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= ( x/(x-6)(x+6) - x-6/x(x+6) ) : 2x-6/x2 + 6x + 6/6-x
=( x2/x(x+6)(x-6) - (x -6 )(x-6)/x(x+6)(x-6) ) : .....
= (12x -36 / x(x+6)(x-6) : 2x-6/ x2 + 6x )+ 6/6-x
=6/x-6 + 6/6-x
= 6-6/ x-6
=0/x-6
câu trước mình thiếu 6/6-x
\(\left(\frac{x}{x^2-36}+\frac{6-x}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
đkxđ: \(x\ne0;x\ne\pm6\)
MTC=x(x+6)(x-6)
\(=\left[\frac{x}{\left(x+6\right)\left(x-6\right)}+\frac{6-x}{x\left(x+6\right)}\right]\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)
\(=\left[\frac{x^2}{x\left(x^2-36\right)}-\frac{\left(x-6\right)^2}{x\left(x^2-36\right)}\right]\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)
\(=\frac{12\left(x-3\right)}{x\left(x+6\right)\left(x-6\right)}\cdot\frac{x\left(x+6\right)}{x\left(x-3\right)}-\frac{x}{x-6}\)
\(=\frac{12}{x\left(x-6\right)}-\frac{x^2}{x\left(x-6\right)}\)
\(=\frac{12-x^2}{x\left(x-6\right)}\)
.....................
a)
DK:tồn tại P \(\hept{\begin{cases}x\ne0\\x\ne-+6\\x\ne3\end{cases}}\)
\(P=\left(\frac{x}{\left(x-6\right)\left(x+6\right)}-\frac{x-6}{x\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}\\ \)
\(P=\left(\frac{x^2-\left(x-6\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}\)
\(P=\left(\frac{x^2-\left(x^2-12x+36\right)}{x\left(x-6\right)\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}\)
\(P=\left(\frac{12\left(x-3\right)}{x\left(x-6\right)\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}=\frac{6}{x-6}\)
b)6/(x-6)=1=> x-6=6=> x=12
c)x-6<0=> x<6
A = \(\left(\frac{x}{x^2-36}-\frac{x-6}{x^2+6x}\right):\frac{2x-6}{x^2+6x}+\frac{x}{6-x}\)
= \(\left[\frac{x}{\left(x-6\right)\left(x+6\right)}-\frac{x-6}{x\left(x+6\right)}\right]:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\left[\frac{x^2}{x\left(x-6\right)\left(x+6\right)}-\frac{\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right]:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\frac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\frac{\left(x-x+6\right)\left(x+x-6\right)}{x\left(x-6\right)\left(x+6\right)}:\frac{2\left(x-3\right)}{x\left(x+6\right)}-\frac{x}{x-6}\)
=
= \(\frac{x\left(2x-6\right)}{x\left(x-6\right)\left(x+6\right)}:\frac{2x-6}{x\left(x+6\right)}-\frac{x}{x-6}\)
= \(\frac{2x-6}{\left(x-6\right)\left(x+6\right)}.\frac{x\left(x+6\right)}{2x-6}\) \(-\frac{x}{x-6}\)
= \(\frac{x}{x-6}-\frac{x}{x-6}\)
= 0
Bài 1:
\(\left(x-y+z\right)^2+\left(z-y\right)^2+\left(x-y+z\right)\left(2y-2z\right)\)
\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z+y-z\right)^2\)
\(=x^2\)
Bài 2:
đk: \(x\ne\left\{0;-1;-2;-3;-4;-5\right\}\)
Xét BT trái ta có:
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+5}\)
\(=\frac{5}{x\left(x+5\right)}=\frac{5}{x^2+5x}\)
GT của biểu thức lớn sẽ là: \(\frac{5}{x^2+5x}\cdot\frac{x^2+5x}{5}=1\) không phụ thuộc vào biến
=> đpcm
Bài 1.
( x - y + z ) + ( z - y )2 + ( x - y + z )( 2y - 2z )
= ( x - y + z ) - 2( x - y + z )( z - y ) + ( z - y )2
= [ ( x - y + z ) - ( z - y ) ]2
= ( x - y + z - z + y )2
= x2
Bài 2. ĐKXĐ tự ghi nhé :))
\(\left(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\right)\times\left(\frac{x^2+5x}{5}\right)\)
\(=\left(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)
\(=\left(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+4}-\frac{1}{x+5}\right)\times\left(\frac{x\left(x+5\right)}{5}\right)\)
\(=\left(\frac{1}{x}-\frac{1}{x+5}\right)\times\frac{x\left(x+5\right)}{5}\)
\(=\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{\left(x+5\right)}\right)\times\frac{x\left(x+5\right)}{5}\)
\(=\frac{x+5-x}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}\)
\(=\frac{5}{x\left(x+5\right)}\times\frac{x\left(x+5\right)}{5}=1\)
=> đpcm
P\(=\frac{\left(x+6\right)^2+\left(x-6\right)^2}{x^2+36}=\frac{\left(x^2+12x+36\right)+\left(x^2-12x+36\right)}{x^2+36}\)
=\(\frac{x^2+12x+36+x^2-12x+36}{x^2+36}=\frac{2x^2+72}{x^2+36}=\frac{2\left(x^2+36\right)}{x^2+36}=2\)
Vì P=2 nên giá trị của P không phụ thuộc vào giá trị của x