3x.(8x - 4)-6x.(4x -3 )=30
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a)
3x(8x-4)-6x(4x-3)=30
=> 6x(4x-2)-6x(4x-3)=30
=> 6x(4x-2-4x+3)=30
=> 6x=30
=> x=5
b)
3x(5-2x)+2x(3x-5)=20
\(15x-6x^2+6x^2-10x=20\)
\(5x=20\)
\(x=4\)
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`@` `\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`
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a: =x^4-3x^5+4x^8
b: =2x^3+2x^2+4x
c: =4x^2+8x-5
d: =2x+3x^2+7x^4
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1) 4x-(2x-5)=21
4x-2x+5=21
2x+5=21
2x=21 -5
2x=16
x=16/2
x=8
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXD: x<>0
\(\dfrac{14x^3+12x^2-14x}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(\dfrac{2x\left(7x^2+6x-7\right)}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(7x^2+6x-7=3x^2-4x+6x-8\)
=>\(7x^2+6x-7=3x^2+2x-8\)
=>\(4x^2+4x+1=0\)
=>\(\left(2x+1\right)^2=0\)
=>2x+1=0
=>x=-1/2(nhận)
b: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)
=>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)
=>\(24x^2-26x-5-24x^2+23x+12=15\)
=>-3x+7=15
=>-3x=8
=>\(x=-\dfrac{8}{3}\)
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1) (x+1)2+2x=x(x+1)+6
⇔x2+2x+1+2x=x2+x+6
⇔x2+2x+1+2x-x2-x-6=0
⇔3x-5=0
⇔x=\(\frac{5}{3}\)
Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{5}{3}\)}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(8x^2+6x^3=2x^2\left(4+3x\right)\)
\(x^3-5x^2-4x+20=x^2\left(x-5\right)-4\left(x-5\right)=\left(x^2-4\right)\left(x-5\right)=\left(x-2\right)\left(x+2\right)\left(x-5\right)\)
\(x^2-9y^2-4x+4=\left(x^2-4x+4\right)-\left(3y\right)^2=\left(x-2\right)^2-\left(3y\right)^2=\left(x-2-3y\right)\left(x-2+3y\right)\)
a: \(8x^2+6x^3=2x^2\left(4+3x\right)\)
b: \(x^3-5x^2-4x+20\)
\(=x^2\left(x-5\right)-4\left(x-5\right)\)
\(=\left(x-5\right)\left(x-2\right)\left(x+2\right)\)
c: \(x^2-4x+4-9y^2\)
\(=\left(x-2\right)^2-9y^2\)
\(=\left(x-2-3y\right)\left(x-2+3y\right)\)
24x2-12x-24x2+18x=30
6x=30
x=5
Ta có \(3x\left(8x-4\right)-6x\left(4x-3\right)=30\)
\(\Rightarrow24x^2-12x-24x^2+18x=30\)
\(\Rightarrow6x=30\)
\(\Rightarrow x=\frac{30}{6}=5\)
Vậy phương trình có tập nghiệm S={5}