a.

$x^2-11=0$

$\Leftrightarrow x^2=11$

$\Leftrightarrow x=\pm \sqrt{11}$

b. $x^2-12x+52=0$

$\Leftrightarrow (x^2-12x+36)+16=0$

$\Leftrightarrow (x-6)^2=-16< 0$ (vô lý)

Vậy pt vô nghiệm.

c.

$x^2-3x-28=0$

$\Leftrightarrow x^2+4x-7x-28=0$

$\Leftrightarrow x(x+4)-7(x+4)=0$

$\Leftrightarrow (x+4)(x-7)=0$

$\Leftrightarrow x+4=0$ hoặc $x-7=0$

$\Leftrightarrow x=-4$ hoặc $x=7$

d.

$x^2-11x+38=0$

$\Leftrightarrow (x^2-11x+5,5^2)+7,75=0$

$\Leftrightarrow (x-5,5)^2=-7,75< 0$ (vô lý)

Vậy pt vô nghiệm

e.

$6x^2+71x+175=0$

$\Leftrightarrow 6x^2+21x+50x+175=0$

$\Leftrightarrow 3x(2x+7)+25(2x+7)=0$

$\Leftrightarrow (3x+25)(2x+7)=0$

$\Leftrightarrow 3x+25=0$ hoặc $2x+7=0$

$\Leftrightarrow x=-\frac{25}{3}$ hoặc $x=-\frac{7}{2}$

ĐKXĐ: \(\left[{}\begin{matrix}x=0\\x\ge3\end{matrix}\right.\)

Với \(x=0\) là nghiệm

Với \(x\ge3\), chia 2 vế cho \(\sqrt{x}\) ta được:

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

\(\Leftrightarrow\sqrt{x+1}+\sqrt{x+2}-\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x+1}+\dfrac{5}{\sqrt{x+2}+\sqrt{x-3}}=0\) (vô nghiệm do vế trái luôn dương)

Vậy pt có nghiệm duy nhất \(x=0\)

Tham khảo:

1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24

ghê thậc, còn cái còn lại thì seo?

1: ĐKXĐ: x>=-1

\(\begin{cases}\sqrt{x+1}=\sqrt{2}(8y^2+8y+1)\left(1\right)\\ 4(x^3-8y^3)-6(x^2+4y^2)+3(x+2y)-1=0\left(2\right)\end{cases}\)

(2): \(4(x^3-8y^3)-6(x^2+4y^2)+3(x+2y)-1=0\)

=>\(8x^3 - 64y^3 - 12x^2 - 48y^2 + 6x + 12y - 2 = 0\)

=>\((8x^3-12x^2+6x-1)-(64y^3+48y^2+12y+1)=0\)

=>\((2x-1)^3-(4y+1)^3=0\)

=>\(\left(2x-1\right)^3=\left(4y+1\right)^3\)

=>2x-1=4y+1

=>2x-4y=2

=>x-2y=1

=>x=2y+1

(1) sẽ tương đương: \(\sqrt{2y + 1 + 1} = \sqrt{2}(8y^2 + 8y + 1)\)

=>\(\sqrt{2(y + 1)}=\sqrt{2}(8y^2+8y+1)\)

=>\(\sqrt{y + 1}=8y^2+8y+1\)

Đặt \(t = 8y^2 + 8y + 1 \ge 0\)

Ta có: \(\sqrt{y + 1}=8y^2+8y+1\)

=>\(\sqrt{y+1}=t\)

=>\(y+1=t^2\)

Phương trình sẽ trở thành:

\(8(t^2-1)^2+8(t^2-1)+1=t\)

=>\(8t^4-8t^2-t+1=0\)

=>\((t-1)(8t^3+8t^2-1)=0\)

=>t-1=0

=>t=1

=>8y(y+1)=0

=>y=0 hoặc y=-1

Khi y=0 thì x=1(nhận)

Khi y=-1 thì x=-1(loại)

1) \(\sqrt[]{9\left(x-1\right)}=21\)

\(\Leftrightarrow9\left(x-1\right)=21^2\)

\(\Leftrightarrow9\left(x-1\right)=441\)

\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)

2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)

\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)

\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)

\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)

\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)

\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)

mà \(\sqrt[]{1-x}\ge0\)

\(\Leftrightarrow pt.vô.nghiệm\)

3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)

\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)

\(\Leftrightarrow2x=50\Leftrightarrow x=25\)

1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))

\(\Leftrightarrow3\sqrt{x-1}=21\)

\(\Leftrightarrow\sqrt{x-1}=7\)

\(\Leftrightarrow x-1=49\)

\(\Leftrightarrow x=49+1\)

\(\Leftrightarrow x=50\left(tm\right)\)

2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))

\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)

\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)

\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý) 

Phương trình vô nghiệm

3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\)) 

\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)

\(\Leftrightarrow2x=50\)

\(\Leftrightarrow x=\dfrac{50}{2}\)

\(\Leftrightarrow x=25\left(tm\right)\)

4) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

5) \(\sqrt{\left(x-3\right)^2}=3-x\)

\(\Leftrightarrow\left|x-3\right|=3-x\)

\(\Leftrightarrow x-3=3-x\)

\(\Leftrightarrow x+x=3+3\)

\(\Leftrightarrow x=\dfrac{6}{2}\)

\(\Leftrightarrow x=3\)