h: \(\sqrt{18x}+\sqrt{32x}-14=0\)

\(\Leftrightarrow7\sqrt{2x}=14\)

hay x=2

\(3T=\left(\sqrt{x^2-6x+19}-\sqrt{x^2-6x+10}\right)\left(\sqrt{x^2-6x+19}+\sqrt{x^2-6x+10}\right)\)

\(=x^2-6x+19-\left(x^2-6x+10\right)=9\)

\(\Rightarrow T=3\)

a) `sqrt(x^2-6x _9) = 4-x`

`<=> sqrt[(x-3)^2] =4-x`

`<=> |x-3| =4-x ( đk :x<=4)`

`<=> |x-3| = |4-x|`

`<=> [(x-3 =4-x),(x-3 = x-4):}`

`<=>[(x = 7/2(t//m)),(0=-1(vl)):}`

Vậy `S = {7/2}`

b) `sqrt(x^2 -9) + sqrt(x^2 -6x +9) =0(đk : x>=3(hoặc) x<=-3)`

`<=>sqrt(x^2 -9) =- sqrt(x^2 -6x +9) `

`<=>(sqrt(x^2 -9))^2 =(- sqrt(x^2 -6x +9))^2`

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

`<=> 6x = 9+9 =18`

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

Vậy `S={3}`

c) `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`

xét `x<1 =>{(|x-1| =1-x ),(|x-2|=2-x):}`

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

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

xét `1<=x<2 => {(|x-1|=x-1),(|x-2|= 2-x):}`

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

`=>1=3 (vl)`

xét `x>=2 => {(|x-1| =x-1),(|x-2|=x-2):}`

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

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

Vậy `S = {0;3}`

a: =>|x-3|=4-x

TH1: x>=3

=>4-x=x-3

=>x=7/2(nhận)

TH2: x<3

=>3-x=4-x(loại)

b: =>căn x-3(căn x+3+căn x-3)=0

=>x-3=0

=>x=3

c: =>|x-1|+|x-2|=3

Th1: x<1

=>1-x+2-x=3

=>x=0(nhận)

TH2: 1<=x<2

=>x-1+2-x=3

=>1=3(loại)

TH3: x>=2

=>x-1+x-2=3

=>x=3

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé