giai PT
a) $\frac{x}{x+3}$ – $\frac{x+1}{2x-6}$ = $\frac{5-2x}{x^2-9}$
b) 11-5x> -8
c) $\frac{x+1}{3}$ – $\frac{2x-1}{15}$ $\geq$ $\frac{4x+3}{5}$
giai PT
a) $\frac{x}{x+3}$ – $\frac{x+1}{2x-6}$ = $\frac{5-2x}{x^2-9}$
b) 11-5x> -8
c) $\frac{x+1}{3}$ – $\frac{2x-1}{15}$ $\geq$ $\frac{4x+3}{5}$
Đáp án:
a) `S \ = \ { \ 3+sqrt(22) \ ; \ 3-sqrt(22) \ }`
b) `S \ = \ { \ x>19/5 \ }`
c) `S \ = \ { \ x<=-1/3 \ }`
Giải thích các bước giải:
a) `x/(x+3) – (x+1)/(2x-6) = (5-2x)/(x^2-9)` ĐKXĐ : `x ne +-3`
`<=> x/(x+3) – (x+1)/(2(x-3)) = (5-2x)/(x^2-9)`
`<=> (2x(x-3)-(x+1)(x+3))/(2(x^2-9)) = (2(5-2x))/(2(x^2-9))`
`=> 2x(x-3)-(x+1)(x+3)=2(5-2x)`
`<=> 2x^2-6x-(x^2+4x+3)=10-4x`
`<=> 2x^2-6x-x^2-4x-3-10+4x=0`
`<=> x^2-6x-13=0`
`<=> x^2-6x+9-22=0`
`<=> (x-3)^2=22`
`<=>`\(\left[ \begin{array}{l}x-3=\sqrt{22}\\x-3=-\sqrt{22}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=3+\sqrt{22}\\x=3-\sqrt{22}\end{array} \right.\)
Vậy `S \ = \ { \ 3+sqrt(22) \ ; \ 3-sqrt(22) \ }`
b) `11-5x> -8`
`<=> 5x > 11-(-8)`
`<=> 5x>19`
`<=> x>19/5`
Vậy `S \ = \ { \ x>19/5 \ }`
c) `(x+1)/3-(2x-1)/15 >= (4x+3)/5`
`<=> (5.(x+1)-(2x-1))/15 >= (3.(4x+3))/15`
`<=> 5x+5-2x+1 >= 12x+9`
`<=> 5x-2x-12x+5+1-9 >= 0`
`<=> -9x-3 >= 0`
`<=> -9x >= 3`
`<=> x<=-1/3`
Vậy `S \ = \ { \ x<=-1/3 \ }`
@Mốc
a) $\frac{x}{x+3}$ – $\frac{x+1}{2x-6}$ = $\frac{5-2x}{x^{2}-9}$ (ĐKXĐ: x$\neq$±3)
<=> $\frac{x}{x+3}$ – $\frac{x+1}{2(x-3)}$ = $\frac{5-2x}{x^{2}-9}$
<=> $\frac{2x(x-3)}{2(x^{2}-9)}$ – $\frac{(x+1)(x+3)}{2(x^{2}-9)}$ – $\frac{2(5 – 2x)}{2(x^{2}-9)}$ = 0
=> $2x^{2}$ – 6x – ($x^{2}$ + 3x + x + 3) – 10 + 4x = 0
<=> $2x^{2}$ – 6x – $x^{2}$ – 3x – x – 3 – 10 + 4x = 0
<=> $x^{2}$ – 6x – 13 = 0
<=> $x^{2}$ – 2.3.x + 9 – 22 = 0
<=> $( x – 3)^{2}$ – 22 = 0
Mà $( x – 3)^{2}$ ≥ 0 ∨ x ∈ R
<=> $( x – 3)^{2}$ – 22 ≥ – 22 ∨ x ∈ R
<=> $( x – 3)^{2}$ – 22 > 0 ∨ x ∈ R
Vậy phương trình vô nghiệm ( x ∈ ∅ )
b) 11 – 5x > – 8
<=> 5x > 19
<=> x > $\frac{19}{5}$
Vậy bất phương trình có nghiệm x > $\frac{19}{5}$.
c) $\frac{x+1}{3}$ – $\frac{2x-1}{15}$ ≥ $\frac{4x+3}{5}$
<=> $\frac{5(x+1)}{15}$ – $\frac{2x-1}{15}$ – $\frac{3(4x+3)}{15}$ ≥ 0
<=> 5x + 5 – 2x + 1 – 12x – 9 ≥ 0
<=> -9x – 3 ≥ 0
<=> -9x ≥ 3
<=> x ≤ $\frac{-1}{3}$
Vậy bất phương trình có nghiệm là x ≤ $\frac{-1}{3}$
#chucbanhoctotnhe;333