......

a: \(M=\left(a+1\right)\cdot\left(a-1\right)=a^2-1\)

b: Để M=a² thì a²-1=a²

=>-1=0(vô lý)

\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)-\left(2x+1\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left(24x^2+16x-9x-6\right)-\left(4x^2+16x+7x+28\right)-\left(10x^2-2x+5x-1\right)=0\)

\(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28-10x^2+2x-5x+1=0\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Leftrightarrow10x^2+11x-30x-33=0\)

\(\Leftrightarrow\left(10x^2+11x\right)-\left(30x+33\right)=0\)

\(\Leftrightarrow x\left(10x+11\right)-3\left(10x+11\right)=0\)

\(\Leftrightarrow\left(10x+11\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}10x+11=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}10x=-11\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11}{10}\\x=3\end{matrix}\right.\)

Vậy \(x=\dfrac{-11}{10}\) hoặc \(x=3\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)=192\)

\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=192\)

\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3-192=0\)

\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-195=0\)

\(\Leftrightarrow\left(x^2+2x-15\right)\left(x^2+2x+13\right)=0\)

=>(x+5)(x-3)=0

=>x=3 hoặc x=-5