b: ĐKXĐ: \(\begin{cases}5x^2+14x+9\ge0\\ x+1\ge0\\ x^2-x-2\ge0\end{cases}\)

=>(5x+9)(x+1)>=0 và x>=-1 và (x-2)(x+1)>=0

=>(x>=-1 hoặc x<=-9/5) và x>=-1 và (x>=2 hoặc x<=-1)

=>x=-1 hoặc x>=2

Ta có: \(\sqrt{5x^2+14x+9}-5\sqrt{x+1}=\sqrt{x^2-x-2}\)

=>\(\sqrt{\left(5x+9\right)\left(x+1\right)}-5\sqrt{x+1}-\sqrt{\left(x-2\right)\left(x+1\right)}=0\)

=>\(\sqrt{x+1}\left(\sqrt{5x+9}-5-\sqrt{x-2}\right)=0\)

TH1: x+1=0

=>x=-1(nhận)

TH2: \(\sqrt{5x+9}-\sqrt{x-2}-5=0\)

=>\(\sqrt{5x+9}-\sqrt{x-2}=5\)

=>\(5x+9+x-2-2\cdot\sqrt{\left(5x+9\right)\left(x-2\right)}=25\)

=>\(2\cdot\sqrt{\left(5x+9\right)\left(x-2\right)}=6x-11-25=6x-36\)

=>\(\sqrt{\left(5x+9\right)\left(x-2\right)}=3x-18\)

=>\(\begin{cases}3x-18\ge0\\ \left(5x+9\right)\left(x-2\right)=\left(3x-18\right)^2\end{cases}\Rightarrow\begin{cases}x\ge6\\ 9x^2-108x+324=5x^2-10x+9x-18\end{cases}\)

=>\(\begin{cases}x\ge6\\ 9x^2-108x+324-5x^2+x+18=0\end{cases}\Rightarrow\begin{cases}x\ge6\\ 4x^2-107x+342=0\end{cases}\)

\(4x^2-107x+342=0\)

\(\Delta=\left(-107\right)^2-4\cdot4\cdot342=5977\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{107-\sqrt{5977}}{2\cdot4}=\frac{107-\sqrt{5977}}{8}\left(loại\right)\\ x=\frac{107+\sqrt{5977}}{2\cdot4}=\frac{107+\sqrt{5977}}{8}\left(nhận\right)\end{array}\right.\)

1,\(K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{x}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{5}-1\right|+\sqrt{5}+1\right)\)\(=\dfrac{1}{\sqrt{2}}\left|\sqrt{5}-1+\sqrt{5}+1\right|=\dfrac{1}{\sqrt{2}}.2\sqrt{5}\)\(=\sqrt{10}\)

2, \(\sqrt{x-3}-2\sqrt{x^2-3x}=0\left(đk:x\ge3\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1-2\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\left(ktm\right)\end{matrix}\right.\)

Vậy pt có nghiệm x=3

3, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\left(đk:x>-\dfrac{5}{7}\right)\)

\(\Leftrightarrow9x-7=7x+5\)

\(\Leftrightarrow x=6\left(tm\right)\)

4, \(x-5\sqrt{x}+4=0\)(đk: \(x\ge0\))

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\) (tm)

Vậy...

1) Bạn tự làm

2) ĐK: \(x\ge3\)

PT \(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\2\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)

  Vậy ...

3) ĐK: \(x>-\dfrac{5}{7}\)

PT \(\Rightarrow9x-7=7x+5\) \(\Leftrightarrow x=6\)

  Vậy ...

4) ĐK: \(x\ge0\)

PT \(\Leftrightarrow x-4\sqrt{x}-\sqrt{x}+4=0\)

      \(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)

      \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)

  Vậy ...

c: ĐKXĐ: \(x^3+3x^2+x-1\ge0\)

=>\(x^3+x^2+2x^2+2x-x-1\ge0\)

=>(x+1)\(\left(x^2+2x-1\right)\ge0\)

=>-1-\(\sqrt2\) <=x<=-1 hoặc \(x\ge-1+\sqrt2\)

\(x^2+5x+2=4\cdot\sqrt{x^3+3x^2+x-1}\)

=>\(x^2-x+6x-6=4\cdot\sqrt{x^3+3x^2+x-1}-8\)

=>(x-1)(x+6)=\(4\cdot\left(\sqrt{x^3+3x^2+x-1}-2\right)=4\cdot\frac{x^3+3x^2+x-1-4}{\sqrt{x^3+3x^2+x-1}+2}\)

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

=>(x-1)(x+6)=4\(\frac{\left(x-1\right)\left(x^2+4x+5\right)}{\sqrt{x^3+3x^2+x-1}+2}\)

=>(x-1)\(\left\lbrack\frac{4\left(x^2+4x+5\right)}{\sqrt{x^3+3x^2+x-1}+2}-x-6\right\rbrack=0\)

=>x-1=0

=>x=1(nhận)

Câu 1:

ĐK: \(x\geq \frac{-3}{2}\)

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

\(\Rightarrow 2x+3=(3-\sqrt{5})^2=14-6\sqrt{5}\)

