Toán a (2x-3).(6-2x)=0 b (x+1 phần 2 ) .(2-3x)=0 c 2x(x^2+1).(x^2-9)=0 21/09/2021 By Bella a (2x-3).(6-2x)=0 b (x+1 phần 2 ) .(2-3x)=0 c 2x(x^2+1).(x^2-9)=0
$\text {a, (2x-3)(6-2x) =0}$ $\text {⇔\(\left[ \begin{array}{l}2x-3 =0\\6-2x =0\end{array} \right.\) }$ $\text {⇔\(\left[ \begin{array}{l}2x=3\\2x=6\end{array} \right.\) }$ $\text {⇔\(\left[ \begin{array}{l}x=\frac{3}{2} \\x=3\end{array} \right.\) }$ $\text {Vậy x=$\frac{3}{2}$ hoặc x=3}$ $\text {b, (x+$\frac{1}{2}$)(2-3x) =0 }$ $\text {⇔\(\left[ \begin{array}{l}x+\frac{1}{2}= 0 \\2-3x =0\end{array} \right.\) }$ $\text {⇔\(\left[ \begin{array}{l}x=\frac{-1}{2} \\3x=2\end{array} \right.\) }$ $\text {⇔\(\left[ \begin{array}{l}x=\frac{-1}{2} \\x=\frac{2}{3} \end{array} \right.\) }$ $\text {Vậy x=$\frac{-1}{2}$ hoặc x=$\frac{2}{3}$ }$ $\text {c, 2x(x²+1)(x²-9) =0}$ $\text {⇔\(\left[ \begin{array}{l}2x =0\\x²+1 =0\\x²-9 =0\end{array} \right.\) }$ $\text {⇔\(\left[ \begin{array}{l}x =0\\x²+1 =0 (vô nghiệm)\\x²=9\end{array} \right.\) }$ $\text {⇔\(\left[ \begin{array}{l}x =0\\x=3\\x=-3\end{array} \right.\) }$ $\text {Vậy x=0 hoặc x=3 hoặc x=-3}$ $\text {Chúc bạn học tốt~}$ $\text {@lamtung2}$ Trả lời
Giải thích các bước giải: `a)``(2x-3).(6-2x)=0`TH`1``2x-3=0``=>2x=0+3``=>2x=3``=>x=3:2``=>x=3/2`TH`2``6-2x=0``=>2x=6-0``=>2x=6``=>x=6:2``=>x=3`Vậy `x\in{3/2;3}``b (x+1 /2 ) .(2-3x)=0`TH`1``x+1/2=0``=>x=0-1/2``=>x=-1/2`TH`2``2-3x=0``=>3x=2-0``=>3x=2``=>x=2:3``=>x=2/3`Vậy `x\in{-1/2;2/3}``c)``2x(x^2+1).(x^2-9)=0`TH`1``2x=0``=>x=0`TH`2``x^2+1=0``=>x^2=0-1``=>x^2=-1`Vì `x^2ge0` mọi `x``=>x` thuộc rỗngTH`3``x^2-9=0``=>x^2=0+9``=>x^2=9``=>x^2=(+-3)^2``=>x=+-3`Vậy `x\in{0;+-3}` Trả lời
$\text {a, (2x-3)(6-2x) =0}$
$\text {⇔\(\left[ \begin{array}{l}2x-3 =0\\6-2x =0\end{array} \right.\) }$
$\text {⇔\(\left[ \begin{array}{l}2x=3\\2x=6\end{array} \right.\) }$
$\text {⇔\(\left[ \begin{array}{l}x=\frac{3}{2} \\x=3\end{array} \right.\) }$
$\text {Vậy x=$\frac{3}{2}$ hoặc x=3}$
$\text {b, (x+$\frac{1}{2}$)(2-3x) =0 }$
$\text {⇔\(\left[ \begin{array}{l}x+\frac{1}{2}= 0 \\2-3x =0\end{array} \right.\) }$
$\text {⇔\(\left[ \begin{array}{l}x=\frac{-1}{2} \\3x=2\end{array} \right.\) }$
$\text {⇔\(\left[ \begin{array}{l}x=\frac{-1}{2} \\x=\frac{2}{3} \end{array} \right.\) }$
$\text {Vậy x=$\frac{-1}{2}$ hoặc x=$\frac{2}{3}$ }$
$\text {c, 2x(x²+1)(x²-9) =0}$
$\text {⇔\(\left[ \begin{array}{l}2x =0\\x²+1 =0\\x²-9 =0\end{array} \right.\) }$
$\text {⇔\(\left[ \begin{array}{l}x =0\\x²+1 =0 (vô nghiệm)\\x²=9\end{array} \right.\) }$
$\text {⇔\(\left[ \begin{array}{l}x =0\\x=3\\x=-3\end{array} \right.\) }$
$\text {Vậy x=0 hoặc x=3 hoặc x=-3}$
$\text {Chúc bạn học tốt~}$
$\text {@lamtung2}$
Giải thích các bước giải:
`a)`
`(2x-3).(6-2x)=0`
TH`1`
`2x-3=0`
`=>2x=0+3`
`=>2x=3`
`=>x=3:2`
`=>x=3/2`
TH`2`
`6-2x=0`
`=>2x=6-0`
`=>2x=6`
`=>x=6:2`
`=>x=3`
Vậy `x\in{3/2;3}`
`b (x+1 /2 ) .(2-3x)=0`
TH`1`
`x+1/2=0`
`=>x=0-1/2`
`=>x=-1/2`
TH`2`
`2-3x=0`
`=>3x=2-0`
`=>3x=2`
`=>x=2:3`
`=>x=2/3`
Vậy `x\in{-1/2;2/3}`
`c)`
`2x(x^2+1).(x^2-9)=0`
TH`1`
`2x=0`
`=>x=0`
TH`2`
`x^2+1=0`
`=>x^2=0-1`
`=>x^2=-1`
Vì `x^2ge0` mọi `x`
`=>x` thuộc rỗng
TH`3`
`x^2-9=0`
`=>x^2=0+9`
`=>x^2=9`
`=>x^2=(+-3)^2`
`=>x=+-3`
Vậy `x\in{0;+-3}`