Question

a) \(2\sqrt{9x-27}\times\dfrac{1}{5}\sqrt{25x-75}\times\dfrac{1}{7}\sqrt{49x-147}=20\)

b) \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)

Answer 1

a: \(\Leftrightarrow6\sqrt{-3}+\dfrac{1}{5}\cdot5\sqrt{x-3}+\dfrac{1}{7}\cdot7\sqrt{x-3}=20\)

\(\Leftrightarrow8\sqrt{x-3}=20\)

\(\Leftrightarrow\sqrt{x-3}=\dfrac{5}{2}\)

=>x-3=25/4

hay x=37/4

b: \(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\)

Trường hợp 1 x<1

Pt sẽ là 1-x+2-x=3

=>3-2x=3

hay x=0(nhận)

Trường hợp 2: 1<=x<2

Pt sẽ là x-1+2-x=3

=>1=3(loại)

Trườg hợp 3: x>=2

Pt sẽ là x-1+x-2=3

=>2x-3=3

hay x=3(nhận)

Answer 2

`a)2\sqrt{9x-27}. 1/5\sqrt{25x-75}. 1/7\sqrt{49x-147}=20`

`<=>2.3\sqrt{x-3}. 1/5 .5\sqrt{x-3}. 1/7 .7\sqrt{x-3}=20`

`<=>6\sqrt{x-3}.\sqrt{x-3}.\sqrt{x-3}=20`

`<=>(\sqrt{x-3})^3=10/3`

`<=>\sqrt{x-3}=\root{3}{\frac{10}{3}}`

`<=>x-3=(\root{3}{\frac{10}{3}})^2`

`<=>x=(\root{3}{\frac{10}{3}})^2+3`

Vậy `S={(\root{3}{\frac{10}{3}})^2+3}`

__________________________________________________________

`b)\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3`

`<=>\sqrt{(x-1)^2}+\sqrt{(x-2)^2}=3`

`<=>|x-1|+|x-2|=3`

Ta có bảng:

\begin{array}{|c|cc|}\hline x&-\infty & &1&&2&&+\infty\\\hline |x-1|& &1-x&0&x-1&|&x-1\\\hline |x-2|& &2-x &|&2-x&0&x-2\\\hline \end{array}

`@` Với `x < 1` khi đó ta có ptr:

    `1-x+2-x=3`

`<=>x=0` (t/m)

`@` Với `1 <= x < 2` khi đó ta có ptr:

    `x-1+2-x=3`

`<=>0x=2` (Vô lí)

`@` Với `x >= 2` khi đó ta có ptr:

    `x-1+x-2=3`

`<=>2x=6`

`<=>x=3` (t/m)

Vậy `S={0;3}`

Answer 3

a).

<=>

\(2\sqrt{9\left(x-3\right)}.\dfrac{1}{5}\sqrt{25\left(x-3\right)}.\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\)

<=>

\(2.3\sqrt{x-3}.\dfrac{1}{5}.5\sqrt{x-3}.\dfrac{1}{7}.7.\sqrt{x-3}=20\)

<=>

\(6\sqrt{x-3}.\sqrt{x-3}.\sqrt{x-3}=20\)

tới đây tự lm đc r nek:v