Đáp án+Giải thích các bước giải:
Bài `4:`
`a) \sqrt{2x+1} = 3 + \sqrt{5}(ĐK: x ≥ (-1)/2)`
`<=> 2x + 1 = 9 + 6\sqrt{5} + 5`
`<=> 2x + 1 = 14 + 6\sqrt{5}`
`<=> 2x = 15 + 6\sqrt{5}`
`<=> x = (15 + 6\sqrt{5})/2(tm)`
Vậy ` x = (15 + 6\sqrt{5})/2(tm)`
`b) \sqrt{\sqrt{3x}-5} = 3 - \sqrt{2}(ĐK: x ≥ (25)/3)`
`<=> \sqrt{3x}-5 = 9 - 6\sqrt{2} + 2`
`<=> \sqrt{3x}= 16 - 6\sqrt{2} `
`<=> 3x = 256 - 192\sqrt{2} + 72`
`<=> 3x = 328-192\sqrt{2}`
`<=> x = (328-192\sqrt{2})/3(tm)`
Vậy ` x = (328-192\sqrt{2})/3`
`c) \sqrt{5-4x} = \sqrt{2} + \sqrt{3}(ĐK: x ≤ 5/4)`
`<=> 5-4x = 2 + 2\sqrt{6} + 3`
`<=> 5 - 4x = 5 + 2\sqrt{6}`
`<=> 5-4x - 5 = 2\sqrt{6}`
`<=> -4x = 2\sqrt{6}`
`<=> x = -(\sqrt{6})/2`
`<=> x = -(\sqrt{3})/(\sqrt{2})`
Vậy `x =-(\sqrt{3})/(\sqrt{2})`
`d) \sqrt{4x + \sqrt{7}} = 1 - \sqrt{3}(` vô nghiệm vì `1 - \sqrt{3} < 0)`
`e) \sqrt{9x} + \sqrt{16x}=14(ĐK: x ≥0)`
`<=> 3\sqrt{x} + 4\sqrt{x} = 14`
`<=> 7\sqrt{x} = 14`
`<=> \sqrt{x} = 2`
`<=> x = 4(tm)`
Vậy `x = 4`
