https://i.imgur.com/CBGVBUp.jpg
Violympic toán 8
Các câu hỏi tương tự
Giải phương trình:
1. \(x^2-3x+2+\left|x-1\right|=0\)
2. \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
GIÚP MK VS!!!
3) frac{x-2}{x-5} -frac{5}{x^2-5x}frac{1}{x}
Leftrightarrow frac{x-2}{x-5}-frac{5}{x.left(x-5right)}frac{1}{x}
Leftrightarrowfrac{left(x-2right).left(x+5right)}{x.left(x-5right)}-frac{5}{x.left(x-5right)}frac{1.left(x+5right)}{x.left(x-5right)}
Leftrightarrow x^2+5x-2x-10-51x+5
Leftrightarrow x^2+5x-2x-1x-10-5-5 0
Leftrightarrow x^2+2x-200
Leftrightarrow x^2+2x-10x-200
Leftrightarrow (x^2 + 2x) - (10x + 20) 0
Leftrightarrow x.(x + 2) - 10.(x + 2) 0
Leftrightarrow
4) frac{x-4}{x+7}-f...
Đọc tiếp
3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\)
\(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)
\(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\)
\(\Leftrightarrow x^2+5x-2x-10-5=1x+5\)
\(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0
\(\Leftrightarrow\) \(x^2+2x-20=0\)
\(\Leftrightarrow x^2+2x-10x-20=0\)
\(\Leftrightarrow\) (x\(^2\) + 2x) - (10x + 20) = 0
\(\Leftrightarrow\) x.(x + 2) - 10.(x + 2) = 0
\(\Leftrightarrow\)
4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)
\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x\left(x+7\right)}\)
\(\Leftrightarrow\frac{\left(x-4\right).\left(x+7\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)
\(\Leftrightarrow\) \(x^2+7x-4x-28-x-7=-7\)
\(\Leftrightarrow x^2+7x-4x-x-28-7+7=0\)
\(\Leftrightarrow\) x\(^2\) + 2x - 28 = 0
\(\Leftrightarrow\) x\(^2\) + 2x - 14x - 28 = 0
\(\Leftrightarrow\) (x\(^2\) + 2x) - (14x + 28) = 0
\(\Leftrightarrow\) x.(x + 2) - 14.(x + 2) = 0
\(\Leftrightarrow\) (x - 14) = 0 hoặc (x + 2) = 0
\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = -2 (Loại)
5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)
\(\Leftrightarrow\) \(\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)
\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)
\(\Leftrightarrow\) \(x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)
\(\Leftrightarrow2x^2-8x+8=0\)
\(\Leftrightarrow\) 2x\(^2\) - 2x - 8x + 8 = 0
\(\Leftrightarrow\) 2x(x - 1) - 8(x - 1) = 0
\(\Leftrightarrow\) 2x - 8 = 0 hoặc x - 1 = 0
\(\Leftrightarrow\) 2x = 8 hoặc x = 1
\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = 1 (Nhận)
Vậy S = {4; 1}
6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)
\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)
\(\Leftrightarrow\) x\(^2\) + x + x + 1 - x\(^2\) + x + x - 1 = 4
\(\Leftrightarrow\) 4x - 4 = 0
\(\Leftrightarrow\) 4 (x - 1) =0
\(\Leftrightarrow\) x - 1 = 0 / 4 = 0
\(\Leftrightarrow\) x = 1 (Nhận)
Vậy S = {1}
7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)
\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x+1\right)}\)
\(\Leftrightarrow x^2+x+x+1-4x=x^2-x-x+1\)
\(\Leftrightarrow\) 0
Vậy S ={\(\varnothing\)}
\(\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x+4}\)
\(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{\left(x-4\right)}{x\left(x+2\right)}=0\)
\(\frac{1}{x-2}\frac{6}{x+3}=\frac{5}{6-x^2-x}\)
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)
giải pt
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
Giải phương trình: \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
giải pt trên
Giải PT
\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\)
Giải phương trình : \(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-12\left(\frac{x-2}{x-4}\right)^2=0\)
11 tháng 3 2020 lúc 15:52
a, (x-1)3 - x(x-1)2 = 5(2-x) - 11(x+2)
b, (x-2)3 + (3x-1)(3x+1) = (x+1)3
c, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)
d, \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
e, \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)