P=( x+2/x^2+2x+1 – x-2/x^2-1 ) : x/(x+1) A,rút gọn P B, tìm x để P=4 C, tính P biết x=5 D, tìm x để P > 1 15/07/2021 Bởi Raelynn P=( x+2/x^2+2x+1 – x-2/x^2-1 ) : x/(x+1) A,rút gọn P B, tìm x để P=4 C, tính P biết x=5 D, tìm x để P > 1
Giải thích các bước giải: $P=(\dfrac{x+2}{x^2+2x+1}-\dfrac{x-2}{x^2-1}):\dfrac{x}{x+1}$ $\to P=(\dfrac{x+2}{(x+1)^2}-\dfrac{x-2}{(x-1)(x+1)}).\dfrac{x+1}{x}$ $\to P=(\dfrac{x+2}{x+1}-\dfrac{x-2}{x-1}).\dfrac{1}{x}$ $\to P=\dfrac{(x+2)(x-1)-(x+1)(x-2)}{(x+1)(x-1)}.\dfrac{1}{x}$ $\to P=\dfrac{2x}{(x+1)(x-1)}.\dfrac{1}{x}$ $\to P=\dfrac{2}{(x+1)(x-1)}.\dfrac{1}{x}$ $\to P=\dfrac{2}{(x+1)(x-1)}$ b.Để $P=4\to \dfrac{2}{(x+1)(x-1)}=4\to \dfrac{2}{x^2-1}=4\to x^2-1=\dfrac{1}{2}\to x^2=\dfrac{3}{2}\to x=\pm\sqrt{\dfrac{3}2}$ c.$x=5\to P=\dfrac{2}{(5+1)(5-1)}=\dfrac{1}{12}$ d.Để $P > 1\to \dfrac{2}{(x+1)(x-1)}>1\to \dfrac{2}{x^2-1}>1\to 0< x^2-1<2$ $\to -\sqrt{3}<x<-1 $ hoặc $1<x<\sqrt{3}$ Bình luận
Giải thích các bước giải: P=(x+2x2+2x+1−x−2x2−1):xx+1P=(x+2×2+2x+1−x−2×2−1):xx+1 →P=(x+2(x+1)2−x−2(x−1)(x+1)).x+1x→P=(x+2(x+1)2−x−2(x−1)(x+1)).x+1x →P=(x+2x+1−x−2x−1).1x→P=(x+2x+1−x−2x−1).1x →P=(x+2)(x−1)−(x+1)(x−2)(x+1)(x−1).1x→P=(x+2)(x−1)−(x+1)(x−2)(x+1)(x−1).1x →P=2x(x+1)(x−1).1x→P=2x(x+1)(x−1).1x →P=2(x+1)(x−1).1x→P=2(x+1)(x−1).1x →P=2(x+1)(x−1)→P=2(x+1)(x−1) b.ĐểP=4→2(x+1)(x−1)=4→2x2−1=4→x2−1=12→x2=32→x=±√32P=4→2(x+1)(x−1)=4→2×2−1=4→x2−1=12→x2=32→x=±32 c.x=5→P=2(5+1)(5−1)=112x=5→P=2(5+1)(5−1)=112 d.Để P>1→2(x+1)(x−1)>1→2x2−1>1→0<x2−1<2P>1→2(x+1)(x−1)>1→2×2−1>1→0<x2−1<2 →−√3<x<−1→−3<x<−1 hoặc 1<x<√31<x<3 Bình luận
Giải thích các bước giải:
$P=(\dfrac{x+2}{x^2+2x+1}-\dfrac{x-2}{x^2-1}):\dfrac{x}{x+1}$
$\to P=(\dfrac{x+2}{(x+1)^2}-\dfrac{x-2}{(x-1)(x+1)}).\dfrac{x+1}{x}$
$\to P=(\dfrac{x+2}{x+1}-\dfrac{x-2}{x-1}).\dfrac{1}{x}$
$\to P=\dfrac{(x+2)(x-1)-(x+1)(x-2)}{(x+1)(x-1)}.\dfrac{1}{x}$
$\to P=\dfrac{2x}{(x+1)(x-1)}.\dfrac{1}{x}$
$\to P=\dfrac{2}{(x+1)(x-1)}.\dfrac{1}{x}$
$\to P=\dfrac{2}{(x+1)(x-1)}$
b.Để $P=4\to \dfrac{2}{(x+1)(x-1)}=4\to \dfrac{2}{x^2-1}=4\to x^2-1=\dfrac{1}{2}\to x^2=\dfrac{3}{2}\to x=\pm\sqrt{\dfrac{3}2}$
c.$x=5\to P=\dfrac{2}{(5+1)(5-1)}=\dfrac{1}{12}$
d.Để $P > 1\to \dfrac{2}{(x+1)(x-1)}>1\to \dfrac{2}{x^2-1}>1\to 0< x^2-1<2$
$\to -\sqrt{3}<x<-1 $ hoặc $1<x<\sqrt{3}$
Giải thích các bước giải:
P=(x+2x2+2x+1−x−2x2−1):xx+1P=(x+2×2+2x+1−x−2×2−1):xx+1
→P=(x+2(x+1)2−x−2(x−1)(x+1)).x+1x→P=(x+2(x+1)2−x−2(x−1)(x+1)).x+1x
→P=(x+2x+1−x−2x−1).1x→P=(x+2x+1−x−2x−1).1x
→P=(x+2)(x−1)−(x+1)(x−2)(x+1)(x−1).1x→P=(x+2)(x−1)−(x+1)(x−2)(x+1)(x−1).1x
→P=2x(x+1)(x−1).1x→P=2x(x+1)(x−1).1x
→P=2(x+1)(x−1).1x→P=2(x+1)(x−1).1x
→P=2(x+1)(x−1)→P=2(x+1)(x−1)
b.ĐểP=4→2(x+1)(x−1)=4→2x2−1=4→x2−1=12→x2=32→x=±√32P=4→2(x+1)(x−1)=4→2×2−1=4→x2−1=12→x2=32→x=±32
c.x=5→P=2(5+1)(5−1)=112x=5→P=2(5+1)(5−1)=112
d.Để P>1→2(x+1)(x−1)>1→2x2−1>1→0<x2−1<2P>1→2(x+1)(x−1)>1→2×2−1>1→0<x2−1<2
→−√3<x<−1→−3<x<−1 hoặc 1<x<√31<x<3