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`a)` `(x-2)^2-(x+3)(x-3)=1`
`=>x^2-4x+4-x^2+9=1`
`=>(x^2-x^2)-4x+(4+9)=1`
`=>-4x+13=1`
`=>-4x=1-13`
`=>-4x=-12`
`=>x=-12:(-4)`
`=>x=3`
Vậy `x=3`
`b)` `(x+2)^3-x^2(x+6)=4`
`=>x^3+6x^2+12x+8-x^3-6x^2=4`
`=>(x^3-x^3)+(6x^2-6x^2)+12x+8=4`
`=>12x=4-8`
`=>12x=-4`
`=>x=-4:12`
`=>x=(-1)/3`
Vậy `x=(-1)/3`
`c)` `(x+5)^2-(x+2)(x-3)=4`
`=>x^2+10x+25-(x^2-3x+2x-6)=4`
`=>x^2+10x+25-x^2+3x-2x+6=4`
`=>(x^2-x^2)+(10x+3x-2x)+(25+6)=4`
`=>11x+31=4`
`=>11x=4-31`
`=>11x=-27`
`=>x=(-27)/11`
Vậy `x=(-27)/11`
`d)` `(x+3)^2-(x+4)(x-1)=7`
`=>x^2+6x+9-(x^2-x+4x-4)=7`
`=>x^2+6x+9-x^2+x-4x+4=7`
`=>(x^2-x^2)+(6x+x-4x)+(9+4)=7`
`=>3x+13=7`
`=>3x=7-13`
`=>3x=-6`
`=>x=-6:3`
`=>x=-2`
Vậy `x=-2`
`e)` `(x-1)^2+(2x+1)^2-3(x-1)(1+x)=5`
`=>x^2-2x+1+4x^2+4x+1-3(x^2-1)=5`
`=>x^2-2x+1+4x^2+4x+1-3x^2+3=5`
`=>(x^2+4x^2-3x^2)-(2x-4x)+(1+1+3)=5`
`=>2x^2+2x+5=5`
`=>2x^2+2x+5-5=0`
`=>2x^2+2x=0`
`=>2x(x+1)=0`
`=>` \(\left[ \begin{array}{l}2x=0\\x+1=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
Vậy `x\in{0;-1}`
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`a)
`(x-2)^2-(x+3).(x-3)=1`
`=>x^2-4x+4-(x^2-3^2)=1`
`=>x^2-4x+4-x^2+9=1`
`=>(x^2-x^2)-4x=1-4-9`
`=>-4.x=-12`
`=>x=(-12):(-4)`
`=>x=3`
Vậy `x=3`
`b)`
`(x+2)^3-x^2.(x+6)=4`
`=>x^3+6x^2+12x+8-x^3-6x^2=4`
`=>(x^3-x^3)+(6x^2-6x^2)+12.x=4-8`
`=>12.x=-4`
`=>x=-1/3`
Vậy `x=-1/3`
`c)`
`(x+5)^2-(x+2).(x-3)=4`
`=>x^2+10x+25-(x^2-3x+2x-6)=4`
`=>x^2+10x+25-(x^2-x-6)=4`
`=>x^2+10x+25-x^2+x+6=4`
`=>(x^2-x^2)+(10x+x)=4-25-6`
`=>11.x=-27`
`=>x=-27/11`
Vậy `x=-27/11`
`d)`
`(x+3)^2-(x+4).(x-1)=7`
`=>x^2+6x+9-(x^2-x+4x-4)=7`
`=>x^2+6x+9-(x^2+3x-4)=7`
`=>x^2+6x+9-x^2-3x+4=7`
`=>(x^2-x^2)+(6x-3x)=7-9-4`
`=>3.x=-6`
`=>x=-2`
Vậy `x=-2`
`e)`
`(x-1)^2+(2x+1)^2-3.(x-1).(1+x)=5`
`=>x^2-2x+1+4x^2+4x+1-3.(x-1).(x+1)=5`
`=>x^2-2x+1+4x^2+4x+1-3.(x^2-1)=5`
`=>x^2-2x+1+4x^2+4x+1-3x^2+3=5`
`=>(x^2+4x^2-3x^2)+(-2x+4x)+(1+1+3-5)=0`
`=>2x^2+2x=0`
`=>2x.(x+1)=0`
`=>[(2x=0),(x+1=0):}`
`=>[(x=0),(x=-1):}`
Vậy `x in {0;-1}`
Hãy giúp mọi người biết câu trả lời này thế nào?
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