a) Rút Gọn `A=((x+1)/(x-1)-(x-1)/(x+1)) : (2x)/(5x-5)` b) Tìm x để `A + 6/(x-2) =-1` 30/07/2021 Bởi Parker a) Rút Gọn `A=((x+1)/(x-1)-(x-1)/(x+1)) : (2x)/(5x-5)` b) Tìm x để `A + 6/(x-2) =-1`
`a) A = ((x+1)/(x-1) – (x-1)/(x+1)) : (2x)/(5x-5) (ĐKXĐ : x \ne +-1)` `= (((x+1).(x+1))/((x-1).(x+1)) – ((x-1).(x-1))/((x+1).(x-1)) ): (2x)/(5.(x-1))` ` = ((x+1)^2 – (x-1)^2)/((x-1).(x+1)) . (5.(x-1))/(2x)` ` = ((x^2 + 2x+1) – (x^2 – 2x+1))/((x-1).(x+1)) . (5.(x-1))/(2x)` ` = (x^2 + 2x + 1 – x^2 + 2x – 1)/((x-1).(x+1)) . (5.(x-1))/(2x)` ` = (4x)/((x-1).(x+1)) . (5.(x-1))/(2x)` ` = ( 2 . 2x)/((x-1).(x+1)) . (5.(x-1))/(2x)` ` = (2.5)/(x+1)` ` = 10/(x+1)` `b)` Ta có : `A + 6/(x-2) = -1` `<=> A = -1 – 6/(x-2)` Để `A + 6/(x+1) = -1` thì `10/(x+1) = -1-6/(x-2)` và `x\ne +-1` `<=> (10.(x-2))/((x+1).(x-2)) = (-1.(x+1).(x-2))/((x+1).(x-2)) – (6.(x+1))/((x+1).(x-2))` `=> 10.(x-2) = -1.(x+1).(x-2) – 6.(x+1)` `<=> 10x – 20 = (-x-1).(x-2) – 6x – 6` `<=> 10x – 20 = -x^2 + 2x – x + 2 – 6x – 6` `<=> 10x + x^2 – 2x + x + 6x = 2 – 6 + 20` `<=> x^2 +15x = 16` `<=> x^2 + 15x – 16 = 0` `<=> x^2 – x + 16x – 16 = 0 ` `<=> x.(x-1) + 16.(x-1) = 0` `<=> (x+16).(x-1) = 0` `<=> x+16=0` hoặc `x-1=0` `+)x+16=0 <=> x =-16 (TMĐKXĐ)` `+) x-1=0 <=> x =1 (KTMĐKXĐ)` Vậy với `x=-16` thì `A + 6/(x-2) = -1` Bình luận
Đáp án+Giải thích các bước giải: `a)` Với `x\ne1;x\ne-1;x\ne0` Ta có: `A=((x+1)/(x-1)-(x-1)/(x+1)):(2x)/(5x-5)` `\to A=((x+1)(x+1)-(x-1)(x-1))/((x-1)(x+1)):(2x)/(5x-5)` `\to A=((x+1)^2-(x-1)^2)/((x-1)(x+1)):(2x)/(5x-5)` `\to A=((x+1+x-1)(x+1-x+1))/((x-1)(x+1)):(2x)/(5(x-1))` `\to A=(2x.2)/((x-1)(x+1)).(5(x-1))/(2x)` `\to A=2.5/(x+1)` `\to A=10/(x+1)` `b)ĐKXĐ:x\ne-1;x\ne1;x\ne2;x\ne0` `A+6/(x-2)=-1` `⇔10/(x+1)+6/(x-2)=-1` `⇔(10(x-2)+6(x+1))/((x+1)(x-2))=-((x+1)(x-2))/((x+1)(x-2))` `⇒10(x-2)+6(x+1)=-(x+1)(x-2)` `⇔10x-20+6x+6=-(x^2-x-2)` `⇔16x-14=-x^2+x+2` `⇔x^2-x-2+16x-14=0` `⇔x^2+15x-16=0` `⇔x^2+16x-x-16=0` `⇔x(x+16)-(x+16)=0` `⇔(x-1)(x+16)=0` \(⇔\left[ \begin{array}{l}x-1=0\\x+16=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=1(ktm)\\x=-16(tm)\end{array} \right.\) Vậy `x=-16` để `A+6/(x-2)=-1` Bình luận
