1.Tìm x,biết:
a)\(x+4 \over 2018\)+\(x+3 \over 2019\)+\(x+2 \over 2020\)=-1
b)\(3 \over 5\)+\(2 \over 5\):x=1
c)3/2x-1/-2=4
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a, 8/x-8 + 11/x-11 = 9/x-9 + 10/ x-10
b, x/x-3 - x/x-5 = x/x-4 - x/x-6
c, 4/x^2-3x+2 - 3/2x^2-6x+1 +1 = 0
d, 1/x-1 + 2/ x-2 + 3/x-3 = 6/x-6
e, 2/2x+1 - 3/2x-1 = 4/4x^2-1
f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3
g, 1/x-1 + 2x^2 -5/x^3 -1 = 4/ x^2 +x+1
1)
ĐK: \(x,y\neq 0\); \(x+y\neq 0\)
\(\frac{x^2-y^2}{6x^2y^2}: \frac{x+y}{12xy}\)
\(=\frac{x^2-y^2}{6x^2y^2}. \frac{12xy}{x+y}=\frac{(x-y)(x+y).12xy}{6x^2y^2(x+y)}=\frac{2(x-y)}{xy}\)
2) ĐK: \(x\neq \frac{\pm 1}{2}; 0; 1\)
\(\frac{5x}{2x+1}: \frac{3x(x-1)}{4x^2-1}=\frac{5x}{2x+1}.\frac{4x^2-1}{3x(x-1)}\)
\(=\frac{5x(2x-1)(2x+1)}{(2x+1).3x(x-1)}=\frac{5(2x-1)}{3(x-1)}\)
3) ĐK: \(x\neq \frac{\pm 1}{2}; 0\)
\(\left(\frac{2x-1}{2x+1}-\frac{2x-1}{2x+1}\right): \frac{4x}{10x-5}=0: \frac{4x}{10x-5}=0\)
4) ĐK: \(x\neq \frac{\pm 1}{3}\)
\(\frac{2}{9x^2+6x+1}-\frac{3x}{9x^2-1}=\frac{2}{(3x+1)^2}-\frac{3x}{(3x-1)(3x+1)}\)
\(=\frac{2(3x-1)}{(3x+1)^2(3x-1)}-\frac{3x(3x+1)}{(3x-1)(3x+1)^2}\)
\(=\frac{6x-2-9x^2-3x}{(3x+1)^2(3x-1)}=\frac{-9x^2+3x-2}{(3x-1)(3x+1)^2}\)
5) ĐK: \(x\neq \pm 1; \frac{-7\pm \sqrt{89}}{4}\)
\(\left(\frac{5}{x^2+2x+1}+\frac{2x}{x^2-1}\right): \frac{2x^2+7x-5}{3x-3}\)
\(=\left(\frac{5}{(x+1)^2}+\frac{2x}{(x-1)(x+1)}\right). \frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{5(x-1)+2x(x+1)}{(x-1)(x+1)^2}. \frac{3(x-1)}{2x^2+7x-5}=\frac{2x^2+7x-5}{(x+1)^2(x-1)}.\frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{3}{(x+1)^2}\)
1) 3(x + 2) = 5x + 8
<=> 3x + 6 = 5x + 8
<=> 3x + 6 - 5x - 8 = 0
<=> -2x - 2 = 0
<=> -2x = 0 + 2
<=> -2x = 2
<=> x = -1
2) 2(x - 1) = 3(3 + x) + 3
<=> 2x - 2 = 9 + x + 3
<=> 2x - 2 = 12 + x
<=> 2x - 2 - 12 - x = 0
<=> x - 14 = 0
<=> x = 0 + 14
<=> x = 14
3) 5 - (x - 6) = 4(3 - 2x)
<=> 5 - x + 6 = 12 - 8x
<=> 11 - x = 12 - 8x
<=> 11 - x - 12 + 8x = 0
<=> -1 + 7x = 0
<=> 7x = 0 + 1
<=> 7x = 1
<=> x = 1/7
\(\text{a) }3x+6=8x+3\)
\(\Leftrightarrow3x-8x=3-6\)
\(\Leftrightarrow-5x=-3\)
\(\Leftrightarrow x=\frac{-3}{-5}=\frac{3}{5}\)
\(\text{Câu b và câu c bạn ghi rõ lại giùm}\)