Question

[Nguyễn Kim Chi]

1) \(\sqrt{x-1}=\sqrt{2x+3}\)

2) \(\sqrt{2x-3}=\sqrt{x-1}\)

3) \(\sqrt{2-x}=\sqrt{3+x}\)

4) \(\sqrt{4x-8}=2\sqrt{x-2}\)

5) \(\sqrt{x^2-5}\sqrt{4x-9}\)

6) \(\sqrt{x-2}=\sqrt{x^2-2x}\)

7) \(\sqrt{x^2-3x}=\sqrt{15-5x}\)

8) \(\sqrt{4x^2-9}=\sqrt{-20-18}\)

Có điều kiện cho từng câu hộ mình nhé!!!

9) \(\sqrt{x-2}=\sqrt{x-2}\)

Answer 1

[Cô Nàng Song Tử]

1) \(\sqrt{x-1}=\sqrt{2x+3}\) ĐK: x ≥ 1; x ≥ \(\dfrac{-3}{2}\) => x ≥ 1

=> x - 1 = 2x + 3

=> x - 2x = 3 + 1

=> -x = 4 => x = -4 (ko TMĐK)

Vậy S = ∅

2) \(\sqrt{2x-3}=\sqrt{x-1}\) ĐK: x ≥ \(\dfrac{3}{2}\); x ≥ 1 => x ≥ \(\dfrac{3}{2}\)

=> 2x - 3 = x - 1

=> 2x - x = -1 + 3

=> x = -2 (ko TMĐK)

Vậy S = ∅

3) \(\sqrt{2-x}=\sqrt{3+x}\) ĐK: x ≥ 2; x ≥ -3 => x ≥ 2

=> 2 - x = 3 + x

=> -x - x = 3 - 2

=> -2x = 1 => x = \(\dfrac{-1}{2}\) (ko TMĐK)

Vậy S = ∅

4) \(\sqrt{4x-8}=2\sqrt{x-2}\) ĐK: x ≥ 2

=> 4x - 8 = 2(x - 2)

=> 4x - 8 = 2x - 4

=> 4x - 2x = -4 + 8

=> 2x = 4 => x = 4 : 2 = 2 (TMĐK)

Vậy S = \(\left\{2\right\}\)

5) \(\sqrt{x^2-5}=\sqrt{4x-9}\) ĐK: \(\left|x\right|=\sqrt{5}\) ; x ≥ \(\dfrac{9}{4}\)

<=> x² - 5 = 4x - 9

<=> x² - 4x - 5 + 9 = 0

<=> x² - 4x - 4 = 0 <=> (x - 2)² =0

=> x = 2 (ko TMĐK)

6) \(\sqrt{x-2}=\sqrt{x^2-2x}\) ĐK: x ≥ 2

=> x - 2 = x² - 2x

=> x - 2 - x² + 2x = 0

=> (x - 2) - x(x - 2) = 0

=> (1- x) . (x - 2) = 0

=> \(\left\{{}\begin{matrix}1-x=0\\x-2=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-0=1\left(loai\right)\\x=0+2=2\left(TMĐK\right)\end{matrix}\right.\)

Vậy S = \(\left\{2\right\}\)

7) \(\sqrt{x^2-3x}-\sqrt{15-5x}=0\) ĐK: x ≥ 3 hoặc x ≤ 0

<=> \(\sqrt{x^2-3x}=\sqrt{15-5x}\)

<=> x² - 3x = 15 - 5x

=> x² - 3x + 5x - 15 = 0

=> x(x -3) + 5(x - 3) = 0

=> (x + 5) . (x - 3) = 0

=> \(\left[{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=0-5=-5\\x=0+3=3\end{matrix}\right.\)(TMĐK)

Vậy S = \(\left\{-5;3\right\}\)

8) \(\sqrt{4x^2-9}=\sqrt{-20x-18}\) ĐK: \(\left|x\right|\text{≥}\dfrac{3}{2}\) hoặc x ≤ \(\dfrac{-9}{10}\)

<=> 4x² - 9 = -20x - 18

<=> 4x² - 9 + 20x + 18 = 0

<=> 4x2 + 20x + 9 =0

<=> 4x2 + 2x + 18x + 9 =0

<=> 2x(2x + 1) + 9(2x + 1) = 0

<=> (2x + 9) . (2x + 1) = 0

=> \(\left[{}\begin{matrix}2x+9=0\\2x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=-9\\2x=-1\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=\dfrac{-9}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy S = \(\left\{\dfrac{-9}{2};\dfrac{-1}{2}\right\}\)

9) \(\sqrt{x-2}=\sqrt{x-2}\) ĐK: x ≥ 2

<=> x - 2 = x - 2

<=> x - x = 2 - 2

=> 0x = 0 với mọi x TMĐK: x ≥ 2

Kết luận: Phương trình vô nghiệm thoả mãn: x ≥ 2

Answer 2

[Đỗ Viết Ngọc Cường]

1,

√(x-1) = √(2x+3)

->(√x-1)^2 = (√2x+3)^2

->x-1=2x+3

->x=-4

2

) (bình phương lên tiếp)

->2x-3=x-1

->x=2

3->9 tự làm nha tương tự