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 $x=-\frac{7}{2}$

f.

$x^2-(\sqrt{2}+\sqrt{8})x+4=0$

$\Leftrightarrow x^2-\sqrt{2}x-2\sqrt{2}x+4=0$

$\Leftrightarrow x(x-\sqrt{2})-2\sqrt{2}(x-\sqrt{2})=0$

$\Leftrightarrow (x-\sqrt{2})(x-2\sqrt{2})=0$

$\Leftrightarrow x-\sqrt{2}=0$ hoặc $x-2\sqrt{2}=0$

$\Leftrightarrow x=\sqrt{2}$ hoặc $x=2\sqrt{2}$

g.

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

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

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

$\Leftrightarrow (x-1)[(1+\sqrt{3})x-\sqrt{3}]=0$

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

$\Leftrightarrow x=1$ hoặc $x=\frac{3-\sqrt{3}}{2}$

\(1,PT\Leftrightarrow2x-1=5\Leftrightarrow x=3\\ 2,\Leftrightarrow x-5=9\Leftrightarrow x=14\\ 3,ĐK:x\ge1\\ PT\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow x=50\left(tm\right)\\ 4,\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)

1:

\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)

\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)

\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)

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) \(\sqrt{\left(x-3\right)^2}=3-x\)

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

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

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

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

\(\Leftrightarrow x=3\)

1) => 9(x-1)=\(21^2\)

=> 9x-9=441

=> 9x=450

=> x=50

2)=>\(\sqrt{1-x}\) + \(\sqrt{4\left(1-x\right)}\)-\(\dfrac{1}{3}\sqrt{16\left(1-x\right)}\)+5=0

=>\(\sqrt{1-x}\)\(\left(1+2-\dfrac{1}{3}.4\right)\)+5=0

=>\(\dfrac{5}{3}\sqrt{1-x}\) +5=0

=>\(\sqrt{1-x}\)=-3

Phuong trinh vo nghiem

c.

ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)

\(\Leftrightarrow x+4-2\sqrt[]{\left(\dfrac{x+2}{x-1}\right)^2\left(\dfrac{x-1}{x+2}\right)}=0\)

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

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

\(\Leftrightarrow x^2+8x+16=\dfrac{4\left(x+2\right)}{x-1}\)

\(\Rightarrow x^3+7x^2+4x-24=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+4x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2+2\sqrt{3}\\x=-2-2\sqrt{3}\left(loại\right)\end{matrix}\right.\)

a.

\(\Leftrightarrow2x^2-11x+21=3\sqrt[3]{4\left(x-1\right)}\)

Do \(2x^2-11x+21=2\left(x-\dfrac{11}{4}\right)^2+\dfrac{47}{8}>0\Rightarrow3\sqrt[3]{4\left(x-1\right)}>0\Rightarrow x-1>0\)

Ta có:

\(VT=2x^2-11x+21-3\sqrt[3]{4x-4}=2\left(x^2-6x+9\right)+x+3-3\sqrt[3]{4\left(x-1\right)}\)

\(=2\left(x-3\right)^2+x+3-3\sqrt[3]{4\left(x-1\right)}\)

\(\Rightarrow VT\ge x+3-3\sqrt[3]{4\left(x-1\right)}=\left(x-1\right)+2+2-3\sqrt[3]{4\left(x-1\right)}\)

\(\Rightarrow VT\ge3\sqrt[3]{\left(x-1\right).2.2}-3\sqrt[3]{4\left(x-1\right)}=0\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\x-1=2\\\end{matrix}\right.\) \(\Leftrightarrow x=3\)

Vậy pt có nghiệm duy nhất \(x=3\)

b.

ĐKXD: \(x\ge-1\)

Phương trình: \(2\left(x+1\right)-\left(3x-2\right)\sqrt[]{x+1}+x^2-x=0\)

Đặt \(\sqrt[]{x+1}=t\ge0\)

\(\Rightarrow2t^2-\left(3x-2\right)t+x^2-x=0\)

\(\Delta=\left(3x-2\right)^2-8\left(x^2-x\right)=\left(x-2\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x-2+x-2}{4}=x-1\\t=\dfrac{3x-2-x+2}{4}=\dfrac{x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x+1}=x-1\left(x\ge1\right)\\\sqrt[]{x+1}=\dfrac{x}{2}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=x^2-2x+1\left(x\ge1\right)\\x+1=\dfrac{x^2}{4}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2+2\sqrt[]{2}\end{matrix}\right.\)

Gọi chiều dài, chiều rộng hình chữ nhật lần lượt là x ; y > 0, m 

Chu vi hình chữ nhật là : \(P=\left(a+b\right).2=46\)

Nếu tăng chiều dài 5m, giảm chiều rộng 3m thì hình chữ nhật có chiều dài gấp 4 lần chiều rộng : \(a+5=4\left(b-3\right)\)

Ta có hệ phương trình sau : \(\left\{{}\begin{matrix}\left(a+b\right).2=46\\a+5=4\left(b-3\right)\end{matrix}\right.\)

giải hệ ta được a = 15 ; b = 8 

Vậy diện tích hình chữ nhật là : \(a.b=15.8=120\)m²

\(1,ĐKx\ge5\)

\(\sqrt{\left(x-5\right)\left(x+5\right)}+2\sqrt{x-5}=3\sqrt{x+5}+6\)

\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)

\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)

\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)

2a,ĐK \(x\ge0;x\ne9\)

,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)

\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(M=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)