x+2/x-2 - 1/x = 2/x^2-2x
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: \(=\left(\dfrac{1}{1-x}-1\right):\left(x+1-\dfrac{2x-1}{x-1}\right)\)
\(=\dfrac{1-1+x}{1-x}:\dfrac{x^2-1-2x+1}{x-1}\)
\(=\dfrac{-x}{x-1}\cdot\dfrac{x-1}{x\left(x-2\right)}=\dfrac{-1}{x-2}\)
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.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: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
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)
giải phương trình:
a, 1+2x-5/6=3-x/4
b, x+3/x+1+x-2/x=2
c, x-2/x+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+2/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
h,2x/2x-1+x/2x+1=1+4/2x-1*2x+1
mình cần bài này gấp ạ
a, \(\frac{1+2x-5}{6}=\frac{3-x}{4}\)
\(\frac{4+8x-20}{24}=\frac{18-6x}{24}\)
\(-16-8x=18-6x\)
\(-16-8x-18+6x=0\)
\(-34-2x=0\)
\(2x=-34\Leftrightarrow x=-17\)
b, \(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)ĐKXĐ : x \(\ne\)-1 ; 0
\(\frac{x^2+3x}{x^2+x}+\frac{x^2-x-2}{x^2+x}=\frac{2x^2+2x}{x^2+x}\)
\(x^2+3x+x^2-x-2=2x^2+2x\)
\(2x^2+2x-2=2x^2+2x\)
\(2x^2+2x-2x^2-2x-2=0\)
\(-2\ne0\) Nên phuwong trình vô nghiệm. (xem lại hộ)
Quy đồng mẫu các phân thức:
1) 7x-1/2x^2+6x; 3-2x/x^2-9
2) 2x-1/x-x^2; x+1/2-4x+2x^2
3) x-1/x^3+1; 2x/x^2-x+1; 2/x+1
4) 7/5x; 4/x-2y; x-y/8y^2-2x^2
5) x/x^3-1; x+1/x^2-x; x-1/x^2+x+1
6) x/x^2-2ax+a^2; x+a/x^2-ax
Quy đồng mẫu các phân thức:
1) 7x-1/2x^2+6x; 3-2x/x^2-9
2) 2x-1/x-x^2; x+1/2-4x+2x^2
3) x-1/x^3+1; 2x/x^2-x+1; 2/x+1
4) 7/5x; 4/x-2y; x-y/8y^2-2x^2
5) x/x^3-1; x+1/x^2-x; x-1/x^2+x+1
6) x/x^2-2ax+a^2; x+a/x^2-ax
1)
\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}\)
\(\dfrac{3-2x}{x^2-9}=\dfrac{3-2x}{\left(x-3\right)\left(x+3\right)}\)
MTC: \(x\left(x-3\right)\left(x+3\right)\)
\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}=\dfrac{\left(x-3\right)\left(7x-12\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-12x-21x+36}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-33x+36}{x\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{3-2x}{x^2-9}=\dfrac{3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{ x\left(3-2x\right)}{x\left(x-3\right)\left(x+3\right)}\dfrac{3x-2x^2}{x\left(x-3\right)\left(x+3\right)}\)
2)
\(\dfrac{2x-1}{x-x^2}=\dfrac{2x-1}{x\left(1-x\right)}\)
\(\dfrac{x+1}{2-4x+2x^2}=\dfrac{x+1}{2\left(1-2x+x^2\right)}=\dfrac{x+1}{2\left(1-x\right)^2}\)
MTC: \(2x\left(1-x\right)^2\)
\(\dfrac{2x-1}{x-x^2}=\dfrac{2x-1}{x\left(1-x\right)}=\dfrac{2\left(1-x\right)\left(2x-1\right)}{2x\left(1-x\right)^2}=\dfrac{\left(2-2x\right)\left(2x-1\right)}{2x\left(1-x\right)^2}=\dfrac{4x-2-4x^2+2x}{2x\left(1-x\right)^2}=\dfrac{6x-2-4x^2}{2x\left(1-x\right)^2}\)
\(\dfrac{x+1}{2-4x+2x^2}=\dfrac{x+1}{2\left(1-2x+x^2\right)}=\dfrac{x+1}{2\left(1-x\right)^2}=\dfrac{ x\left(x+1\right)}{2x\left(1-x\right)^2}=\dfrac{x^2+x}{2x\left(1-x\right)^2}\)
Phần còn lại nhé :v
3.
\(x^3+1=\left(x+1\right)\left(x^2-x+1\right)\)
\(x^2-x+1=x^2-x+1\)
\(x+1=x+1\)
MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(\dfrac{x-1}{x^3+1}=\dfrac{x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\dfrac{2x}{x^2-x+1}=\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\dfrac{2}{x+1}=\dfrac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
4.
\(5x\)
\(x-2y=x-2y=-\left(2y-x\right)\)
\(8y^2-2x^2=2\left(4y^2-x^2\right)=2\left(2y-x\right)\left(2y+x\right)\)
MTC: \(-10x\left(2y-x\right)\left(2y+x\right)\)
\(\dfrac{7}{5x}=\dfrac{7\left(2y-x\right)\left(2y+x\right)-2}{5x\left(2y-x\right)\left(2y+x\right)-2}=\dfrac{-14\left(2y-x\right)\left(2y+x\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)
\(\dfrac{4}{x-2y}=\dfrac{4\left(2y-x\right)\left(2y+x\right)10x}{-\left(2y-x\right)\left(2y+x\right)10x}=\dfrac{40x\left(2y-x\right)\left(2y+x\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)
\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{\left(x-y\right)-5x}{2\left(2y-x\right)\left(2y+x\right)-5x}=\dfrac{-5x\left(x-y\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)
5.
\(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(x^2-x=x\left(x-1\right)\)
\(x^2+x+1\)
MTC: \(x\left(x-1\right)\left(x^2+x+1\right)\)
\(\dfrac{x}{x^3-1}=\dfrac{x.x}{\left(x-1\right)\left(x^2+x+1\right)x}=\dfrac{x^2}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{x+1}{x^2-x}=\dfrac{\left(x+1\right)\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{x-1}{x^2+x+1}=\dfrac{x\left(x-1\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x\left(x-1\right)^2}{x\left(x-1\right)\left(x^2+x+1\right)}\)
6.
\(x^2-2ax+a^2=\left(x-a\right)^2\)
\(x^2-ax=x\left(x-a\right)\)
MTC: \(x\left(x-a\right)^2\)
\(\dfrac{x}{x^2-2ax+a^2}=\dfrac{x.x}{\left(x-a\right)^2x}=\dfrac{x^2}{x\left(x-a\right)^2}\)
\(\dfrac{x+a}{x^2-ax}=\dfrac{\left(x+a\right)\left(x-a\right)}{x\left(x-a\right)\left(x-a\right)}=\dfrac{x^2-a^2}{x\left(x-a\right)^2}\)
a) (2x+1)^2-2x-1=2b) (x^2-3x)^2+5(x^2-3x)+6=0c) (x^2-x-1)(x^2-x)-2=0d) (5-2x)^2+4x-10=8e) (x^2+2x+3)(x^2+2x+1)=3f) x(x-1)(x^2-x+1)-6=0
sửa lại chút: a) (2x+1)^2-2x-1=2 b) (x^2-3x)^2+5(x^2-3x)+6=0 c) (x^2-x-1)(x^2-x)-2=0 d) (5-2x)^2+4x-10=8 e) (x^2+2x+3)(x^2+2x+1)=3 f) x(x-1)(x^2-x+1)-6=0
a) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1}{2};-1\right\}\)
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 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