`a) A = ((x+1)/(x-1) – (x-1)/(x+1)) : (2x)/(5x-5) (ĐKXĐ : x \ne +-1)`
`= (((x+1).(x+1))/((x-1).(x+1)) – ((x-1).(x-1))/((x+1).(x-1)) ): (2x)/(5.(x-1))`
` = ((x+1)^2 – (x-1)^2)/((x-1).(x+1)) . (5.(x-1))/(2x)`
` = ((x^2 + 2x+1) – (x^2 – 2x+1))/((x-1).(x+1)) . (5.(x-1))/(2x)`
` = (x^2 + 2x + 1 – x^2 + 2x – 1)/((x-1).(x+1)) . (5.(x-1))/(2x)`
` = (4x)/((x-1).(x+1)) . (5.(x-1))/(2x)`
` = ( 2 . 2x)/((x-1).(x+1)) . (5.(x-1))/(2x)`
` = (2.5)/(x+1)`
` = 10/(x+1)`
`b)`
Ta có : `A + 6/(x-2) = -1`
`<=> A = -1 – 6/(x-2)`
Để `A + 6/(x+1) = -1` thì `10/(x+1) = -1-6/(x-2)` và `x\ne +-1`
`<=> (10.(x-2))/((x+1).(x-2)) = (-1.(x+1).(x-2))/((x+1).(x-2)) – (6.(x+1))/((x+1).(x-2))`
`=> 10.(x-2) = -1.(x+1).(x-2) – 6.(x+1)`
`<=> 10x – 20 = (-x-1).(x-2) – 6x – 6`
`<=> 10x – 20 = -x^2 + 2x – x + 2 – 6x – 6`
`<=> 10x + x^2 – 2x + x + 6x = 2 – 6 + 20`
`<=> x^2 +15x = 16`
`<=> x^2 + 15x – 16 = 0`
`<=> x^2 – x + 16x – 16 = 0 `
`<=> x.(x-1) + 16.(x-1) = 0`
`<=> (x+16).(x-1) = 0`
`<=> x+16=0` hoặc `x-1=0`
`+)x+16=0 <=> x =-16 (TMĐKXĐ)`
`+) x-1=0 <=> x =1 (KTMĐKXĐ)`
Vậy với `x=-16` thì `A + 6/(x-2) = -1`
Đáp án+Giải thích các bước giải:
`a)`
Với `x\ne1;x\ne-1;x\ne0`
Ta có:
`A=((x+1)/(x-1)-(x-1)/(x+1)):(2x)/(5x-5)`
`\to A=((x+1)(x+1)-(x-1)(x-1))/((x-1)(x+1)):(2x)/(5x-5)`
`\to A=((x+1)^2-(x-1)^2)/((x-1)(x+1)):(2x)/(5x-5)`
`\to A=((x+1+x-1)(x+1-x+1))/((x-1)(x+1)):(2x)/(5(x-1))`
`\to A=(2x.2)/((x-1)(x+1)).(5(x-1))/(2x)`
`\to A=2.5/(x+1)`
`\to A=10/(x+1)`
`b)ĐKXĐ:x\ne-1;x\ne1;x\ne2;x\ne0`
`A+6/(x-2)=-1`
`⇔10/(x+1)+6/(x-2)=-1`
`⇔(10(x-2)+6(x+1))/((x+1)(x-2))=-((x+1)(x-2))/((x+1)(x-2))`
`⇒10(x-2)+6(x+1)=-(x+1)(x-2)`
`⇔10x-20+6x+6=-(x^2-x-2)`
`⇔16x-14=-x^2+x+2`
`⇔x^2-x-2+16x-14=0`
`⇔x^2+15x-16=0`
`⇔x^2+16x-x-16=0`
`⇔x(x+16)-(x+16)=0`
`⇔(x-1)(x+16)=0`
\(⇔\left[ \begin{array}{l}x-1=0\\x+16=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=1(ktm)\\x=-16(tm)\end{array} \right.\)
Vậy `x=-16` để `A+6/(x-2)=-1`