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ĐKXĐ \(x\ne0,-1,-2,...,-100\)
\(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+...+\frac{1}{x^2+199x+9900}=\frac{25}{51}\)
\(\Leftrightarrow\frac{1}{x\left(x+1\right)}+\frac{1}{x^2+x+2x+2}+...+\frac{1}{x^2+99x+100x+9900}=\frac{25}{51}\)
\(\Leftrightarrow\frac{1}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)+2\left(x+1\right)}+....+\frac{1}{x\left(x+99\right)+100\left(x+99\right)}=\frac{25}{51}\)
\(\Leftrightarrow\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+99\right)\left(x+100\right)}=\frac{25}{21}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+99}-\frac{1}{x+100}=\frac{25}{21}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+100}=\frac{25}{21}\)
\(\Leftrightarrow\frac{x+100-x}{x\left(x+100\right)}=\frac{25}{21}\)
\(\Leftrightarrow\frac{100}{x\left(x+100\right)}=\frac{25}{21}\)
\(\Leftrightarrow25x^2+2500x=2100\)
\(\Leftrightarrow x^2+100x-84=0\)
\(\Leftrightarrow x^2+2.x.50+50^2-50^2-84=0\)
\(\Leftrightarrow\left(x+50\right)^2-2584=0\)
\(\Leftrightarrow\left(x+50-2\sqrt{646}\right)\left(x+50+2\sqrt{646}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-50+2\sqrt{646}\\x=-50-2\sqrt{646}\end{cases}}\)
Vậy ...
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
\(\frac{1}{\left(x-1\right)^3}+\frac{1}{\left(x+1\right)^3}+\frac{1}{x^3}-\frac{1}{3x\left(x^2+2\right)}=0\)
\(\Leftrightarrow\frac{x\left(2x^2+6\right)}{\left(x^2-1\right)^3}+\frac{2x^2+6}{3x^3\left(x^2+2\right)}=0\)
\(\Leftrightarrow\frac{x}{\left(x^2-1\right)^3}+\frac{1}{3x^3\left(x^2+2\right)}=0\)
\(\Leftrightarrow4x^6+3x^4+3x^2-1=0\)
Đặt \(x^2=a\)
\(\Rightarrow4a^3+3a^2+3a-1=0\)
\(\Leftrightarrow\left(4a-1\right)\left(a^2+a+1\right)=0\)
\(\Leftrightarrow4a=1\)
\(\Rightarrow4x^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
a) \(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}\)
Đặt \(x^2-2x+3=t\left(t\ge2\right)\), khi đó phương trình trở thành:
\(\frac{1}{t-1}+\frac{2}{t}=\frac{6}{t+1}\)
\(\Leftrightarrow\frac{t\left(t+1\right)+t^2-1}{\left(t-1\right)t\left(t+1\right)}=\frac{6t\left(t-1\right)}{\left(t-1\right)t\left(t+1\right)}\)
\(\Leftrightarrow t\left(t+1\right)+t^2-1=6t\left(t-1\right)\)
\(\Leftrightarrow2t^2+t-1=6t^2-6t\)
\(\Leftrightarrow-4t^2+7t-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=\frac{7+\sqrt{33}}{8}\\t=\frac{7-\sqrt{33}}{8}\end{cases}}\left(ktmđk\right)\)
Vậy phương trình vô nghiệm.
A . 3x + 2(x + 1) = 6x - 7
<=> 3x + 2x + 2 = 6x -7
<=> 5x - 6x = -7 - 2
<=> -x = -9
<=> x =9
B . \(\frac{x+3}{5}\).< \(\frac{5-x}{3}\)
=> 3(x +3) < 5(5 -x)
<=> 3x+9 < 25 - 5x
<=> 3x + 5x < 25 - 9
<=> 8x < 16
<=> x < 2
C . \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2-3x-4}\)=\(\frac{2}{x-4}\)
<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{x^2+x-4x-4_{ }}\)= \(\frac{2}{x-4}\)
<=> \(\frac{5}{x+1}\)+ \(\frac{2x}{\left(x+1\right)\left(x-4\right)}\)= \(\frac{2}{x-4}\)
<=> 5(x - 4) + 2x = 2(x +1)
<=> 5x - 20 + 2x = 2x + 2
<=>7x - 2x = 2 + 20
<=> 5x = 22
<=> x =\(\frac{22}{5}\)
\(pt\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2+3x-1}{x^3-1}=\frac{3x^2}{x^3-1}\)
\(\Rightarrow x^2+3x-1=3x^2\Leftrightarrow3x-1=2x^2\Leftrightarrow2x^2-3x+1=0\Leftrightarrow x^2-\frac{3}{2}x+\frac{1}{2}=0\)
đến đây là pt bậc 2