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Theo đề bài ta có :
\(\frac{x\left(3-x\right)}{x+1}\cdot\left(x+\frac{\left(3-x\right)}{x+1}\right)=2\)
=> \(\frac{\left(3x-x^2\right)}{x+1}\cdot\frac{\left(3-x+x^2+x\right)}{x+1}=2\)
=> \(\left(3x-x^2\right)\left(x^2+3\right)=2\left(x+1\right)^2\)
=> \(3x^3+9x-x^4-3x^2=2x^2+4x+2\)
=> \(3x^3+\left(9x-4x\right)+\left(-3x^2-2x^2\right)-x^4-2=0\)
=> \(3x^3+5x-5x^2-x^4-2=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x^3-1\right)=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)+2\left(x-1\right)\left(x^2+x+1\right)=0\)
=> \(5x\left(1-x\right)+x^3\left(1-x\right)-2\left(1-x\right)\left(x^2+x+1\right)=0\)
=> \(\left(1-x\right)\left(5x+x^3-2x^2-2x-2\right)=0\)
=> \(\left(1-x\right)\left(3x+x^3-2x^2-2\right)=0\)
=> \(\left(1-x\right)\left(x^3-x^2-x^2+x+2x-2\right)=0\)
=> \(\left(1-x\right)\left(x^2\left(x-1\right)-x\left(x-1\right)+2\left(x-1\right)\right)=0\)
=> \(\left(1-x\right)\left(x-1\right)\left(x^2-x+2\right)=0\)
Ta Thấy :
\(\left(x^2-x+2\right)=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\)
=> \(\hept{\begin{cases}1-x=0\\x-1=0\end{cases}}\)
=> x = 1
1) \(2x\left(x-5\right)+\left(x-2\right)\left(x+3\right)=2x^2-10x+x^2+3x-2x-6=3x^2-9x-6\)
2) \(\left(2x-5\right)\left(1-x\right)-\left(x-3\right)\left(-2x\right)=2x-2x^2-5+5x+2x^2-6x=x-5\)
3) \(\left(4x-3\right)\left(4x-3\right)-\left(3x+2\right)\left(3x-2\right)=\left(4x-3\right)^2-9x^2+4=16x^2-24x+9-9x^2+4\)
\(=7x^2-24x+13\)
4) \(\left(2x-1\right)\left(2x+1\right)\left(2x+1\right)-4\left(x^2+1\right)=\left(2x-1\right)[\left(2x+1\right)^2]-4x^2-4\)
\(=\left(2x-1\right)\left(4x^2+4x+4\right)-4x^2-4=8x^3+8x^2+8x-4x^2-4x-4-4x^2-4=8x^3+4x-8\)
5) \(3x\left(2x-8\right)-\left(2-6x\right)\left(5+x\right)=6x^2-24x-10-2x+30x+6x^2=12x^2+4x-10\)
6) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+6x+12x-24+8=-16\)
7) \(\left(x+2\right)\left(x^2-2x+4\right)-x^2\left(x-2\right)-2x^2=x^3+8-x^3+2x^2-2x^2=8\)
mình cũng giải câu này rồi nhưng ko biết đúng ko
2)x+6/x-5 + x-5/x+6 = 2x2+23x+61/x2+x-30
dkxd:x khắc 5;x khác-6
mc:(x-5)(x+6)
2x2+2x+61 =2x2+23x+61
2x=23x
2x=0 suy ra x=0
23x=0 suyra x=0
s={0}
3)6/x-5 + x+2/x-8 = 18/9(x-5)(8-x) - 1
dkxd: x khác 5 ; x khác -8
mc(x-5)(x-8)
3x+x2-58 =36x-x2+264
3x-58=36x+264
3x-58=0 suy ra x=58/3
36x+264=0 suy ra x=-22/8
s={58/3;-22/3}
\(3-\left(x-1\right)=2-2\left(x-3\right)\)
\(3-x+1=2-2x+6\)
\(4-x=8-2x\)
\(4-x-8+2x=0\)
\(x-4=0\)
\(x=4\)
3-(x-1)=2-2(x-3)=>3-2=x-1-2(x-3)=>1=x-1-2x+6
=>1=-x+5=>-x=1-5=-4=>x=4
Chúc bạn học tốt nhớ k cho mik nha.
