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a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x
=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x
=8x.5(2x+1)(2x−1)(2x+1).4x=102x−18x.5(2x+1)(2x−1)(2x+1).4x=102x−1
b) (1x2+x−2−xx+1):(1x+x−2)(1x2+x−2−xx+1):(1x+x−2)
=(1x(x+1)+x−2x+1):1+x2−2xx(1x(x+1)+x−2x+1):1+x2−2xx
=1+x(x−2)x(x+1).xx2−2x+11+x(x−2)x(x+1).xx2−2x+1
=(x2−2x+1)xx(x+1)(x2−2x+1)=1x+1(x2−2x+1)xx(x+1)(x2−2x+1)=1x+1
c) 1x−1−x3−xx2+1.(1x2−2x+1+11−x2)1x−1−x3−xx2+1.(1x2−2x+1+11−x2)
=1x−1−x3−xx2+1.[1(x−1)2−1(x−1)(x+1)]
a) (2x+12x−1−2x−12x+1):4x10x−5(2x+12x−1−2x−12x+1):4x10x−5
= 0 - 0
= 0
b) (1x2+x−2−xx+1):(1x+x−2);(1x2+x−2−xx+1):(1x+x−2)
= (x-xx+1) : (2x-2) : (x-xx+1) : (2x-2)
c) 1x−1−x3−xx2+1.(1x2−2x+1+11−x2)
= -2x-1-xx2+1. (14 - 4x)
= -x2-1-xx2+14-4x
= -6x-xx2+13
a) \(G=\left(\dfrac{x+1}{x-1}+\dfrac{x}{x+1}+\dfrac{x}{1-x^2}\right):\left(\dfrac{x+1}{x-1}+\dfrac{1-x}{x+1}\right)\)
\(=\left(\dfrac{x+1}{x-1}+\dfrac{x}{x+1}+\dfrac{x}{\left(1-x\right)\cdot\left(1+x\right)}\right):\dfrac{\left(x+1\right)^2+\left(x-1\right)\cdot\left(1-x\right)}{\left(x-1\right)\cdot\left(x+1\right)}\)
\(=\left(\dfrac{x+1}{x-1}+\dfrac{x}{x+1}+\dfrac{x}{-\left(x-1\right)\cdot\left(1+x\right)}\right):\dfrac{\left(x+1\right)^2+\left(x-1\right)\cdot\left(-\left(x-1\right)\right)}{\left(x-1\right)\cdot\left(x+1\right)}\)
\(=\left(\dfrac{x+1}{x-1}+\dfrac{x}{x+1}-\dfrac{x}{\left(x-1\right)\cdot\left(1+x\right)}\right):\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\cdot\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2+x\cdot\left(x-1\right)-x}{\left(x-1\right)\cdot\left(x+1\right)}:\dfrac{2\cdot2x}{\left(x-1\right)\cdot\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2+x^2-x-x}{\left(x-1\right)\cdot\left(x+1\right)}:\dfrac{4x}{\left(x-1\right)\cdot\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2+x^2-2x}{\left(x-1\right)\cdot\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\cdot\left(x+1\right)}{4x}\)
\(=\left(\left(x+1\right)^2+x^2-2x\right)\cdot\dfrac{1}{4x}\)
\(=\dfrac{\left(x+1\right)^2+x^2-2x}{4x}\)
\(=\dfrac{x^2+2x+1+x^2-2x}{4x}\)
\(=\dfrac{2x^2+1}{4x}\)
a, sửa đề : \(C=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{1}{2-x}\)ĐK : \(x\ne-3;2\)
\(=\frac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-12-x}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b, Ta có : \(x^2-x=2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\Leftrightarrow x=-1;x=2\)
Kết hợp với giả thiết vậy x = -1
Thay x = -1 vào biểu thức C ta được : \(\frac{-1-4}{-1-2}=-\frac{5}{-3}=\frac{5}{3}\)
c, Ta có : \(C=\frac{1}{2}\Rightarrow\frac{x-4}{x-2}=\frac{1}{2}\Rightarrow2x-8=x-2\Leftrightarrow x=6\)( tm )
d, \(C>1\Rightarrow\frac{x-4}{x-2}>1\Rightarrow\frac{x-4}{x-2}-1>0\Leftrightarrow\frac{x-4-x+2}{x-2}>0\Leftrightarrow\frac{-2}{x-2}>0\)
\(\Rightarrow x-2< 0\Leftrightarrow x< 2\)vì -2 < 0
e, tự làm nhéee
f, \(C< 0\Rightarrow\frac{x+4}{x+2}< 0\)
mà x + 4 > x + 2
\(\hept{\begin{cases}x+4>0\\x+2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-4\\x< -2\end{cases}\Leftrightarrow-4< x< -2}}\)
Vì \(x\inℤ\Rightarrow x=-3\)( ktmđk )
Vậy ko có x nguyên để C < 0
g, Ta có : \(\frac{x+4}{x+2}=\frac{x+2+2}{x+2}=1+\frac{2}{x+2}\)
Để C nguyên khi \(x+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x + 2 | 1 | -1 | 2 | -2 |
| x | -1 | -3 | 0 | -4 |
h, Ta có : \(D=C\left(x^2-4\right)=\frac{x+4}{x+2}.\frac{\left(x-2\right)\left(x+2\right)}{1}=x^2+2x-8\)
\(=\left(x+1\right)^2-9\ge-9\)
Dấu ''='' xảy ra khi x = -1
Vậy GTNN D là -9 khi x = -1