Giải phương trình 20(\(\dfrac{x-2}{x-1}\))\(^2\)-5(\(\dfrac{x+2}{x-1}\))\(^2\)+48\(\dfrac{x^2-4}{x^2-1}\)=0 ta đc nghiệm x\(_1\)và x\(_2\)với x\(_1\)<x\(_2\) Tính 3x\(_1\)-x\(_2\)
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) \(20\left(\frac{x-2}{x+1}\right)^2-5\left(\frac{x+2}{x-1}\right)^2+48\cdot\frac{x^2-4}{x^2-1}=0\)
TÌm x:
( x + 2)4 + (x -4) 4 = 162
x2 - 5x - 21 = 0
x2 - (x+1)2 = 0
(x - 1).(x+2) - x-2= 0
x2 - 4x +3 = 0
3x ( x-4) +12x - 48 = 0
(2x-1).(5-3x) = (x+2).(5-3x)
3) \(\frac{1-x}{x+1}-\frac{3+2x}{x+1}=0\)
13) \(\frac{x+2}{x}-\frac{x^2+5x+4}{x\left(x+2\right)}=\frac{x}{x+2}\)
14) \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{20}{\left(x+1\right)\left(2-x\right)}\)
16) \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
17) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
18) \(\frac{x-1}{x}+\frac{1}{x+1}=\frac{2x-1}{2x^2+2}\)
19) \(\frac{2}{x+1}-\frac{3x+1}{\left(x+1\right)}=\frac{1}{\left(x+1\right)\left(x-2\right)}\)
20) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\)
Câu 1: Cho a, b là bình phương của 2 số nguyên lẻ liên tiếp. Chứng minh: ab – a – b + 1 chia hết 48
Câu 2: Tìm tất cả các số nguyên x y, thỏa mãn x > y > 0: x^3 + 7y = y^3 +7x
Câu 3: Giải phương trình : (8x – 4x^2 – 1)(x^2 + 2x + 1) = 4(x^2 + x + 1)
k, x3 - x2 - 17x - 15 = 0
l, x3 +4x2+x- 6=0
m, x4+2x3-13x2 -14x+ 24 =0
n, \(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
i, (x-4) (x-5) (x-6) (x-7) = 1680
p, \(\frac{1}{x^2-5x-6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)
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 ={\(\varnothing\)}
Câu 1 : Giải phương trình
a. 5(x-3)-4=2(x-1)
b. 5-(6-x)=4(3-2x)
c. (3x+5)(2x+1)=(6x-2)(x-3)
d. (x+2)2 + 2(x-4)=(x-4)(x-2)
Bài 2 : Giải phương trình
a) x/3 - 5x/6 - 15x/12 = x/4 - 5
b) 8x-3/4 - 3x-2/2 = 2x-1/2 + x+3/4
c) x-1/2 - x+1/15 - 2x-13/6 = 0
d) 3(3-x)/8 + 2(5-x)/3 = 1-x/2 - 2
e) 3(5x-2)/4 - 2 = 7x/3 - 5(x-7)
Bài 3 Giải phương trình
a) (5x-4)(4x+6)=0
b) (x-5)(3-2x)(3x+4)=0
c) (2x+1)(x2+2)=0
d) (8x-4)(x2+2x+2)=0
Bài 4 Giải phương trình
a) (x-2)(2x+3)=(x-1)(x-2)
b) (2x+5)(x-4)=(x-5)(4-x)
c) 9x2 -1 =(3x+1)(2x-3)
d) (x+2)2=9(x2-4x+4)
e)4(2x+7)2 -9(x+3)2 =0
Bài 5 Giải phương trình
a) (9x2 -4)(x+1)=(3x+2)(x2 -1)
b) (x-1)2 -1+x2 =(1-x)(x+3)
c) x4 +x3 3+x+1=0
Bài 1) giải phương trình sau a) 4x-20=0
b)3-2x = 3(x+1)-x-2
C) x+2/2008 + x+3/2007 + x+4/2006 + x+2028/6 = 0
d) 2x(x+3)+5(x+3)= 0
bài2 tìm điều kiện xác định của phương trình sau: 2 /x - 1 = 1 /x + l
bài6 giải phương trình
1/x-1 + 1/x-2 = 1/ x+2 + 1/x+1.