a/

ĐKXĐ: \(x\ge\frac{1}{4}\)

\(\Leftrightarrow\sqrt{5x+1}\le\sqrt{4x-1}+3\sqrt{x}\)

\(\Leftrightarrow5x+1\le13x-1+6\sqrt{x\left(4x-1\right)}\)

\(\Leftrightarrow3\sqrt{x\left(4x-1\right)}\ge1-4x\)

Do \(x\ge\frac{1}{4}\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP\le0\end{matrix}\right.\) BPT luôn đúng

Vậy nghiệm của BPT đã cho là \(x\ge\frac{1}{4}\)

b/

ĐKXĐ: \(\left[{}\begin{matrix}x\ge\frac{-5+2\sqrt{5}}{5}\\x\le\frac{-5-2\sqrt{5}}{5}\end{matrix}\right.\)

Đặt \(\sqrt{5x^2+10x+1}=t\ge0\Rightarrow x^2+2x=\frac{t^2-1}{5}\)

BPT trở thành:

\(t\ge7-\frac{t^2-1}{5}\Leftrightarrow t^2+5t-36\ge0\)

\(\Rightarrow\left[{}\begin{matrix}t\le-9\left(l\right)\\t\ge4\end{matrix}\right.\)

\(\Rightarrow\sqrt{5x^2+10x+1}\ge4\)

\(\Leftrightarrow5x^2+10x-15\ge0\)

\(\Rightarrow\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)

c/

ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow x^2-4+1-\sqrt{x-1}+2-\sqrt{2x}< 0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\frac{x-2}{1+\sqrt{x-1}}-\frac{2\left(x-2\right)}{2+\sqrt{2x}}< 0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-\frac{1}{1+\sqrt{x-1}}-\frac{2}{2+\sqrt{2x}}\right)< 0\)

\(\Leftrightarrow\left(x-2\right)\left(x+\frac{\sqrt{x+1}}{1+\sqrt{x-1}}+\frac{\sqrt{2x}}{2+\sqrt{2x}}\right)< 0\)

\(\Leftrightarrow x-2< 0\Rightarrow x< 2\) (phần trong ngoặc to luôn dương)

Vậy nghiệm của BPT là \(1\le x< 2\)

a. ĐKXĐ: $x\geq 2$ hoặc $x=1$

PT $\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}$

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

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)

b.

PT $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$

$\Leftrightarrow |x-2|=|2x-3|$

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

c. ĐKXĐ: $x=2$ hoặc $x\geq 3$

PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$

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

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

d.

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

$\Leftrightarrow |2x-1|=|x-3|$

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

a: Ta có: \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)

\(\Leftrightarrow x^2-3x+2=x-1\)

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

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)

\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)

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

c: Ta có: \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)

\(\Leftrightarrow x^2-5x+6=x-2\)

\(\Leftrightarrow x^2-6x+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

a/ Với x = - 1 thì BĐT đúng.

Xét \(x\ne-1\)

Ta có: \(x^3+\left(3x^2-4x-4\right)\sqrt{x+1}\le0\)

\(\Leftrightarrow x^3+3x^2\sqrt{x+1}-4\sqrt{\left(x+1\right)^3}\le0\)

 \(\Leftrightarrow\frac{x^3}{\sqrt{\left(x+1\right)^3}}+3.\frac{x^2}{\sqrt{\left(x+1\right)^2}}-4\le0\)

Đặt \(\frac{x}{\sqrt{x+1}}=t\)thì ta có bpt thành

\(t^3+3t^2-4\le0\)

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

Tới đây thì đơn giản rồi b làm tiếp nhé.

Câu b còn lại mình nghĩ chỉ cần bình phương rồi chuyển cái chứa căn sang 1 bên không chứa căn sang 1 bên. Sau đó bình phương thêm 1 lần nữa rồi đặt nhân tử chung là ra :)

uk thank nha 

\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy pt vô nghiệm

\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)

\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

a) \(\sqrt{x-5}=\sqrt{3-x}\)

⇔\(\left(\sqrt{x-5}\right)^2=\left(\sqrt{3-x}\right)^2\)

⇔\(x-5=3-x\)

⇔\(x=4\)

b) \(\sqrt{4-5x}=\sqrt{2-5x}\)

⇔\(\left(\sqrt{4-5x}\right)^2=\left(\sqrt{2-5x}\right)^2\)

⇔\(4-5x=2-5x\)

⇔\(2=0\) (Vô lí)

1.

