MM
4 tháng 10 2018 lúc 22:48
Violympic toán 8
Các câu hỏi tương tự
Chứng minh biểu thức sau không phụ thuộc vào giá trị của biến :
\(A=x.\left(5x-3\right)-x^2.\left(x-1\right)+x.\left(x^2-6x\right)-10+3x+x.\left(x^2+x+1\right)-x^2.\left(x+1\right)-x+5\)
\(B=3.\left(2x-1\right)-5.\left(x-3\right)+6.\left(3x-4\right)-19x+x.\left(3x+12\right)-\left(7x-20\right)+x^2.\left(2x-3\right)-x.\left(2x^2+5\right)\)
Giaỉ các phương trình sau:
a) \(\left(x^2+11x+12\right)\left(x^2+9x+20\right)\left(x^2+13x+42\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)
b) \(20\left(\frac{x-2}{x+1}\right)^2-5\left(\frac{x+2}{x-1}\right)^2+48\cdot\frac{x^2-4}{x^2-1}=0\)
Cho biểu thức
A=\(\left[\frac{3\left(x+2\right)}{2\left(x^3+x^2+x+1\right)}+\frac{2x^2-x-10}{2\left(x^3-x^2+x-1\right)}\right]:\left[\frac{5}{x^2+1}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x+1\right)}\right]\)
Chứng minh:
a) \(\left(x^{50}+x^{10}+1\right)⋮\left(x^{20}+x^{10}+1\right)\)
b) \(\left(x^{10}-10x+9\right)⋮\left(x^2+1\right)\)
c) \(\left(x+1\right)^{4n+2}+\left(x-1\right)^{4n+2}⋮\left(x^2+1\right)\)
Bài 1:cho phương trình
a,left(x-1right)^3-xleft(x-1right)^25xleft(2-xright)-11left(x+2right)
b,dfrac{left(x+10right)left(x+4right)}{12}-dfrac{left(x+4right)left(2-xright)}{4}dfrac{left(x+10right)left(x-2right)}{3}
c,dfrac{2left(x-3right)}{7}+dfrac{x-5}{3}dfrac{13x+4}{21}
d,dfrac{2x-1}{5}-dfrac{x-2}{3}dfrac{x+7}{5}
e,left(x-2right)^3+left(3x-1right)left(3x+1right)left(x+1right)^3
Đọc tiếp
Bài 1:cho phương trình
a,\(\left(x-1\right)^3-x\left(x-1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)
b,\(\dfrac{\left(x+10\right)\left(x+4\right)}{12}-\dfrac{\left(x+4\right)\left(2-x\right)}{4}=\dfrac{\left(x+10\right)\left(x-2\right)}{3}\)
c,\(\dfrac{2\left(x-3\right)}{7}+\dfrac{x-5}{3}=\dfrac{13x+4}{21}\)
d,\(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{5}\)
e,\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
1) left(dfrac{-3}{4}right)^{3x+1}dfrac{81}{256} 6) left(8x-1right)^{2n-4}5^{2n-4}
2) 172.x^2-dfrac{7^9}{98^3}dfrac{1}{2^3} 7) left(dfrac{1}{2x}-5right)^{20}+left(y^2-dfrac{1}{4}right)^{10}0
3) left(x-dfrac{2}{9}right)^3left(dfrac{2}{3}right)^6
4) left(x+2right)^2+left(y-dfrac{1}{10}right)^20
5) left(x-7right)^{n+1}-left(x-7right)^{n+11}0
Giúp mk với!!...
Đọc tiếp
1) \(\left(\dfrac{-3}{4}\right)^{3x+1}=\dfrac{81}{256}\) 6) \(\left(8x-1\right)^{2n-4}=5^{2n-4}\)
2) \(172.x^2-\dfrac{7^9}{98^3}=\dfrac{1}{2^3}\) 7) \(\left(\dfrac{1}{2x}-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
3) \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
4) \(\left(x+2\right)^2+\left(y-\dfrac{1}{10}\right)^2=0\)
5) \(\left(x-7\right)^{n+1}-\left(x-7\right)^{n+11}=0\)
Giúp mk với!!!!!
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 ={\(\varnothing\)}
Bài 1: Rút gọn biểu thức
a. \(\left(x-3\right)\left(x+7\right)-\left(x+5\right)\left(x-1\right)\)
b. \(x^2\left(x-4\right)\left(x+4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
Bài 2: Tìm x
a. \(x^2-25-\left(x+5\right)=0\)
b. \(3x\left(x-2\right)-x+2=0\)
c. \(x\left(x-4\right)-2x+8=0\)
Tìm x, biết:
a) \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)
b) \(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)