(x mu 2 -81)(x mu 2-36)=0; 15-(x+30)=16-(-x+5)-x 09/11/2021 Bởi Kennedy (x mu 2 -81)(x mu 2-36)=0; 15-(x+30)=16-(-x+5)-x
`(x^2-81)(x^2-36)=0` `<=>` \(\left[ \begin{array}{l}x^2-81=0\\x^2-36=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=±9\\x=±6\end{array} \right.\) Vậy `x∈{9;-9;6;-6}` `b)` `15-(x+30)=16-(-x+5)-x` `<=>15-x-30=16+x-5-x` `<=>-15-x=11` `<=>-x=11+15` `<=>-x=26` `<=>x=-26` Vậy `x=-26` Bình luận
`a)(x^2-81)(x^2-36)=0` `→` \(\left[ \begin{array}{l}x^2-81=0\\x^2-36=0\end{array} \right.\) `→` \(\left[ \begin{array}{l}x^2=81\\x^2=36\end{array} \right.\) `→` \(\left[ \begin{array}{l}x=±9\\x=±6\end{array} \right.\) Vậy `x∈{±9;±6}` `b)15-(x+30)=16-(-x+5)-x` `→15-x-30=16+x-5-x` `→-x-x+x=16-5-15+30` `→-x=26` `→x=-26` Vậy `x=-26` Bình luận
`(x^2-81)(x^2-36)=0`
`<=>` \(\left[ \begin{array}{l}x^2-81=0\\x^2-36=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=±9\\x=±6\end{array} \right.\)
Vậy `x∈{9;-9;6;-6}`
`b)` `15-(x+30)=16-(-x+5)-x`
`<=>15-x-30=16+x-5-x`
`<=>-15-x=11`
`<=>-x=11+15`
`<=>-x=26`
`<=>x=-26`
Vậy `x=-26`
`a)(x^2-81)(x^2-36)=0`
`→` \(\left[ \begin{array}{l}x^2-81=0\\x^2-36=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x^2=81\\x^2=36\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=±9\\x=±6\end{array} \right.\)
Vậy `x∈{±9;±6}`
`b)15-(x+30)=16-(-x+5)-x`
`→15-x-30=16+x-5-x`
`→-x-x+x=16-5-15+30`
`→-x=26`
`→x=-26`
Vậy `x=-26`