Toán Lớp 8: Có 2 ý thôi lm nhanh nha!
Bài 1: Giải pt:
a) $\dfrac{4}{x-1}$ – $\dfrac{5}{x-2}$ = -3
b) 3x – $\dfrac{1}{x-2}$ = $\dfrac{x-1}{2-x}$
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Toán Lớp 8: Có 2 ý thôi lm nhanh nha!
Bài 1: Giải pt:
a) $\dfrac{4}{x-1}$ – $\dfrac{5}{x-2}$ = -3
b) 3x – $\dfrac{1}{x-2}$ = $\dfrac{x-1}{2-x}$
Comments ( 2 )
Giải đáp+Lời giải và giải thích chi tiết:
a) 4/(x-1)-5/(x-2)=-3 ĐK: x\ne1; x\ne2
<=>(4(x-2))/((x-1)(x-2))-(5(x-1))/((x-1)(x-2))=(-3(x-1)(x-2))/((x-1)(x-2))
=>4(x-2)-5(x-1)=-3(x-1)(x-2)
<=>4x-8-5x+5=-3x^2+6x+3x-6
<=>-3x^2+9x-4x+5x-6+8-5=0
<=>-3x^2+10x-3=0
<=>3x^2-10x+3=0
<=>3x^2-x-9x+3=0
<=>x(3x-1)-3(3x-1)=0
<=>(3x-1)(x-3)=0
<=>\(\left[ \begin{array}{l}3x-1=0\\x-3=0\end{array} \right.\)
<=>\(\left[ \begin{array}{l}x=\dfrac{1}{3} (TM)\\x=3 (TM)\end{array} \right.\)
Vậy S={1/3; 3}
b) 3x-1/(x-2)=(x-1)/(2-x) ĐK: x\ne2
<=>(3x(2-x))/(2-x)+1/(2-x)=(x-1)/(2-x)
=>3x(2-x)+1=x-1
<=>6x-3x^2+1=x-1
<=>3x^2+x-6x-1-1=0
<=>3x^2-5x-2=0
<=>3x^2-6x+x-2=0
<=>3x(x-2)+(x-2)=0
<=>(x-2)(3x+1)=0
<=>\(\left[ \begin{array}{l}x-2=0\\3x+1=0\end{array} \right.\)
<=>\(\left[ \begin{array}{l}x=2 (loại)\\x=\dfrac{-1}{3}\end{array} \right.\)
Vậy S={-1/3}
Giải đáp+Lời giải và giải thích chi tiết:
a) 4/(x-1) – 5/(x-2) = -3(ĐK: x ∉ {1;2})
<=> (4(x-2))/((x-1)(x-2)) – (5(x-1))/((x-1)(x-2)) = (-3(x-1)(x-2))/((x-1)(x-2))
=> 4x – 8 – 5x + 5 = -3(x^2 – 3x +2)
<=> -x – 3 = -3x^2 + 9x – 6
<=> 3x^2 – 10x + 3 =0
<=> 3x^2 – x – 9x + 3 =0
<=> x(3x-1) -3(3x-1)=0
<=> (x-3)(3x-1)=0
<=>\(\left[ \begin{array}{l}x-3=0\\3x-1=0\end{array} \right.\)
<=>\(\left[ \begin{array}{l}x=3(tm)\\x=\dfrac{1}{3}(tm)\end{array} \right.\)
Vậy S = {3; 1/3}
b) 3x -1/(x-2) = (x-1)/(2-x)(ĐK: x \ne 2)
<=> (3x(2-x))/(2-x) + 1/(2-x) = (x-1)/(2-x)
=> 6x – 3x^2 + 1 = x – 1
<=> 3x^2 – 5x – 2=0
<=> 3x^2 + x – 6x – 2=0
<=> (x-2)(3x +1)=0
<=>\(\left[ \begin{array}{l}x-2=0\\3x+1=0\end{array} \right.\)
<=>\(\left[ \begin{array}{l}x=2(ktm)\\x=-\dfrac{1}{3}(tm)\end{array} \right.\)
Vậy S = {1/3}