|2x+1|=x-1
Những câu hỏi liên quan
cho M= (√x+1√2x+1+√2x+√x√2x−1−1)÷(1+√x√2x+1−√2x+√x√2x−1)
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1) |2x - 1| 52) |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)
Đọc tiếp
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)
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
Rút gọn :
1. (2x-5)(3x+1)-(x-3)^2+(2x+5)^2-(3x+1)^3
2. (2x-1)(2x+1)-3x-2)(2x+3)-(x-1)^3+(2x+3)^3
3. (x-2)(x^2+2x+4)-(3x-2)^3+(3x-4)^2
4. (7x-1)(8x+2)-(2x-7)^2-(x-4)^3-(3x+1)^3
5. (5x-1)(5x+1)-(x+3)(x^2-3x+9)-(2x+4)^2-(3x-4)^2+(2x-5)^3
6. (4x-1)(x+2)-(2x+5)^2-(3x-7)^2+(2x+3)^3=(3x-1)^3
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
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(\(\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}\)-\(\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\)-1):(1+\(\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}\)_\(\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\))
b1:
[4+2x]=-3x [3x-1]+2=x
[x+15]+1=3x [2x-5]+x=2
b2:
[2x-5]=x+1 [3x-2]-1=x
[3x-7]=2x+1 [2x-1]+1=x
Pleft(frac{sqrt{x}+1}{sqrt{2x}+1}+frac{sqrt{2x}+sqrt{x}}{sqrt{2x}-1}-1right):left(1+frac{sqrt{x}+1}{sqrt{2x}+1}-frac{sqrt{2x}-sqrt{x}}{sqrt{2x}-1}right)frac{left(sqrt{x}+1right)left(sqrt{2x}-1right)}{left(sqrt{2x}+1right)left(sqrt{2x}-1right)}+frac{left(sqrt{2x}+sqrt{x}right)left(sqrt{2x}+1right)}{MTC}-frac{2x-1}{MTC}frac{xsqrt{2}-sqrt{x}+sqrt{2x}-1+2x+sqrt{2x}+xsqrt{2}+sqrt{x}-2x+1}{MTC}frac{2xsqrt{2}+2sqrt{2x}}{MTC}
Đọc tiếp
P=\(\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}-\sqrt{x}}{\sqrt{2x}-1}\right)\)
=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)}{\left(\sqrt{2x}+1\right)\left(\sqrt{2x}-1\right)}+\frac{\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}+1\right)}{MTC}-\frac{2x-1}{MTC}\)
=\(\frac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}-2x+1}{MTC}\)
=\(\frac{2x\sqrt{2}+2\sqrt{2x}}{MTC}\)
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
tính giá trị biểu thức a) (2x+3).(2x-3)-(2x+1)2 tại x1/2 b) (x-2)2 - (x-1).(x+1)-x.(1-x) tai x5 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 x2022
Đọc tiếp
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
giải phương trình sau
1/ ( x-1) (2x+1) =0
2/ x (2x-1) (3x+15) =0
3/ (2x-6) (3x+4) x=0
4/ (2x-10)(x^2+1)=0
5/ (x^2+3) (2x-1) =0
6/ (3x-1) (2x^2 +1)=0
1/ ( x-1) (2x+1) =0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-0,5\end{matrix}\right.\)
2/ x (2x-1) (3x+15) =0
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-5\end{matrix}\right.\)
3/ (2x-6) (3x+4).x=0
\(\Rightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)
4/ (2x-10)(x2+1)=0
\(\Rightarrow\left[{}\begin{matrix}2x-10=0\\x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x^2=-1\left(loại\right)\end{matrix}\right.\)
5/ (x2+3) (2x-1) =0
\(\Rightarrow\left[{}\begin{matrix}x^2+3=0\\2x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x^2=-3\left(loại\right)\\x=0,5\end{matrix}\right.\)
6/ (3x-1) (2x2 +1)=0
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\2x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x^2=-0,5\left(loại\right)\end{matrix}\right.\)
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1: Ta có: \(\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
2: Ta có: \(x\left(2x-1\right)\left(3x+15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-5\end{matrix}\right.\)
3: Ta có: \(\left(2x-6\right)\left(3x+4\right)x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)
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4: Ta có: \(\left(2x-10\right)\left(x^2+1\right)=0\)
mà \(x^2+1>0\forall x\)
nên 2x-10=0
hay x=5
5: Ta có: \(\left(x^2+3\right)\left(2x-1\right)=0\)
mà \(x^2+3>0\forall x\)
nên 2x-1=0
hay \(x=\dfrac{1}{2}\)
6: Ta có: \(\left(3x-1\right)\left(2x^2+1\right)=0\)
mà \(2x^2+1>0\forall x\)
nên 3x-1=0
hay \(x=\dfrac{1}{3}\)
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