Đáp án:
Giải thích các bước giải:
`1)\sqrt{2x-1}=3(x>=1/2)`
`=>2x-1=9`
`=>2x=10`
`=>x=5(tm)`
Vậy `x=5`
`2)\sqrt{5x+9}-4=3(x>=-9/5)`
`=>\sqrt{5x+9}=7`
`=>5x+9=49`
`=>5x=40`
`=>x=8(tm)`
Vậy `x=8`
`3)2\sqrt{2x-3}=2(x>=3/2)`
`=>\sqrt{2x-3}=1`
`=>2x-3=1`
`=>2x=4`
`=>x=2(tm)`
Vậy `x=2`
`4)2+\sqrt{x-1}=3(x>=1)`
`=>\sqrt{x-1}=1`
`=>x-1=1`
`=>x=2`
Vậy `x=2`
`5)3+\sqrt{7x+4}=8(x>=-4/7)`
`=>\sqrt{7x+4}=5`
`=>7x+4=25`
`=>7x=21`
`=>x=3(tm)`
Vậy `x=3`
`6)11-\sqrt{2x-5}=8(x>=5/2)`
`=>\sqrt{2x-5}=3`
`=>2x-5=9`
`=>2x=14`
`=>x=7(tm)`
Vậy `x=7`
`7)\sqrt{x-1}=\sqrt{1-x}`
`=>{(x-1>=0),(1-x>=0):}`
`=>{(x>=1),(x<=1):}`
`=>x=1`
Vậy `x=1`
`8)\sqrt{2x-5}=\sqrt{1-x}`
`=>{(2x-5>=0),(1-x>=0):}`
`=>`$\begin{cases} 2x≥5 \\ x≤1 \end{cases}$
`=>`$\begin{cases} x≥\dfrac{5}{2} \\ x≤1 \end{cases}$ (vô lý)
Vậy pt này vô nghiệm.
`9)\sqrt{36x^2+12x+1}=2`
`=>\sqrt{(6x+1)^2}=2`
`=>|6x+1|=2`
`=>[(6x+1=2),(-6x-1=2):}`
`=>[(6x=1),(6x=-3):}`
`=>`$\left[\begin{matrix} x=\dfrac{1}{6} \\ x=\dfrac{-1}{2} \end{matrix}\right.$
Vậy `S={1/6;-1/2}`
`10)\sqrt{36x^2+12x+1}=7`
`=>\sqrt{(6x+1)^2}=7`
`=>|6x+1|=7`
`=>[(6x+1=7),(6x+1=-7):}`
`=>[(6x=6),(6x=-8):}`
`=>`$\left[\begin{matrix} x=1 \\ x=\dfrac{-4}{3} \end{matrix}\right.$
Vậy `S={1;-4/3}`