\(\Rightarrow x=\frac{11-6\sqrt{5}}{2}\)

Câu 2: ĐK: \(x\geq 0\)

\(\sqrt{5+\sqrt{7x}}=2+\sqrt{7}\)

\(\Rightarrow 5+\sqrt{7x}=(2+\sqrt{7})^2=11+4\sqrt{7}\)

\(\Rightarrow \sqrt{7x}=6+4\sqrt{7}\)

\(\Rightarrow 7x=(6+4\sqrt{7})^2\Rightarrow x=\frac{(6+4\sqrt{7})^2}{7}\)

Câu 3: ĐK: \(x\geq 0\)

\((\sqrt{x}-2)(5-\sqrt{x})=4-x\)

\(\Leftrightarrow 5\sqrt{x}-x-10+2\sqrt{x}=4-x\)

\(\Leftrightarrow 7\sqrt{x}=14\Rightarrow \sqrt{x}=2\Rightarrow x=4\)

Câu 4: ĐK: \(x\ge 1\)

Sửa đề \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)

\(\Leftrightarrow \frac{\sqrt{x-1}}{2}-\frac{3}{2}\sqrt{9}.\sqrt{x-1}+24\sqrt{\frac{1}{64}}\sqrt{x-1}=-17\)

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

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

\(\Leftrightarrow -\sqrt{x-1}=-17\Rightarrow \sqrt{x-1}=17\Rightarrow x=17^2+1=290\)

a.

ĐKXĐ: \(x\ge-5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+6=0\\\sqrt{x+5}+4=3x+5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\\sqrt{x+5}=3x+1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{3}\\x+5=9x^2+6x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{3}\\9x^2+5x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=\dfrac{4}{9}\end{matrix}\right.\)

b. Bạn coi lại đề, pt này nghiệm rất xấu

c.

ĐKXĐ: \(1\le x\le7\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

Tìm x để căn có nghĩa ak mn giúp e với ak

\(a,ĐK:\dfrac{3}{x+7}\ge0\Leftrightarrow x+7>0\left(3>0;x+7\ne0\right)\Leftrightarrow x>-7\\ b,ĐK:\dfrac{-2}{5-x}\ge0\Leftrightarrow5-x< 0\left(2-< 0;5-x\ne0\right)\Leftrightarrow x>5\\ c,ĐK:x^2-7x+10\ge0\Leftrightarrow\left(x-5\right)\left(x-2\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le2\end{matrix}\right.\)

\(d,ĐK:x^2-8x+10\ge0\Leftrightarrow\left(x-4-\sqrt{6}\right)\left(x-4+\sqrt{6}\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-4-\sqrt{6}\ge0\\x-4+\sqrt{6}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-4-\sqrt{6}\le0\\x-4+\sqrt{6}\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge4+\sqrt{6}\\x\ge4-\sqrt{6}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le4+\sqrt{6}\\x\le4-\sqrt{6}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4+\sqrt{6}\\x\le4-\sqrt{6}\end{matrix}\right.\)

\(e,ĐK:9x^2+1\ge0\Leftrightarrow x\in R\left(9x^2+1\ge1>0\right)\)

a: 

ĐKXĐ: x>=5/2

\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)

=>\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\cdot\sqrt{2x-5}}=14\)

=>\(\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)

=>\(\sqrt{2x-5}+1+\sqrt{2x-5}+3=14\)

=>\(2\sqrt{2x-5}+4=14\)

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

=>2x-5=25

=>2x=30

=>x=15

b: \(x^2-4x=\sqrt{x+2}\)

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

=>x^4-8x^3+16x^2-x-2=0 và x^2-4x>=0

=>(x^2-5x+2)(x^2-3x-1)=0 và x^2-4x>=0

=>\(\left[{}\begin{matrix}x=\dfrac{5+\sqrt{17}}{2}\\x=\dfrac{3-\sqrt{13}}{2}\end{matrix}\right.\)

a) ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\0>x\ge-1\end{matrix}\right.\). Để pt có nghiệm => x>0=> \(x\ge1\) pt<=> \(x-\sqrt{1-\dfrac{1}{x}}=\sqrt{x-\dfrac{1}{x}}.Bìnhphương2vetaco\left(x-\sqrt{1-\dfrac{1}{x}}\right)^2=x-\dfrac{1}{x}\)\(\Leftrightarrow x^2+1-\dfrac{1}{x}-2x\sqrt{1-\dfrac{1}{x}}=x-\dfrac{1}{x}\Leftrightarrow x^2-x+1=2\sqrt{x^2-x}\Leftrightarrow\left(\sqrt{x^2-x}-1\right)^2=0\Leftrightarrow x^2-x=1\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

b) ĐKXĐ\(0\le x\le1\) pt \(\Leftrightarrow\left(\sqrt{x^2+x}+\sqrt{x-x^2}\right)^2=\left(x+1\right)^2\Leftrightarrow2x+2x.\sqrt{1-x^2}=x^2+2x+1\Leftrightarrow x^2-2x\sqrt{1-x^2}+1-x^2+x^2=0\Leftrightarrow\left(x-\sqrt{1-x^2}\right)^2+x^2=0\)