`a)sqrt{1-4x+4x^2}+5=x-2`

`<=>\sqrt{(2x-1)^2}=x-2-5`

`<=>|2x-1|=x-7(x>=7)`

`<=>[(2x-1=x-7),(2x-1=7-x):}`

`<=>[(x=-6(ktm)),(3x=8):}`

`<=>x=8/3(ktm)`

Vậy PTVN

`b)3sqrt{12+4x}+4/7sqrt{147+49x}=3/2sqrt{48+16x}+4(x>=-3)`

`<=>6sqrt{x+3}+4sqrt{x+3}=6sqrt{x+3}+4`

`<=>4sqrt{x+3}=4`

`<=>sqrt{x+3}=1<=>x+3=1`

`<=>x=-2(tm)`

Vậy `S={-2}`

a) \(\sqrt{1-4x+4x^2}+5=x-2\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\Leftrightarrow\left|1-2x\right|=x-7\left(1\right)\)TH1: \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow1-2x=x-7\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)(không thỏa đk)

TH2: \(1-2x< 0\Leftrightarrow x>\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow2x-1=x-7\Leftrightarrow x=-6\)(không thỏa đk)

Vậy \(S=\varnothing\)

b) \(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}=6\sqrt{3+x}+4\Leftrightarrow4\sqrt{3+x}=4\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\)

a. \(\sqrt{1-4x+4x^2}+5=x-2\)

\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\)

\(\Leftrightarrow\left|1-2x\right|-x=-7\)

\(\Leftrightarrow\left[{}\begin{matrix}1-2x-x=-7\\2x-1-x=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\x=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-6\end{matrix}\right.\)

b. ĐKXĐ: \(x\ge-3\)
\(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\)

\(\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}-6\sqrt{3+x}=4\)

\(\Leftrightarrow4\sqrt{3+x}=4\) \(\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\) ( thỏa mãn đk )

a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$

$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$

$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$

Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}=\frac{12}{5}$

$\Leftrightarrow x=5,76$ (thỏa mãn)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 2\sqrt{x-2}=8$

$\Leftrightarrow \sqrt{x-2}=4$

$\Leftrightarrow x=4^2+2=18$ (tm)

b. ĐKXĐ: $x^2\geq 5$

PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$

$\Leftrightarrow \sqrt{x^2-5}=0$

$\Leftrightarrow x=\pm \sqrt{5}$

6x^3 + x + 4 = 11x^2
<=>6x3-11x2+x+4=0
<=>6x3+3x2-14x2-7x+8x+4=0
<=>3x2(2x+1)-7x(2x+1)+4(2x+1)=0
<=>(2x+1)(3x2-7x+4)=0
<=>(2x+1)(3x2-3x-4x+4)=0
<=>(2x+1)(3x-4)(x-1)=0
<=>2x+1=0 hoặc 3x-4=0 hoặc x-1=0
<=>x\(\in\){-1/2;1;4/3}
b)x^6 - 14x^4 + 49x^2 = 36
<=>x6-14x4+49x2-36=0
<=>x6-x4-13x4+13x2+36x2-36=0
<=>x4(x2-1)-13x2(x2-1)+36(x2-1)=0
<=>(x2-1)(x4-13x2+36)=0
<=>(x+1)(x-1)(x4-9x2-4x2+36)=0
<=>(x+1)(x-1)[x2(x2-9)-4(x2-9)]=0
<=>(x-1)(x+1)(x2
-9)(x2-4)=0
<=>(x-1)(x+1)(x+3)(x-3)(x+2)(x-2)=0
<=>x\(\in\){-3;-2;-1;1;2;3}

p/s: kham khảo

6x^3 + x + 4 = 11x^2

<=>6x³-11x²+x+4=0

<=>6x³+3x²-14x²-7x+8x+4=0

<=>3x²(2x+1)-7x(2x+1)+4(2x+1)=0

<=>(2x+1)(3x²-7x+4)=0

<=>(2x+1)(3x²-3x-4x+4)=0

<=>(2x+1)(3x-4)(x-1)=0

<=>2x+1=0 hoặc 3x-4=0 hoặc x-1=0

<=>x\(\in\){-1/2;1;4/3}

b)x^6 - 14x^4 + 49x^2 = 36

<=>x⁶-14x⁴+49x²-36=0

<=>x⁶-x⁴-13x⁴+13x²+36x²-36=0

<=>x⁴(x²-1)-13x²(x²-1)+36(x²-1)=0

<=>(x²-1)(x⁴-13x²+36)=0

<=>(x+1)(x-1)(x⁴-9x²-4x²+36)=0

<=>(x+1)(x-1)[x²(x²-9)-4(x²-9)]=0

<=>(x-1)(x+1)(x²-9)(x²-4)=0

<=>(x-1)(x+1)(x+3)(x-3)(x+2)(x-2)=0

<=>x\(\in\){-3;-2;-1;1;2;3}

phù.mệt