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a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
`@` `\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`
a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)
=-4/3x^2+8/3-10/3
=-4/3x^2-2/3
d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)
\(=3x^2+9x+22+\dfrac{68}{x-3}\)
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
Bài 1:
a, \(\left(5x^3-4x\right):\left(-2x\right)\)
\(=5x^3:\left(-2x\right)+\left(-4x\right):\left(-2x\right)\)
\(=\dfrac{-5}{2}x^2+2\)
b, \(\left(-2x^5-4x^3+3x^2\right):2x^2\)
\(=-2x^5:2x^2+\left(-4x^3\right):2x^2+3x^2:2x^2\)
\(=-x^3-2x+\dfrac{3}{2}\)
c, \(\left(-5x^3+15x^2+18x\right):\left(-5x\right)\)
\(=-5x^3:\left(-5x\right)+15x^2:\left(-5x\right)+18x:\left(-5x\right)\)
\(=-x^2-3x-\dfrac{18}{5}\)
d, \(\left(-15x^6-24x^3\right):\left(-3x^2\right)\)
\(=-15x^6:\left(-3x^2\right)+\left(-24x^3\right):\left(-3x^2\right)\)
\(=5x^4+8x\)
Bài 2:
a, \(\left(x^2-2x+1\right):\left(x-1\right)\)
\(=\left(x^2-x-x+1\right):\left(x-1\right)\)
\(=\left[x\left(x-1\right)-\left(x-1\right)\right]:\left(x-1\right)\)
\(=\left(x-1\right)^2:\left(x-1\right)=x-1\)
b, \(\left(6x^3-2x^2-9x+3\right):\left(3x-1\right)\)
\(=\left[2x^2\left(3x-1\right)-3\left(3x-1\right)\right]:\left(3x-1\right)\)
\(=\left(3x-1\right)\left(2x^2-3\right):\left(3x-1\right)=2x^2-3\)
c, \(\left(x^3-4x^2-x+12\right):\left(x-3\right)\) (sửa đề)
\(=\left(x^3-3x^2-x^2+3x-4x+12\right):\left(x-3\right)\)
\(=\left[x^2\left(x-3\right)-x\left(x-3\right)-4\left(x-3\right)\right]:\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x-4\right):\left(x-3\right)=x^2-x-4\)
d, \(\left(4x^4+14x^3+21x-9\right):\left(2x^2-3\right)\)
\(=\left(4x^4-6x^2+14x^3-21x+6x^2-9+42x\right):\left(2x^2-3\right)\)
\(=\left[2x^2\left(2x^2-3\right)+7x\left(2x^2-3\right)+3\left(2x^2-3\right)+42x\right]:\left(2x^2-3\right)\)
\(=\left[\left(2x^2-3\right)\left(2x^2+7x+3\right)+42x\right]:\left(2x^2-3\right)\)
\(=2x^2+7x+3+42:\left(2x^2-3\right)\)
$\text{#}Toru$