a.

ĐKXĐ: \(x\ge3\)

(Tốt nhất bạn kiểm tra lại đề cái căn đầu tiên của \(\sqrt{x-3}\) là căn bậc 2 hay căn bậc 3). Vì nhìn ĐKXĐ thì thấy căn bậc 2 là không hợp lý rồi đó

Pt tương đương:

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

Do \(x\ge3\Rightarrow x-2>0\Rightarrow\left(x+1\right)\left(x-2\right)>0\)

\(\Rightarrow\sqrt{x-3}+\sqrt[3]{x^2+1}+\left(x+1\right)\left(x-2\right)>0\)

Pt vô nghiệm

b.

ĐKXĐ: \(x\ge-\dfrac{3}{2}\)

Pt: \(2x+3-\sqrt{2x+3}-\left(4x^2-6x+2\right)=0\)

Đặt \(\sqrt{2x+3}=t\ge0\) ta được:

\(t^2-t-\left(4x^2-6x+2\right)=0\)

\(\Delta=1+4\left(4x^2-6x+2\right)=\left(4x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t_1=\dfrac{1+4x-3}{2}=2x-1\\t_2=\dfrac{1-4x+3}{2}=2-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=2x-1\left(x\ge\dfrac{1}{2}\right)\\\sqrt{2x+3}=2-2x\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4x^2-4x+1\left(x\ge\dfrac{1}{2}\right)\\2x+3=4x^2-8x+4\left(x\le1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{17}}{4}\\x=\dfrac{5-\sqrt{21}}{4}\end{matrix}\right.\)

c.

ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow2\left(x^2-4x+4\right)+2\left(x+1\right)-5\left(x-2\right)\sqrt{x+1}=0\)

\(\Leftrightarrow2\left(x-2\right)^2-5\left(x-2\right)\sqrt{x+1}+2\left(x+1\right)=0\)

Đặt \(\left\{{}\begin{matrix}x-2=a\\\sqrt{x+1}=b\end{matrix}\right.\) ta được:

\(2a^2-5ab+2b^2=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=x-2\left(x\ge2\right)\\\sqrt{x+1}=2x-4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-4x+4\\x+1=4x^2-16x+16\end{matrix}\right.\) (\(x\ge2\))

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\) (đã loại nghiệm)

a, ĐKXĐ : \(x\ge\dfrac{1}{2}\)

 PT <=> 2x - 1 = 5

<=> x = 3 ( TM )

Vậy ...

b, ĐKXĐ : \(x\ge5\)

PT <=> x - 5 = 9

<=> x = 14 ( TM )

Vậy ...

c, PT <=> \(\left|2x+1\right|=6\)

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

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

Vậy ...

d, PT<=> \(\left|x-3\right|=3-x\)

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

Vậy phương trình có vô số nghiệm với mọi x \(x\le3\)

e, ĐKXĐ : \(-\dfrac{5}{2}\le x\le1\)

PT <=> 2x + 5 = 1 - x

<=> 3x = -4

<=> \(x=-\dfrac{4}{3}\left(TM\right)\)

Vậy ...

f ĐKXĐ : \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)

PT <=> \(x^2-x=3-x\)

\(\Leftrightarrow x=\pm\sqrt{3}\) ( TM )

Vậy ...

a) \(\sqrt{2x-1}=\sqrt{5}\)          (x \(\ge\dfrac{1}{2}\))

<=> 2x - 1 = 5

<=> x = 3 (tmđk)

Vậy S = \(\left\{3\right\}\)

b) \(\sqrt{x-5}=3\)           (x\(\ge5\))

<=> x - 5 = 9

<=> x = 4 (ko tmđk)

Vậy x \(\in\varnothing\)

c) \(\sqrt{4x^2+4x+1}=6\)          (x \(\in R\))

<=> \(\sqrt{\left(2x+1\right)^2}=6\)

<=> |2x + 1| = 6

<=> \(\left[{}\begin{matrix}\text{2x + 1=6}\\\text{2x + 1}=-6\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)(tmđk)

Vậy S = \(\left\{\dfrac{5}{2};\dfrac{-7}{2}\right\}\)

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

Ta có: \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}\ge2\sqrt{\sqrt{\frac{2x-1}{x+1}}\cdot\sqrt{\frac{x+1}{2x-1}}}=2\) (BĐT Cô-si)

Mà \(\sqrt{\frac{2x-1}{x+1}}+\sqrt{\frac{x+1}{2x-1}}=2\) (theo đề bài)

Suy ra dấu bằng phải xảy ra \(\Rightarrow\sqrt{\frac{2x-1}{x+1}}=\sqrt{\frac{x+1}{2x-1}}\) \(\Leftrightarrow\frac{2x-1}{x+1}=\frac{x+1}{2x-1}\) \(\Leftrightarrow\left(2x-1\right)^2=\left(x+1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+1\\2x-1=-x-1\end{matrix}\right.\Leftrightarrow\) \(x=2\) (tmđkxđ) hoặc \(x=0\) (không tmđkxđ)

Vậy \(S=\left\{2\right\}\).

Bạn đừng quên tự tìm ĐKXĐ cho câu a nhé bạn.

c) \(x+\frac{1}{x}+4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+6=0\) ĐKXĐ: \(x>0\)

Vì \(x>0\Rightarrow x+\frac{1}{x}+4\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)+6>0\)

Vậy \(S=\varnothing\).

a)ĐK:\(\begin{cases}25x^2-9 \ge 0\\5x+3 \ge 0\\\end{cases}\)

`<=>` \(\begin{cases}(5x-3)(5x+3) \ge 0\\5x+3 \ge 0\\\end{cases}\)

`<=>` \(\begin{cases}\left[ \begin{array}{l}x\ge \dfrac35\\x \le -\dfrac35\end{array} \right.\\\end{cases}\)

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x \ge \dfrac35\end{array} \right.\)

`pt<=>\sqrt{5x+3}(\sqrt{5x-3}-2)=0`

`<=>` \(\left[ \begin{array}{l}5x+3=0\\\sqrt{5x-3}=2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\5x-3=4\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x=7/5\end{array} \right.\) 

`b)sqrt{x-3}/sqrt{2x+1}=2`

ĐK:\(\begin{cases}x-3 \ge 0\\2x+1>0\\\end{cases}\)

`<=>x>=3`

`pt<=>sqrt{x-3}=2sqrt{2x+1}`

`<=>x-3=8x+4`

`<=>7x=7`

`<=>x=1(l)`

`c)sqrt{x^2-2x+1}+sqrt{x^2-4x+4}=3`

`<=>sqrt{(x-1)^2}+sqrt{(x-2)^2}=3`

`<=>|x-1|+|x-2|=3`

`**x>=2`

`pt<=>x-1+x-2=3`

`<=>2x=6`

`<=>x=3(tm)`

`**x<=1`

`pt<=>1-x+2-x=3`

`<=>3-x=3`

`<=>x=0(tm)`

`**1<=x<=2`

`pt<=>x-1+2-x=3`

`<=>=-1=3` vô lý

Vậy `S={0,3}`

Bạn xem lại đề câu b và c nhé !

a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)

\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)

\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ

\(\Rightarrow x\ge2\) thỏa mãn đề.

d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)

Pt tương đương :

\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )

e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)

\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)

Phương trình (1) tương đương :

\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )