Question

[....]

Phương pháp 4. Bình phương hai vế để làm mất căn

a \(\sqrt{2x-3}=x-3\)

b \(\sqrt{10-x}+\sqrt{x+3}=5\)

c \(\sqrt{x+3}-\sqrt{x-4}=1\)

Answer 1

[๖ۣۜDũ๖ۣۜN๖ۣۜG]

a) \(\sqrt{2x-3}=x-3\) (ĐK: \(x\ge\dfrac{3}{2}\))

<=> \(\left\{{}\begin{matrix}x\ge3\\2x-3=\left(x-3\right)^2\left(1\right)\end{matrix}\right.\)

(1) <=> \(2x-3=x^2-6x+9\)

<=> \(x^2-8x+12=0\)

<=> (x-2)(x-6) = 0 <=> \(\left[{}\begin{matrix}x=2\left(l\right)\\x=6\left(c\right)\end{matrix}\right.\)

KL: Phương trình có nghiệm duy nhất x = 6
b) \(\sqrt{10-x}+\sqrt{x+3}=5\) (ĐK: \(-3\le x\le10\))

<=> \(\left(\sqrt{10-x}+\sqrt{x+3}\right)^2=25\)

<=> \(10-x+x+3+2\sqrt{\left(10-x\right)\left(x+3\right)}=25\)

<=> \(\sqrt{\left(10-x\right)\left(x+3\right)}=6\)

<=> (10-x)(x+3) = 36

<=> 7x - x² + 30 = 36

<=> x² -7x + 6 = 0

<=> (x-1)(x-6) = 0

<=> \(\left[{}\begin{matrix}x=1\left(c\right)\\x=6\left(c\right)\end{matrix}\right.\)

KL: Phương trình có nghiệm S = {1;6}

c) \(\sqrt{x+3}-\sqrt{x-4}=1\) (ĐK: \(x\ge4\))

<=> \(\sqrt{x+3}=\sqrt{x-4}+1\)

<=> \(x+3=x-4+1+2\sqrt{x-4}\)

<=> \(\sqrt{x-4}=3\)

<=> x-4 = 9 <=> x = 13 (c)

KL: Phương trình có nghiệm duy nhất x = 13

Answer 2

[Trần Ái Linh]

a) ĐK: `x≥3`

`\sqrt(2x-3)=x-3`

`<=>2x-3=(x-3)^2`

`<=>2x-3=x^2-6x+9`

`<=>x^2-8x+12=0`

`<=>` \(\left[{}\begin{matrix}x=6\left(TM\right)\\x=2\left(L\right)\end{matrix}\right.\)

Vậy `x=2`.

b) ĐK: `-3<=x<=10`

`\sqrt(10-x)+\sqrt(x-3)=5`

`<=>10-x+x-3+2\sqrt((10-x)(x-3))=25`

`<=>2\sqrt((10-x)(x-3))=18`

`<=>\sqrt((10-x)(x-3))=9`

`<=>(10-x)(x-3)=81`

`<=>-x^2+13x-30=81`

`<=>x^2-13x+111=0` (VN)