2x-1/x(x+1)=1/x
Chứng minh đẳng thức:
a, (x^2-2x/2x^2+8-2x^2/8-4x+2x^2-x^3)(1-1/x-2/x^2)=x+1/2x
b, [2/3x-2/x+1(x+1/3x-x-1)]:x-1/x=2x/x-1
c, [2/(x+1)^3(1/x+1)+1/x^2+2x+1(1/x^2+1)]:x-1/x^3=x/x-1
1) |2x - 1| = 5
2) |2x - 1| = |x + 5|
3) |3x + 1| = x - 2
4) |3 - 2x| = x + 2
5) |2x - 1| = 5 - x
6) |- 3x| = x - 2
7) |2 - 3x| = 2x + 1
8) |2x - 1| + |4x ^ 2 - 1| = 0
9) (2x + 5)/(x + 3) + 1 = 4/(x ^ 2 + 2x - 3) - (3x - 1)/(1 - x)
10) (x - 1)/(x + 3) - x/(x - 3) = (7x - 3)/(9 - x ^ 2)
11) 5 + 96/(x ^ 2 - 16) = (2x - 1)/(x + 4) + (3x - 1)/(x - 4)
12) (2x)/(2x - 1) + x/(2x + 1) = 1 + 4/((2x - 1)(2x + 1))
13) (x + 2)/(x - 2) - 1/x = 2/(x ^ 2 - 2x)
14) x/(2x - 6) + x/(2x + 2) = (2x + 4)/(x ^ 2 - 2x - 3)
thực hiện phép tính
a, 2x-1/2x+1 :(2x-1+2-4x/2x+1)
b,(1/1-x -1) : (x- 1-2x/1-x +1)
c,(1/x + x-2/x^2 -4 -2+x/x^2+2x):(x^2+2x+1)/x^2
d,{1/x^2 + 1/y^2 + 2/ x+y.(1/x + 1/y )} : x^3+y^3/x^2+y^2
cho M= (√x+1√2x+1+√2x+√x√2x−1−1)÷(1+√x√2x+1−√2x+√x√2x−1)
Bạn ghi lại đề đi bạn, khó nhìn quá
a, (2+x/2-x - 4x^2/x^2-4 - 2-x/2+x):x^2-3x/2x^2-x^3
a, 2x-1/2x+1 :(2x-1+2-4x/2x+1)
b,(1/1-x -1) : (x- 1-2x/1-x +1)
c,(1/x + x-2/x^2 -4 -2+x/x^2+2x):(x^2+2x+1)/x^2
d,{1/x^2 + 1/y^2 + 2/ x+y.(1/x + 1/y )} : x^3+y^3/x^2+y^2
đề là thực hiện phép tính
a: \(=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{2x^2-x^3}{x^2-3x}\)
\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(=\dfrac{-4x^2-8x}{x+2}\cdot\dfrac{-x}{x-3}\)
\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)
b: \(=\dfrac{2x-1}{2x+1}:\left(2x-1+\dfrac{2-4x}{2x+1}\right)\)
\(=\dfrac{2x-1}{2x+1}:\dfrac{4x^2-1+2-4x}{2x+1}\)
\(=\dfrac{2x-1}{4x^2-4x+1}=\dfrac{1}{2x-1}\)
c: \(=\left(\dfrac{1}{1-x}-1\right):\left(x+1-\dfrac{2x-1}{x-1}\right)\)
\(=\dfrac{1-1+x}{1-x}:\dfrac{x^2-1-2x+1}{x-1}\)
\(=\dfrac{-x}{x-1}\cdot\dfrac{x-1}{x\left(x-2\right)}=\dfrac{-1}{x-2}\)
Thực hiện phép trừ :
\(\dfrac{2x}{x-1}-\dfrac{x}{x-1}-\dfrac{1}{x-1}\)
Cách thực hiện nào sau đây sai ?
(A) \(\dfrac{2x}{x-1}-\dfrac{x}{x-1}-\dfrac{1}{x-1}=\left(\dfrac{2x}{x-1}-\dfrac{x}{x-1}\right)-\dfrac{1}{x-1}=......................\)
(B) \(\dfrac{2x}{x-1}-\dfrac{x}{x-1}-\dfrac{1}{x-1}=\dfrac{2x}{x-1}-\left(\dfrac{x}{x-1}-\dfrac{1}{x-1}=..............\right)\)
(C) \(\dfrac{2x}{x-1}-\dfrac{x}{x-1}-\dfrac{1}{x-1}=\dfrac{2x}{x-1}-\left(\dfrac{x}{x-1}+\dfrac{1}{x-1}\right)=........\)
(D) \(\dfrac{2x}{x-1}-\dfrac{x}{x-1}-\dfrac{1}{x-1}=\dfrac{2x}{x-1}+\dfrac{-x}{x-1}+\dfrac{-1}{x-1}=.........\)
Giải phương trình sau
C x-2/c+2+3/x-2=x^2-11/x^2-4
D 2/x+1-1/x-2=3x-11/(x+1)(x-2)
E x+3/x-2-1/x=2/x^2+2x
F x+2/x-2-1/x=2/x(x-2)
G 3x-1/x-1-2x+5/x-3=1
H 2x/2x-1+x/2x+1=1+4/(2x-1)(2x+1)
1.giải phương trình :
1)1 + 2/x-1 + 1/x+3=x^2+2x-7/x^2+2x-3
2)x/x^2+5x+6=2/x^2+3x+2 (x=3)
3)1/x^2+9x+20 - 1/x^2+8x+12=x^2-2x-33/x^2+8x+15 (x=-5,7)
4)x+5/3x-6 - 1/2=2x-3/2x-4 (x=25/7)
5)x-1/x^3+1 + 2x+3/x^2-x+1=2x+4/x+1 - 2(x=0)
1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)
ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
1) √(2x-1) <= 8-2x
2) √[(x+1)(4-x)] > x-2
3) √(x-2x^2+1) > 1-x
4) √(x+5) - √(x+4) > √(x+3)
5) √(5x-1) - √(x-1) > √(2x-4)
6) √(x+3) >= √(2x-8) + √(7-x)
7) √(x+2) - √(3-x) < √(5-2x)
8) √(x+1) > 3 - √(x+4)
9) √(5x-1) - √(4x-1)<= 3√x
10) { {√[2(x^2-16)]} / √(x-3) }+ √(x-3) > (7-x) / √(x-3)
Giúp mình 10 câu này với ạaa
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tính giá trị biểu thức a) (2x+3).(2x-3)-(2x+1)2 tại x=1/2 b) (x-2)2 - (x-1).(x+1)-x.(1-x) tai x=5 c) (2x-1)2 + (x+1)2 +2.(2x-1).(x-1) tại x= -2 d) (2x-1)2 -2.(2x+3).(2x+5) +(2x+5)2 tại x=2022