Đáp án+Giải thích các bước giải:
a,
`\sqrt{x^2-4x+5}=x-1(x≥1)`
`⇒x^2-4x+5=x^2-2x+1`
`⇔-2x+4x=5-1`
`⇔2x=4`
`⇔x=2(tm)`
Vậy `S={2}`
b,
`\sqrt{2x^2-4x-1}=x-2(x≥\frac{2+\sqrt{6}}{2})`
`⇒2x^2-4x-1=x^2-4x+4`
`⇔2x^2-x^2=4+1`
`⇔x^2=5`
`x≥\frac{2+\sqrt{6}}{2}> -\sqrt{5}`
`⇒x=\sqrt{5}(tm)`
Vậy `S={\sqrt{5}}`
c,
`\sqrt{4x-8}+\sqrt{9x-18}-\sqrt{16x-32}=3(x≥2)`
`⇔\sqrt{4(x-2)}+\sqrt{9(x-2)}-\sqrt{16(x-2)}=3`
`⇔2\sqrt{x-2}+3\sqrt{x-2}-4\sqrt{x-2}=3`
`⇔\sqrt{x-2}=3`
`⇔x-2=9`
`⇔x=11(tm)`
Vậy `S={11}`
d,
`3\sqrt{2x-3}+2\sqrt{8x-12}=\sqrt{18x-27}+9(x≥\frac{3}{2})`
`⇔3\sqrt{2x-3}+2\sqrt{8x-12}-\sqrt{18x-27}=9`
`⇔3\sqrt{2x-3}+2\sqrt{4(2x-3)}-\sqrt{9(2x-3)}=9`
`⇔3\sqrt{2x-3}+4\sqrt{2x-3}-3\sqrt{2x-3}=9`
`⇔4\sqrt{2x-3}=9`
`⇔\sqrt{2x-3}=\frac{9}{4}`
`⇔2x-3=\frac{81}{16}`
`⇔2x=\frac{129}{16}`
`⇔x=\frac{129}{32}(tm)`
Vậy `S={\frac{129}{32}}`
e,
`\sqrt{9x-9}+\sqrt{4x-4}-\sqrt{16x-16}=\sqrt{5-x}(1≤x≤5)`
`⇔\sqrt{9(x-1)}+\sqrt{4(x-1)}-\sqrt{16(x-1)}=\sqrt{5-x}`
`⇔3\sqrt{x-1}+2\sqrt{x-1}-4\sqrt{x-1}=\sqrt{5-x}`
`⇔\sqrt{x-1}=\sqrt{5-x}`
`⇔x-1=5-x`
`⇔x+x=5+1`
`⇔2x=6`
`⇔x=3(tm)`
Vậy `S={3}`
f,
`\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}(x≥1)`
`⇔\frac{1}{3}\sqrt{x-1}+2\sqrt{4(x-1)}-12\sqrt{\frac{1}{25}(x-1)}=\frac{29}{15}`
`⇔\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}`
`⇔\frac{29}{15}\sqrt{x-1}=\frac{29}{15}`
`⇔\sqrt{x-1}=1`
`⇔x-1=1`
`⇔x=2(tm)`
Vậy `S={2}`
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