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Sửa lại đề \(\frac{x+1}{x-1}+\frac{x-2}{x+2}+\frac{x-3}{x+3}+\frac{x+4}{x-4}=-4\)
ĐK \(x\ne\left\{1;-2;-3;4\right\}\)
\(\Leftrightarrow\left(\frac{x+1}{x-1}+1\right)+\left(\frac{x-2}{x+2}+1\right)+\left(\frac{x-3}{x+3}+1\right)+\left(\frac{x+4}{x-4}+1\right)=0\)
\(\Leftrightarrow\frac{2x}{x-1}+\frac{2x}{x+2}+\frac{2x}{x+3}+\frac{2x}{x-4}=0\)
\(\Leftrightarrow2x\left(\frac{1}{x-1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x-4}\right)=0\Leftrightarrow x=0\)vì \(\frac{1}{x-1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x-4}\ne0\)
Vậy pt có nghiệm \(x=0\)
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\(\int^{1+\frac{4}{x-1}-\frac{3}{y-1}=2}_{\frac{4}{x-1}+\frac{2}{y-1}=6}\Leftrightarrow\int^{\frac{1}{y-1}=1}_{\frac{1}{x-1}=1}\Leftrightarrow x=y=2\)
( đặt ẩn phụ cho dẽ làm nhá)
Bạn viết sai dấu y+1 = y-1 ( theo bạn nhẩm 2;2)
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a) (2x + 1)(3x - 2) = (5x - 8)(2x + 1)
<=> 6x2 - x - 2 = 10x2 - 11x - 8
<=> 6x2 - 10x2 - x + 11x -2 + 8 = 0
<=> -4x2 + 10x + 6 = 0
<=> -2 (2x2 - 5x - 3) = 0
<=> 2x2 - 5x - 3 = 0
<=> 2x2 - 6x + x - 3 = 0
<=> x (2x + 1) - 3 (2x + 1) = 0
<=> (x - 3) (2x + 1) = 0
* x - 3 = 0 => x = 3
* 2x + 1 = 0 => x = -1/2
S = {-1/2; 3}
b) 4x2 – 1 = (2x +1)(3x -5)
<=> 4x2 – 1 - (2x +1)(3x -5) = 0
<=> (2x - 1) (2x + 1) - (2x + 1)(3x - 5) = 0
<=> (2x + 1) (2x - 1 - 3x + 5) = 0
<=> (2x + 1) (-x + 4) = 0
* 2x + 1 = 0 <=> x = -1/2
* -x + 4 = 0 <=> x = 4
S = {-1/2; 4}
c) (x + 1)2 = 4(x2 – 2x + 1)
<=> (x + 1)2 - 4(x2 – 2x + 1) = 0
<=> (x + 1)2 - 4(x2 – 1)2 = 0
* (x + 1)2 = 0 <=> x = -1
* 4(x2 - 1)2 = 0 <=> x = 1 và x = -1
S = {-1; 1}
d) 2x3 + 5x2 – 3x = 0
<=> x (2x2 + 5x - 3) = 0
<=> x (2x2 + 6x - x - 3) = 0
<=> x [x(2x - 1) + 3 (2x - 1)] = 0
<=> x (2x - 1) (x + 3) = 0
* x = 0
* 2x - 1 = 0 <=> x = 1/2
* x + 3 = 0 <=> x = -3
S = { -3; 0; 1/2}
\(\frac{1}{x^2+5x+4}+\frac{1}{x^2+11x+28}+\frac{1}{x^2+17x+70}=\frac{3}{4x-2}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+10\right)}=\frac{3}{4x-2}\)
\(\Leftrightarrow3x^2+21x+36=0\)
\(\Leftrightarrow x=-3\)
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\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{x^2+2x-3}=1\)
\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{\left(x+1\right)^2-4}=1\)
\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{\left(x+1+2\right)\left(x+1-2\right)}=1\)
\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{\left(x+3\right)\left(x-1\right)}=1\)
ĐKXĐ: x \(\ne\) 1 và x \(\ne\) - 3
\(\left(3x-1\right)\left(x+3\right)-\left(2x-5\right)\left(x-1\right)+4=\left(x+3\right)\left(x-1\right)\)
3x2 + 9x - x - 3 - 2x2 + 2x + 5x - 5 + 4 = x2 - x + 3x - 3
3x2 + 9x - x - 3 - 2x2 + 2x + 5x - 5 + 4 - x2 + x - 3x + 3 = 0
13x - 1 = 0
x = \(\frac{1}{13}\)
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Điều kiện: x khác (-3,-2,1,4)
PT <=>
\(1+\frac{2}{x-1}+1-\frac{4}{x+2}+1-\frac{6}{x+3}+1+\frac{8}{x-4}=4\)
<=> \(\frac{1}{x-1}-\frac{2}{x+2}-\frac{3}{x+3}+\frac{4}{x-4}=0\)
<=> (x+2)(x+3)(x-4)-2(x-1)(x+3)(x-4)-3(x-1)(x+2)(x-4)+4(x-1)(x+2)(x+3)=0
<=> (x3+x2-14x-24)-2(x3 - 2x2-11x+12) - 3(x3 - 3x2- 6x+8) + 4(x3+4x2 + x-6) = 0
<=> x3+x2-14x-24-2x3 + 4x2+22x-24 - 3x3 + 9x2+ 18x-24 + 4x3+16x2 + 4x-24 = 0
<=> 30x2 + 30x -96=0
<=> 5x2 + 5x -16 = 0
Giải ra được: \(\orbr{\begin{cases}x_1=\frac{-5-\sqrt{345}}{10}\\x_2=\frac{-5+\sqrt{345}}{10}\end{cases}}\)