a,x mũ 2 = x mũ 5 b, 5 mũ x + 5 mũ x + 2 = 650 c, 32 mũ x .16 mũ x = 1024 16/07/2021 Bởi Valentina a,x mũ 2 = x mũ 5 b, 5 mũ x + 5 mũ x + 2 = 650 c, 32 mũ x .16 mũ x = 1024
`a)x^2=x^5` `⇒x^2-x^5=0` `⇒x^2.1-x^2.x^3=0` `⇒x^2.(1-x^3)=0` Trường hợp 1: `x^2=0` `⇒x=0` Trường hợp 2: `1-x^3=0` `⇒-x^3=0-1` `⇒-x^3=-1` `⇒x^3=1` `x=0` Vậy x ∈ {0;1} `b)5^x+5^(x+2)=650` `⇒5^x.1+5^x+25=650` `⇒5^x.(1+25)=650` `⇒5^x=650:26` `⇒5^x=25` `⇒5^x=5^2` `⇒x=2` Vậy `x=2` `c)32^x.16^x=1024` `⇒(2^5)^x+(2^4)^x=1024` `⇒2^(5x)+2^(4x)=2^10` `⇒2^(5x+4x)=2^10` `⇒2^(9x)=2^10` `⇒9x=10` `⇒x=10/9` Vậy `x=10/9` Bình luận
Giải thích các bước giải: `a)``x^2=x^5``=>x^2-x^5=0``=>x^2 .1-x^2 .x^3=0``=>x^2 .(1-x^3)=0`TH`1`:`x^2=0``=>x=0`TH`2``1-x^3=0``=>x^3=1-0``=>x^3=1``=>x=1`Vậy `x\in{0;1}``b)``5^x +5^(x+2)=650``=>5^x .1+5^x .25=650``=>5^x .(1+25)=650``=>5^x .26=650``=>5^x =650:26``=>5^x=25``=>5^x=5^2``=>x=2`Vậy `x=2``c)``32^x .16^x=1024``=>(2^5)^x .(2^4)^x=2^10``=>2^(5x) .2^(4x)=2^10``=>2^(5x+4x)=2^10``=>2^(9x)=2^10``=>9x=10``=>x=10/9`Vậy `x=10/9` Bình luận
`a)x^2=x^5`
`⇒x^2-x^5=0`
`⇒x^2.1-x^2.x^3=0`
`⇒x^2.(1-x^3)=0`
Trường hợp 1:
`x^2=0`
`⇒x=0`
Trường hợp 2:
`1-x^3=0`
`⇒-x^3=0-1`
`⇒-x^3=-1`
`⇒x^3=1`
`x=0`
Vậy x ∈ {0;1}
`b)5^x+5^(x+2)=650`
`⇒5^x.1+5^x+25=650`
`⇒5^x.(1+25)=650`
`⇒5^x=650:26`
`⇒5^x=25`
`⇒5^x=5^2`
`⇒x=2`
Vậy `x=2`
`c)32^x.16^x=1024`
`⇒(2^5)^x+(2^4)^x=1024`
`⇒2^(5x)+2^(4x)=2^10`
`⇒2^(5x+4x)=2^10`
`⇒2^(9x)=2^10`
`⇒9x=10`
`⇒x=10/9`
Vậy `x=10/9`
Giải thích các bước giải:
`a)`
`x^2=x^5`
`=>x^2-x^5=0`
`=>x^2 .1-x^2 .x^3=0`
`=>x^2 .(1-x^3)=0`
TH`1`:
`x^2=0`
`=>x=0`
TH`2`
`1-x^3=0`
`=>x^3=1-0`
`=>x^3=1`
`=>x=1`
Vậy `x\in{0;1}`
`b)`
`5^x +5^(x+2)=650`
`=>5^x .1+5^x .25=650`
`=>5^x .(1+25)=650`
`=>5^x .26=650`
`=>5^x =650:26`
`=>5^x=25`
`=>5^x=5^2`
`=>x=2`
Vậy `x=2`
`c)`
`32^x .16^x=1024`
`=>(2^5)^x .(2^4)^x=2^10`
`=>2^(5x) .2^(4x)=2^10`
`=>2^(5x+4x)=2^10`
`=>2^(9x)=2^10`
`=>9x=10`
`=>x=10/9`
Vậy `x=10/9`