a: =>2x+5=25
=>2x=20
hay x=10
c: =>3x+1=10
=>3x=9
hay x=3
b: \(\Leftrightarrow\sqrt{x-7}=-3\)(loại)
d: =>16-7x=121
=>7x=-105
hay x=-15
\(\sqrt{2x+5}=5\)
\(\sqrt{2x+5}=5^2\)
\(2x+5=25\)
`2x=25-5`
`2x=20`
`x=20:2`
`x=10`
Vậy.....
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\(\sqrt{x-7}+3=0\)
\(\sqrt{x-7}=0-3\)
\(\sqrt{x-7}=-3\)
Mà : \(\sqrt{x-7}\ge0\forall x\)
`=> x ∈ ∅`
Vậy không có giá trị nào thỏa mãn yêu cầu đề bài
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\(\sqrt{3x+1}=\sqrt{10}\)
\(3x+1=10\)
`3x=10-1`
`3x=9`
`x=9:3`
`x=3`
Vậy `x=3`
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\(\sqrt{16-7x}=11\)
\(\sqrt{16-7x}=11^2\)
\(16-7x=121\)
`7x=-105`
`=> x = -105 : 7`
`x=-15`
Vậy `x=-15`
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`#LeMichael`
