d. \(\sqrt{9x^2+12x+4}=4\)

<=> \(\sqrt{\left(3x+2\right)^2}=4\)

<=> \(|3x+2|=4\)

<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)

\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)

\(\Leftrightarrow x=1\)

a) Điều kiện xác định \(16x+8\ge0\Leftrightarrow x\ge-\frac{1}{2}.\)

Theo bất đẳng thức Cô-Si cho 4 số ta được 

\(4\sqrt[4]{16x+8}=4\sqrt[4]{2\cdot2\cdot2\cdot\left(2x+1\right)}\le2+2+2+2x+1=2x+7\)

Do vậy mà \(4x^3+4x^2-5x+9\le2x+7\Leftrightarrow\left(2x-1\right)^2\left(x+2\right)\le0\).

Vì \(x\ge-\frac{1}{2}\to x+2>0\to\left(2x-1\right)^2\le0\to x=\frac{1}{2}.\) 

b. Ta viết phương trình dưới dạng sau đây  \(9x^4-21x^3+27x^2+16x+16=0\Leftrightarrow3x^2\left(3x^2-7x+7\right)+4\left(x+2\right)^2=0\)

Vì \(3x^2-7x+7=\frac{36x^2-2\cdot6x\cdot7+49+35}{12}=\frac{\left(6x-7\right)^2+35}{12}>0\) nên vế trái dương, suy ra phương trinh vô nghiệm.

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

a) $\sqrt{4x + 20} - 3\sqrt{5 + x} + \frac{4}{3}\sqrt{9x + 45} = 6$

Điều kiện xác định:

$5 + x \ge 0 \Leftrightarrow x \ge -5$

Biến đổi phương trình:

$\sqrt{4(x + 5)} - 3\sqrt{x + 5} + \frac{4}{3}\sqrt{9(x + 5)} = 6$

$\Leftrightarrow 2\sqrt{x + 5} - 3\sqrt{x + 5} + \frac{4}{3} \cdot 3\sqrt{x + 5} = 6$

$\Leftrightarrow 2\sqrt{x + 5} - 3\sqrt{x + 5} + 4\sqrt{x + 5} = 6$

$\Leftrightarrow (2 - 3 + 4)\sqrt{x + 5} = 6$

$\Leftrightarrow 3\sqrt{x + 5} = 6$

$\Leftrightarrow \sqrt{x + 5} = 2$

$\Leftrightarrow x + 5 = 4$

$\Leftrightarrow x = -1$ (Thỏa mãn điều kiện $x \ge -5$)

Vậy tập nghiệm của phương trình là $S = \{-1\}$.

`a)\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=12`     `ĐK: x >= 0`

`<=>\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=12`

`<=>12\sqrt{3x}=12`

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

`<=>3x=1<=>x=1/3` (t/m)

`b)5\sqrt{9x+9}-2\sqrt{4x+4}+\sqrt{x+1}=36`   `ĐK: x >= -1`

`<=>15\sqrt{x+1}-4\sqrt{x+1}+\sqrt{x+1}=36`

`<=>12\sqrt{x+1}=36`

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

`<=>x+1=9`

`<=>x=8` (t/m)

\(\Leftrightarrow\frac{6x+5}{12x+9}-\frac{3x-7}{12x-9}=\frac{4x^2+10x-7}{16x^2-9}.\)

\(\Leftrightarrow\frac{\left(6x+5\right)\left(12x-9\right)-\left(3x-7\right)\left(12x+9\right)}{\left(3.4.x\right)^2-\left(3.3\right)^2}=\frac{4x^2+10x-7}{16x^2-9}\)

\(\Leftrightarrow\frac{72x^2+6x-45-\left(36x^2-57x-63\right)}{3^2\left(16x^2-9\right)}=\frac{4x^2+10x-7}{16x^2-9}\)

ĐK: \(16x^2-9\ne0\Leftrightarrow x^2\ne\left(\frac{3}{4}\right)^2\Rightarrow x\ne\pm\frac{3}{4}\)

\(\Leftrightarrow72x^2+6x-45-36x^2+57x+63=36x^2+90x-63\)

\(\Leftrightarrow27x=81\Leftrightarrow x=3\)

a: ĐKXĐ: x<>4

Ta có: \(1+\frac{14}{\left(\left.x-4\right.\right)^2}=\frac{-9}{x-4}\)

=>\(\frac{\left(x-4\right)^2+14}{\left(x-4\right)^2}=\frac{-9\left(x-4\right)}{\left(x-4\right)^2}\)

=>\(\left(x-4\right)^2+14=-9\left(x-4\right)\)

=>\(\left(x-4\right)^2+9\left(x-4\right)+14=0\)

=>(x-4+2)(x-4+7)=0

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

=>x=2(nhận) hoặc x=-3(nhận)

b: ĐKXĐ: x∉{1/2;-1/2}

Ta có: \(\frac{1+8x}{1+2x}-\frac{2x}{2x-1}+\frac{12x^2-9}{1-4x^2}=0\)

