Tìm \(x\), biết :
a) 2(\(3x-2\))-3(\(x-2\))=-1
b) 2(\(3-3x^2\))\(\div3x\)(\(2x-1\))=9
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VT
0
16 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
26 tháng 10 2021
a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)
\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)
\(\Leftrightarrow-11x^2+23x=0\)
\(\Leftrightarrow x\left(-11x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)
NT
0
NH
11 tháng 11 2020
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
N
2 tháng 11 2018
\(x^2-3x+2.\left(x-3\right)=0\)
\(x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
\(x.\left(x-3\right)-3x+9=0\)
\(x.\left(x-3\right)-3.\left(x-3\right)=0\)
\(\left(x-3\right)^2=0=>x=3\)
2 tháng 11 2018
a,\(x^2-3x+2\left(x-3\right)=0.\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
8 tháng 10 2020
a) \(\left(x+3\right)^2-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-2x^2=54\)
=> x2 + 6x + 9 - x(9x2 + 6x + 1) + (2x)3 + 13 - 2x2 = 54
=> x2 + 6x + 9 - 9x3 - 6x2 - x + 8x3 + 1 - 2x2 = 54
=> (-9x3 + 8x3) + (x2 - 6x2 - 2x2) + (6x - x) + (9 + 1) = 54
=> -x3 - 7x2 + 5x + 10 = 54
=> -(x3 + 7x2 - 5x - 10) = 54
=> phương trình vô nghiệm
b) (x + 3)3 - (x - 3)(x2 + 3x + 9) + 6(x + 1)2 + 3x = -33
=> x3 + 9x2 + 27x + 27 - (x3 - 33) + 6(x2 + 2x + 1) + 3x = -33
=> x3 + 9x2 + 27x + 27 - x3 + 27 + 6x2 + 12x + 6 + 3x = -33
=> (x3 - x3) + (9x2 + 6x2) + (27x + 12x + 3x) + (27 + 27 + 6) = -33
=> 15x2 + 42x + 60 = -33
=> 15x2 + 42x + 60 + 33 = 0
=> 15x2 + 42x + 93 = 0
=> 3(5x2 + 14x + 31) = 0
=> 5x2 + 14x + 31 = 0
=> không tìm được x
22 tháng 9 2021
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
\(2\left(3x-2\right)-3\left(x-2\right)=-1\)
\(6x-4-3x+6=-1\)
\(3x+2=-1\)
\(3x=-1-2\)
\(3x=-3\)
\(x=-1\)
\(2\left(3-3x^2\right):3x\left(2x-1\right)=9\)
\(6-6x^2:6x^2-3x=9\)
\(6-x^2-3x=9\)
\(-x^2-3x+6=9\)
\(-x^2-3x=5\)
\(-x\left(x+3\right)=5\)
\(x=-5;x=2\)