$x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{(x-3)^2}=x+3+|x-3|$

$=x+3+(3-x)=6$

2.

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

$=|x+2|-|x|=x+2-(-x)=2x+2$
3.

$\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}$

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

$=|\sqrt{x^2-1}+1|+|\sqrt{x^2-1}-1|$

$=\sqrt{x^2-1}+1+|\sqrt{x^2-1}-1|$

4.

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

$=\frac{|x-1|}{x-1}=\frac{x-1}{x-1}=1$

5.

$|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=2-x+\frac{\sqrt{(x-2)^2}}{x-2}$
$=2-x+\frac{|x-2|}{x-2}|=2-x+\frac{2-x}{x-2}=2-x+(-1)=1-x$

6.

$2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\frac{\sqrt{(x-5)^2}}{x-5}$

$=2x-1-\frac{|x-5|}{x-5}$

1: ĐKXĐ: x>1/2

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

x^2-2x+1>=0

=>x^2>=2x-1

=>\(\dfrac{x}{\sqrt{2x-1}}>=1\)

Dấu = xảy ra khi x=1

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

=>x^4-4x+3>=0

=>x^4>=4x-3

=>\(\dfrac{x}{\sqrt[4]{4x-3}}>=1\)

=>VT>=2

Dấu = xảy ra khi x=1

2: 4x-1=x+x+2x-1

5x-2=x+2x-1+2x-1

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

=>\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\dfrac{9}{\sqrt{x}+\sqrt{x}+\sqrt{2x-1}}\)

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

=>\(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}< =\sqrt{3\left(4x-1\right)}\)

=>\(\dfrac{2}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\dfrac{3\sqrt{3}}{\sqrt{4x-1}}\)

Tương tự, ta cũng có: \(\dfrac{1}{\sqrt{x}}+\dfrac{2}{\sqrt{2x-1}}>=\dfrac{3\sqrt{3}}{\sqrt{5x-2}}\)

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

Dấu = xảy ra khi x=1

\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)

1.ĐK: \(x\ge\dfrac{1}{4}\)

bpt\(\Leftrightarrow5x+1+4x-1-2\sqrt{20x^2-x-1}< 9x\)

\(\Leftrightarrow2\sqrt{20x^2-x-1}>0\)

\(\Leftrightarrow20x^2-x-1>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{-1}{5}\\x>\dfrac{1}{4}\end{matrix}\right.\)

2.ĐK: \(-2\le x\le\dfrac{5}{2}\)

bpt\(\Leftrightarrow x+2+3-x-2\sqrt{-x^2+x+6}< 5-2x\)

\(\Leftrightarrow2x< 2\sqrt{-x^2+x+6}\)

\(\Leftrightarrow x^2< -x^2+x+6\)

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

\(\Leftrightarrow\dfrac{-3}{2}< x< 2\)

3. ĐK: \(\left\{{}\begin{matrix}12+x-x^2\ge0\\x\ne11\\x\ne\dfrac{9}{2}\end{matrix}\right.\)

.bpt\(\Leftrightarrow\sqrt{12+x-x^2}\left(\dfrac{1}{x-11}-\dfrac{1}{2x-9}\right)\ge0\)

\(\Leftrightarrow\sqrt{-x^2+x+12}.\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)

\(\Rightarrow\dfrac{x+2}{\left(x-11\right)\left(2x-9\right)}\ge0\)

\(\Leftrightarrow\dfrac{x+2}{2x^2-31x+99}\ge0\)

*Xét TH1: \(\left\{{}\begin{matrix}x+2\ge0\\2x^2-31x+99>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x< \dfrac{9}{2}\\x>11\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2\le x< \dfrac{9}{2}\\x>11\end{matrix}\right.\)

*Xét TH2: \(\left\{{}\begin{matrix}x+2\le0\\2x^2-31x+99< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\\dfrac{9}{2}< x< 11\end{matrix}\right.\)\(\Rightarrow\dfrac{9}{2}< x< 11\)