=>\(\frac{8x+1}{2x+1}-\frac{2x}{2x-1}-\frac{12x^2-9}{\left(2x-1\right)\left(2x+1\right)}=0\)

=>\(\frac{\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9}{\left(2x-1\right)\left(2x+1\right)}=0\)

=>\(\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9=0\)

=>\(16x^2-8x+2x-1-4x^2-2x-12x^2+9=0\)

=>-8x+8=0

=>-8x=-8

=>x=1(nhận)

c: ĐKXĐ: x∉{3;1}

\(\frac{1}{2x-6}-\frac{3x-5}{x^2-4x+3}=\frac12\)

=>\(\frac{1}{2\left(x-3\right)}-\frac{3x-5}{\left(x-1\right)\left(x-3\right)}=\frac12\)

=>\(\frac{x-1}{2\left(x-3\right)\left(x-1\right)}-\frac{2\left(3x-5\right)}{2\left(x-1\right)\left(x-3\right)}=\frac{x^2-4x+3}{2\left(x-1\right)\left(x-3\right)}\)

=>\(x^2-4x+3=x-1-2\left(3x-5\right)=x-1-6x+10=-5x+9\)

=>\(x^2+x-6=0\)

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

=>x=-3(nhận) hoặc x=2(nhận)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(3\sqrt{18x}-5\sqrt{8x}+4\sqrt{50x}=38\)

\(\Leftrightarrow9\sqrt{2x}-10\sqrt{2x}+20\sqrt{2x}=38\)

\(\Leftrightarrow19\sqrt{2x}=38\)

\(\Leftrightarrow\sqrt{2x}=2\)

\(\Leftrightarrow2x=4\)

hay x=2(thỏa ĐK)

b) ĐKXĐ: \(x\ge0\)

Ta có: \(3\sqrt{12x}-2\sqrt{27x}+4\sqrt{3x}=8\)

\(\Leftrightarrow6\sqrt{3x}-6\sqrt{3x}+4\sqrt{3x}=8\)

\(\Leftrightarrow\sqrt{3x}=2\)

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

c) ĐKXĐ: \(x\ge5\)

Ta có: \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\)

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

\(\Leftrightarrow x-5=4\)

hay x=9

a)

\(3.3\sqrt{2x}-5.2\sqrt{2x}+4.5.\sqrt{2x}=38\\ \Leftrightarrow19\sqrt{2x}=38\\ \Leftrightarrow\sqrt{2x}=2\\ \Leftrightarrow x=2\)

b)

\(3.2.\sqrt{3x}-2.3.\sqrt{3x}+4.\sqrt{3x}=8\\ \Leftrightarrow4\sqrt{3x}=8\\ \Leftrightarrow\sqrt{3x}=2\\\Leftrightarrow x=\dfrac{2^2}{3}=\dfrac{4}{3} \)

c)

\(\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\)

a) Ta có: \(x^3-9x^2+19x-11=0\)

\(\Leftrightarrow x^3-x^2-8x^2+8x+11x-11=0\)

\(\Leftrightarrow x^2\left(x-1\right)-8x\left(x-1\right)+11\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-8x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-8x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{5}+4\\x=-\sqrt{5}+4\end{matrix}\right.\)

Vậy: \(S=\left\{1;\sqrt{5}+4;-\sqrt{5}+4\right\}\)

a) - 3 =0

<=> (√x-3)(√x+3)-3√x-3=0

<=> (√x-3)(√x+3-3)=0

<=> (√x-3)√x=0

<=> √x-3=0

<=>x=9

b)=x - 3

<=> √(2x -3)² =x-3

<=> 2x-3=x-3

<=>2x-x=-3+3

<=>x=0

c)=3x-1

<=> √(x+3)² =3x-1

<=> x+3=3x-1

<=> -2x=-4

<=>  x=2

Nhớ cho mình 1 tim nha bạn

Lời giải:

a. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow \sqrt{(x-3)(x+3)}-3\sqrt{x-3}=0$

$\Leftrightarrow \sqrt{x-3}(\sqrt{x+3}-3)=0$

$\Leftrightarrow \sqrt{x-3}=0$ hoặc $\sqrt{x+3}-3=0$

$\Leftrightarrow \sqrt{x-3}=0$ hoặc $\sqrt{x+3}=3$

$\Leftrightarrow x=3$ hoặc $x=6$ (tm)

b.

PT \(\Rightarrow \left\{\begin{matrix} x-3\geq 0\\ 4x^2-12x+9=(x-3)^2=x^2-6x+9\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 3x^2-6x=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 3x(x-2)=0\end{matrix}\right.\)

$\Rightarrow$ không có giá trị $x$ nào thỏa mãn 

Vậy pt vô nghiệm.

c.

PT \(\Rightarrow \left\{\begin{matrix} 3x-1\geq 0\\ x^2+6x+9=(3x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ x^2+6x+9=9x^2-6x+1\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ 8x^2-12x-8=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ 4(x-2)(2x+1)=0\end{matrix}\right.\Leftrightarrow x=2\)