5. Tìm x, biết: a) 2x(x – 5) – x(3 + 2x) = 26 b) 3x(x + 2) – 2(x + 2) = 0 12/07/2021 Bởi Katherine 5. Tìm x, biết: a) 2x(x – 5) – x(3 + 2x) = 26 b) 3x(x + 2) – 2(x + 2) = 0
`\text{a)}` `2x(x-5) – x(3 +2x) = 26` `-> x(2x – 10) – x(3+2x) = 26` `-> x[ (2x-10) – (3+2x) ] = 26` `-> x [ (2x – 2x) + (-10-3) ] = 26` `-> x . (-13) = 26` `-> x = 26/{-13}` `-> x = -2` Vậy `x = -2` `\text{b)}` `3x(x+2) – 2(x+2) = 0` `-> (x+2)(3x-2) =0` `->` \(\left[ \begin{array}{l}x+2 =0\\3x -2 =0\end{array} \right.\) `->` \(\left[ \begin{array}{l}x=-2\\x = \dfrac{2}{3} \end{array} \right.\) Vậy `x \in {-2 ; 2/3}` Bình luận
Đáp án: `a)\ x=-2` `b)\ x∈{-2;(2)/(3)}` Giải thích các bước giải: `a)` `2x(x-5)-x(3+2x)=26` `->2x^{2}-10x-3x-2x^{2}=26` `->-13x=26` `->x=-2` `b)` `3x(x+2)-2(x+2)=0` `->(x+2)(3x-2)=0` `->` \(\left[ \begin{array}{l}x+2=0\\3x-2=0\end{array} \right.\) `->` \(\left[ \begin{array}{l}x=-2\\x=\dfrac{2}{3}\end{array} \right.\) Vậy `x∈{-2;(2)/(3)}` Bình luận
`\text{a)}`
`2x(x-5) – x(3 +2x) = 26`
`-> x(2x – 10) – x(3+2x) = 26`
`-> x[ (2x-10) – (3+2x) ] = 26`
`-> x [ (2x – 2x) + (-10-3) ] = 26`
`-> x . (-13) = 26`
`-> x = 26/{-13}`
`-> x = -2`
Vậy `x = -2`
`\text{b)}`
`3x(x+2) – 2(x+2) = 0`
`-> (x+2)(3x-2) =0`
`->` \(\left[ \begin{array}{l}x+2 =0\\3x -2 =0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=-2\\x = \dfrac{2}{3} \end{array} \right.\)
Vậy `x \in {-2 ; 2/3}`
Đáp án:
`a)\ x=-2`
`b)\ x∈{-2;(2)/(3)}`
Giải thích các bước giải:
`a)`
`2x(x-5)-x(3+2x)=26`
`->2x^{2}-10x-3x-2x^{2}=26`
`->-13x=26`
`->x=-2`
`b)`
`3x(x+2)-2(x+2)=0`
`->(x+2)(3x-2)=0`
`->` \(\left[ \begin{array}{l}x+2=0\\3x-2=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}x=-2\\x=\dfrac{2}{3}\end{array} \right.\)
Vậy `x∈{-2;(2)/(3)}`