a, \(x\left(x+1\right)-x\left(x-5\right)=6\Leftrightarrow x^2+x-x^2+5x=6\)
\(\Leftrightarrow x=1\)
b, \(4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)
c, \(x^2-\frac{1}{4}=0\Leftrightarrow\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\Leftrightarrow x=\pm\frac{1}{2}\)
d, \(5x^2=20x\Leftrightarrow5x^2-20x=0\Leftrightarrow5x\left(x-4\right)=0\Leftrightarrow x=0;4\)
e, \(4x^2-9-x\left(2x-3\right)=0\Leftrightarrow4x^2-9-2x^2=3x\Leftrightarrow2x^2-9-3x=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{3}{2};3\)
f, \(4x^2-25=\left(2x-5\right)\left(2x+7\right)\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow-2\left(2x+5\right)=0\Leftrightarrow x=-\frac{5}{2}\)
a) x( x + 1 ) - x( x - 5 ) = 6
⇔ x2 + x - x2 + 5x = 6
⇔ 6x = 6
⇔ x = 1
b) 4x2 - 4x + 1 = 0
⇔ ( 2x - 1 )2 = 0
⇔ 2x - 1 = 0
⇔ x = 1/2
c) x2 - 1/4 = 0
⇔ ( x - 1/2 )( x + 1/2 ) = 0
⇔ \(\orbr{\begin{cases}x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}}\Leftrightarrow x=\pm\frac{1}{2}\)
d) 5x2 = 20x
⇔ 5x2 - 20x = 0
⇔ 5x( x - 4 ) = 0
⇔ \(\orbr{\begin{cases}5x=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
e) 4x2 - 9 - x( 2x - 3 ) = 0
⇔ ( 2x - 3 )( 2x + 3 ) - x( 2x - 3 ) = 0
⇔ ( 2x - 3 )( 2x + 3 - x ) = 0
⇔ ( 2x - 3 )( x + 3 ) = 0
⇔ \(\orbr{\begin{cases}2x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-3\end{cases}}\)
f) 4x2 - 25 = ( 2x - 5 )( 2x + 7 )
⇔ ( 2x - 5 )( 2x + 5 ) - ( 2x - 5 )( 2x + 7 ) = 0
⇔ ( 2x - 5 )( 2x + 5 - 2x - 7 ) = 0
⇔ ( 2x - 5 )(-2) = 0
⇔ 2x - 5 = 0
⇔ x = 5/2
`#3107.101107`
\(x(x+5)(x-5) - (x+2)(x^2-2x+4)=5\)
`<=> x(x^2 - 25) - (x^3 + 2^3) = 5`
`<=> x^3 - 25x - x^3 - 8 = 5`
`<=> -25x - 8 = 5`
`<=> -25x = 13`
`<=> x = -13/25`
Vậy, `x = -13/25`
_____
\((x+1)^3 - (x-1)^3 -6(x-1)^2 = -19\)
`<=> x^3 + 3x^2 + 3x + 1 - (x^3 - 3x^2 + 3x - 1) - 6(x^2 - 2x + 1) = -19`
`<=> x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 - 6x^2 + 12x - 6 = -19`
`<=> (x^3 - x^3) + (3x^2 + 3x^2 - 6x^2) + (3x - 3x + 12x) + (1 + 1 - 6) = -19`
`<=> 12x - 4 = -19`
`<=> 12x = -15`
`<=> x = -15/12 = -5/4`
Vậy, `x = -5/4.`
________
`@` Sử dụng các hđt:
`1)` `A^2 + B^2 = (A - B)(A + B)`
`2)` `A^3 + B^3 = (A + B)(A^2 - AB + B^2)`
`3)` `(A - B)^3 = A^3 - 3A^2B + 3AB^2 - B^3`
`4)` `(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3`
`5)` `(A - B)^2 = A^2 - 2AB + B^2.`
a: \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=5\)
=>\(x\left(x^2-25\right)-x^3-8=5\)
=>\(x^3-25x-x^3-8=5\)
=>-25x=13
=>\(x=-\dfrac{13}{25}\)
b: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)
=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-19\)
=>\(6x^2+2-6x^2+12x-6=-19\)
=>12x-4=-19
=>12x=-15
=>x=-5/4