tìm x,biết a,2x(x-3)-(3-x)=0 b,3x(x+5)-6(x+5)=0 c,x^4-x^2=0 07/07/2021 Bởi Julia tìm x,biết a,2x(x-3)-(3-x)=0 b,3x(x+5)-6(x+5)=0 c,x^4-x^2=0
Đáp án: `a)x=3` hoặc `x=-1/2` `b)x=-5` hoặc `x=2` `c)“x=0` hoặc `x=-1` hoặc `x=1` Giải thích các bước giải: `a)2x(x-3)-(3-x)=0` `⇔2x(x-3)+(x-3)=0` `⇔(x-3)(2x+1)=0` `⇔`\(\left[ \begin{array}{l}x-3=0\\2x+1=0\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=3\\x=\dfrac{-1}{2}\end{array} \right.\) Vậy `x=3` hoặc `x=-1/2` `b)3x(x+5)-6(x+5)=0` `⇔(x+5)(3x-6)=0` `⇔`\(\left[ \begin{array}{l}x+5=0\\3x-6=0\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.\) Vậy `x=-5` hoặc `x=2` `c)x^4-x²=0` `⇔x²(x²-1)=0` `⇔x²(x+1)(x-1)=0` `⇔`\(\left[ \begin{array}{l}x^2=0\\x+1=0\\x-1=0\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=0\\x=-1\\x=1\end{array} \right.\) Vậy `x=0` hoặc `x=-1` hoặc `x=1` Bình luận
`a,` `2x(x-3)-(3-x)=0` `<=>2x(x-3)+(x-3)=0` `<=>(x-3)(2x+1)=0` `<=>x-3=0` or `2x+1=0` `<=>x=3` or `x=-1/2` Vậy `S={3; -1/2}` `b,` `3x(x+5)-6(x+5)=0` `<=>(x+5)(3x-6)=0` `<=>x+5=0` or `3x-6=0` `<=>x=-5` or `x=2` Vậy `S={2;-5}` `c,` `x^4-x^2=0` `<=>(x^2)^2-x^2=0` `<=>(x^2-x)(x^2+x)=0` `<=>x^2-x=0` or `x^2+x=0` `<=>x(x-1)=0` or `x(x+1)=0` `<=>x=0; x=1` or `x=0; x=-1` Vậy `S={0;1;-1}` Bình luận
Đáp án:
`a)x=3` hoặc `x=-1/2`
`b)x=-5` hoặc `x=2`
`c)“x=0` hoặc `x=-1` hoặc `x=1`
Giải thích các bước giải:
`a)2x(x-3)-(3-x)=0`
`⇔2x(x-3)+(x-3)=0`
`⇔(x-3)(2x+1)=0`
`⇔`\(\left[ \begin{array}{l}x-3=0\\2x+1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3\\x=\dfrac{-1}{2}\end{array} \right.\)
Vậy `x=3` hoặc `x=-1/2`
`b)3x(x+5)-6(x+5)=0`
`⇔(x+5)(3x-6)=0`
`⇔`\(\left[ \begin{array}{l}x+5=0\\3x-6=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-5\\x=2\end{array} \right.\)
Vậy `x=-5` hoặc `x=2`
`c)x^4-x²=0`
`⇔x²(x²-1)=0`
`⇔x²(x+1)(x-1)=0`
`⇔`\(\left[ \begin{array}{l}x^2=0\\x+1=0\\x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=-1\\x=1\end{array} \right.\)
Vậy `x=0` hoặc `x=-1` hoặc `x=1`
`a,`
`2x(x-3)-(3-x)=0`
`<=>2x(x-3)+(x-3)=0`
`<=>(x-3)(2x+1)=0`
`<=>x-3=0` or `2x+1=0`
`<=>x=3` or `x=-1/2`
Vậy `S={3; -1/2}`
`b,`
`3x(x+5)-6(x+5)=0`
`<=>(x+5)(3x-6)=0`
`<=>x+5=0` or `3x-6=0`
`<=>x=-5` or `x=2`
Vậy `S={2;-5}`
`c,`
`x^4-x^2=0`
`<=>(x^2)^2-x^2=0`
`<=>(x^2-x)(x^2+x)=0`
`<=>x^2-x=0` or `x^2+x=0`
`<=>x(x-1)=0` or `x(x+1)=0`
`<=>x=0; x=1` or `x=0; x=-1`
Vậy `S={0;1;-1}`