giúp mk vs ạ, mk đang cần gấp.
giải các phương trình sau:
a, 1/x+1 - x-1/x = 3x+1/x(x+1)
b, (x+2)^2/2x-3 - 1/1= x^2+10/2x-3
làm nhanh mk tick ạ.
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12 tháng 2 2018
a, (x+3)*(y+2)=1
=> x+3 và y+2 là ước của 1
Ta có bảng sau:
| x+3 | -1 | 1 |
| x | -4 | 2 |
| y+2 | -1 | 1 |
| y | -3 | 1 |
Vậy...
28 tháng 8 2020
Ít thôi -..-
a) ( 3x + 2 )( 2x + 9 ) - ( x + 3 )( 6x + 1 ) = ( x + 1 )2 - ( x + 2 )( x - 2 )
<=> 6x2 + 31x + 18 - ( 6x2 + 19x + 3 ) = x2 + 2x + 1 - ( x2 - 4 )
<=> 6x2 + 31x + 18 - 6x2 - 19x - 3 = x2 + 2x + 1 - x2 + 4
<=> 12x + 15 = 2x + 5
<=> 12x - 2x = 5 - 15
<=> 10x = -10
<=> x = -1
b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )
<=> 2x2 - 5x - 12 + x2 - 7x + 10 = 3x2 - 17x + 20
<=> 3x2 - 12x - 2 = 3x2 - 17x + 20
<=> 3x2 - 12x - 3x2 + 17x = 20 + 2
<=> 5x = 22
<=> x = 22/5
c) ( x + 2 )3 - ( x - 2 )3 - 12x( x - 1 ) = -8
<=> x3 + 6x2 + 12x + 8 - ( x3 - 6x2 + 12x - 8 ) - 12x2 + 12x = -8
<=> x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
<=> 12x + 16 = -8
<=> 12x = -24
<=> x = -2
d) ( 3x - 1 )2 - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )2 = 16
<=> 9x2 - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x2 - 2x + 1 ) = 16
<=> 9x2 - 11x - 3 - x2 + 2x - 1 = 16
<=> 8x2 - 9x - 4 = 16
<=> 8x2 - 9x - 4 - 16 = 0
<=> 8x2 - 9x - 20 = 0
( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm
2 là nghiệm vô tỉ =) )
28 tháng 8 2020
a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)2 - (x + 2)(x - 2)
=> 3x(2x + 9) + 2(2x + 9) - x(6x + 1) - 3(6x + 1) = x2 + 2x + 1 - x(x - 2) - 2(x - 2)
=> 6x2 + 27x + 4x + 18 - 6x2 - x - 18x - 3 = x2 + 2x + 1 - x2 + 2x - 2x + 4
=> (6x2 - 6x2) + (27x + 4x - x - 18x) + (18 - 3) = (x2 - x2) + (2x + 2x - 2x) + (1 + 4)
=> 12x + 15 = 2x + 5
=> 12x + 15 - 2x - 5 = 0
=> 10x + 10 = 0
=> 10x = -10 => x = -1
b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)
=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 = 3x2 - 12x - 5x + 20
=> (2x2 + x2) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x2 - 17x + 20
=> 3x2 - 12x - 2 = 3x2 - 17x + 20
=> 3x2 - 12x - 2 - 3x2 + 17x - 20 = 0
=> (3x2 - 3x2) + (-12x + 17x) + (-2 - 20) = 0
=> 5x - 22 = 0
=> 5x = 22 => x = 22/5
c) (x + 2)3 - (x - 2)3 - 12x(x - 1) = -8
=> x3 + 6x2 + 12x + 8 - (x3 - 6x2 + 12x - 8) - 12x2 + 12x = -8
=> x3 + 6x2 + 12x + 8 -x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
=> (x3 - x3) + (6x2 + 6x2 - 12x2) + (12x - 12x + 12x) + (8 + 8) = -8
=> 12x + 16 = -8
=> 12x = -24
=> x = -2
Còn bài cuối làm nốt
20 tháng 8 2021
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
BA
30 tháng 7 2019
\(\left|2x-\frac{1}{2}\right|+1=3x\)
\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=3x-1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{2}=3x-1\\2x-\frac{1}{2}=1-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1+\frac{1}{2}\\2x+3x=1+\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)
18 tháng 9 2021
a) \(2x^2-2x-x^2+6=0\)
\(\Leftrightarrow x^2-2x+1+5=0\)
\(\Leftrightarrow\left(x-1\right)^2=-5\) ( vô lý)
Vậy không có x thoả mãn \(2x.\left(x-1\right)-x^2+6=0\)
b) \(x^4-2x^2.\left(3+2x^2\right)+3x^2.\left(x^2+1\right)=-3\)
\(\Leftrightarrow x^4-6x^2-4x^4+3x^4+3x^2+3=0\)
\(\Leftrightarrow3-3x^2=0\)
\(\Leftrightarrow3x^2=3\Leftrightarrow x^2=1\) \(\Leftrightarrow x\in\left\{-1;1\right\}\)
Vậy \(x\in\left\{-1;1\right\}\)
c) \(\left(x+1\right).\left(x^2-x+1\right)-2x=x.\left(x-2\right).\left(x+2\right)\)
\(\Leftrightarrow x^3+1-2x-x.\left(x^2-4\right)=0\)
\(\Leftrightarrow x^3+1-2x-x^3+4x=0\)
\(\Leftrightarrow1+2x=0\Leftrightarrow x=\dfrac{-1}{2}\)
Vậy x=\(\dfrac{-1}{2}\)
d) \(\left(x+3\right).\left(x^2-3x+9\right)-x.\left(x-2\right).\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x.\left(x^2-4\right)-15=0\)
\(\Leftrightarrow x^3-27-x^3+4x-15=0\)
\(\Leftrightarrow4x-42=0\)
\(\Leftrightarrow x=10,5\)
Vậy x=10,5
20 tháng 7 2020
Bài 2: Tìm x
a) Ta có: (x-2)(x-1)=x(2x+1)+2
\(\Leftrightarrow x^2-3x+2=2x^2+x+2\)
\(\Leftrightarrow x^2-3x+2-2x^2-x-2=0\)
\(\Leftrightarrow-x^2-4x=0\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy: S={0;-4}
b) Ta có: \(\left(x+2\right)\left(x+2\right)-\left(x-2\right)\left(x-2\right)=8x\)
\(\Leftrightarrow x^2+4x+4-\left(x^2-4x+4\right)-8x=0\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4-8x=0\)
\(\Leftrightarrow0x=0\)
Vậy: S={x|\(x\in R\)}
c) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-3x^2+3x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x-3=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
d) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x+20=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\frac{10}{3}\)
Vậy: \(S=\left\{-\frac{10}{3}\right\}\)
e) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow x^3+5x^2+3x^2+2x+10-x^3-8x^2=27\)
\(\Leftrightarrow2x=27-10=17\)
hay \(x=\frac{17}{2}\)
Vậy: \(S=\left\{\frac{17}{2}\right\}\)
\(a)\)
\(\frac{1}{x+1}-\frac{x-1}{x}=\frac{3x+1}{x\left(x+1\right)}\)
\(\Leftrightarrow x-x^2+1=3x+1\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b)\)
\(\frac{\left(x+2\right)^2}{2x-3}-\frac{1}{1}=\frac{x^2+10}{2x-3}\)
\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)
\(\Leftrightarrow x^2+2x+1=x^2+10\)
\(\Leftrightarrow2x-9=0\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{2}{9}\)