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HT
2
BN
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`
12 tháng 8 2021
\(a/\)
\(4x-4y+x^2-2xy+y^2\)
\(=\left(4x-4y\right)+\left(x^2-2xy+y^2\right)\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+x-y\right)\)
\(b/\)
\(x^4-4x^3-8x^2+8x\)
\(=\left(x^4+8x\right)-\left(4x^3+8x^2\right)\)
\(=x\left(x^3+8\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4-4x\right)\)
\(=x\left(x+2\right)\left(x^2-6x-4\right)\)
\(d/\)
\(x^4-x^2+2x-1\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2+x-1\right)\left(x^2-x+1\right)\)
\(e/\)(Xem lại đề)
\(x^4+x^3+x^2+2x+1\)
\(=\left(x^4+x^3\right)+\left(x^2+2x+1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^3+x+1\right)\)
\(f/\)
\(x^3-4x^2+4x-1\)
\(=x\left(x^2-4x+4\right)-1^2\)
\(=x\left(x-2\right)^2-1\)
\(=[\sqrt{x}\left(x-2\right)]^2-1\)
\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)
12 tháng 8 2021
\(c/\)
\(x^3+x^2-4x-4\)
\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(2x-4\right)\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+2\right)\)
\(=\left(x-2\right)[\left(x^2+x\right)+\left(2x+2\right)]\)
\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
12 tháng 3 2022
a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
NT
1