\(\sqrt{2x-1}\ge0\)

\(\Rightarrow BPT\ge0\) khi

\(3-2x-x^2\ge0\)

\(\Leftrightarrow x^2+2x-3\le0\)

\(\Leftrightarrow\left(x+1\right)^2-4\le0\)

\(\Leftrightarrow\left(x+1\right)^2\le4\)

\(\Leftrightarrow x+1\le2\)

\(\Rightarrow x\le1\)

a: ĐKXĐ: x>=-9/2 và x<>0

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

=>\((3 - \sqrt{9 + 2x})^2 = \frac{4x^2}{(3 + \sqrt{9 + 2x})^2}\)

\(\frac{2x^2}{(3 - \sqrt{9 + 2x})^2} < x + 21\)

=>\(\frac{2x^2 \cdot (3 + \sqrt{9 + 2x})^2}{4x^2} < x + 21\)

=>\(\frac{(3 + \sqrt{9 + 2x})^2}{2} < x + 21\)

=>\((3 + \sqrt{9 + 2x})^2 < 2x + 42\)

=>\(9 + 6\sqrt{9 + 2x} + (9 + 2x) < 2x + 42\)

=>\(18 + 2x + 6\sqrt{9 + 2x} < 2x + 42\)

=>\(6\sqrt{9 + 2x} < 24\)

=>\(\sqrt{2x+9}<4\)

=>2x+9<16

=>2x<7

=>x<7/2

Kết hợp ĐKXĐ, ta được: -9/2<=x<7/2 và x<>0

a,ĐK: x\(\ge\)1

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

⇔\(\sqrt{\left(\sqrt{x-1}-1\right)^2}\)=\(\sqrt{2}\)

⇔\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{2}\)

TH1:\(\sqrt{x-1}\)-1≥0⇒\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{x-1}\)-1   bn tự giải ra nha

TH2:\(\sqrt{x-1}\)-1<0⇒\(\left|\sqrt{x-1}-1\right|\)=1-\(\sqrt{x-1}\)    bn tự lm nha

`sqrt{x-2}-2>=sqrt{2x-5}-sqrt{x+1}`

`đk:x>=5/2`

`bpt<=>\sqrt{x-2}+\sqrt{x+1}>=\sqrt{2x-5}+2`

`<=>x-2+x+1+2\sqrt{(x-2)(x+1)}>=2x-5+4+4\sqrt{2x-5}`

`<=>2x-1+2\sqrt{(x-2)(x+1)}>=2x-1+4\sqrt{2x-5}`

`<=>2\sqrt{(x-2)(x+1)}>=4\sqrt{2x-5}`

`<=>sqrt{x^2-x-2}>=2sqrt{2x-5}`

`<=>x^2-x-2>=4(2x-5)`

`<=>x^2-x-2>=8x-20`

`<=>x^2-9x+18>=0`

`<=>(x-3)(x-6)>=0`

`<=>` \(\left[ \begin{array}{l}x \ge 6\\x \le 3\end{array} \right.\) 

Kết hợp đkxđ:

`=>` \(\left[ \begin{array}{l}x \ge 6\\\dfrac52 \le x \le 3\end{array} \right.\) 

a: \(2x^2-11x+21=3\cdot\sqrt[3]{4x-4}\)

=>\(2x^2-6x-5x+15=3\cdot\sqrt[3]{4x-4}-6\)

=>\(\left(x-3\right)\left(2x-5\right)=3\cdot\frac{4x-4-8}{\sqrt[3]{\left(4x-4\right)^2}+2\cdot\sqrt[3]{4x-4}+4}\)

=>\(\left(x-3\right)\left(2x-5\right)-3\cdot\frac{4x-12}{\sqrt[3]{\left(4x-4\right)^2}+2\cdot\sqrt[3]{4x-4}+4}=0\)

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

=>x-3=0

=>x=3

a) ĐK: x>=-2

=> \(\sqrt{x+5}+\sqrt{x+2}>0\)

Nhân liên hợp:

\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)

<=> \(\left(x+5-x-2\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)

<=> \(3\left(1+\sqrt{x^2+7x+10}\right)=3\)

<=>1+\(\sqrt{\left(x+5\right)\left(x+2\right)}=1\)

<=> \(\sqrt{\left(x+5\right)\left(x+2\right)}=0\)

<=> (x+5) (x+2) =0

<=> x=-5 hoac x=-2

-Do x>= -2.

Vay x=-2

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