Bài 4:
`a) A=x^2+2x+16`
`\qquadA=(x^2+2x+1)+15`
`\qquadA=(x+1)^2+15`
Thay `x=\sqrt{2}-1` vào A ta có:
`\qquadA=(\sqrt{2}-1+1)^2+15=(\sqrt{2})^2+15=2+15=17`
Vậy `A=17` khi `x=\sqrt{2}-1`
`b) B=x^2+12x-14`
`\qquad B=(x^2+12x+36)-50`
`\qquad B=(x+6)^2-50`
Thay `x=5\sqrt{2}-6` vào B ta có:
`\qquad B=(5\sqrt{2}-6+6)^2-50=(5\sqrt{2})^2-50=50-50=0`
Vậy `B=0` khi `x=5\sqrt{2}-6`
Bài 5:
`a) \sqrt{(2x-3)/(x-1)}=2` ĐKXĐ: `x\ne 1; x>1` hoặc `x<=3/2`
`<=> (2x-3)/(x-1)=4`
`<=> 2x-3=4(x-1)`
`<=> 2x-3=4x-4`
`<=> 4x-2x=-3+4`
`<=> 2x=1`
`<=> x=1/2(\text{tm})`
Vậy `S={1/2}`
`b) \sqrt{2x-3}/\sqrt{x-1}=2` ĐKXĐ: `x>=3/2`
`<=> \sqrt{2x-3}=2\sqrt{x-1}`
`<=> 2x-3=4(x-1)`
`<=> 2x-3=4x-4`
`<=> 4x-2x=-3+4`
`<=> 2x=1`
`<=> x=1/2(\text{ktm})`
Vậy `S=∅`
