ĐKXĐ: \(-2\le x\le3\)

\(\dfrac{\sqrt{-x^2+x+6}}{2x+5}-\dfrac{\sqrt{-x^2+x+6}}{x-4}\ge0\)

\(\Leftrightarrow\sqrt{-x^2+x+6}\left(\dfrac{1}{2x+5}-\dfrac{1}{x-4}\right)\ge0\)

\(\Leftrightarrow\dfrac{\left(-x-9\right)\sqrt{x^2+x+6}}{\left(2x+5\right)\left(x-4\right)}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+6=0\\\dfrac{-x-9}{\left(2x+5\right)\left(x-4\right)}\ge0\end{matrix}\right.\) \(\Leftrightarrow-2\le x\le3\)

Hoặc có thể biện luận như sau:

Ta có: \(\left\{{}\begin{matrix}2x+5>0;\forall x\in\left[-2;3\right]\\x-4< 0;\forall x\in\left[-2;3\right]\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{-x^2+x+6}}{2x+5}\ge0\\\dfrac{\sqrt{-x^2+x+6}}{x-4}\le0\end{matrix}\right.\) ; \(\forall x\in\left[-2;3\right]\)

Do đó nghiệm của BPT là \(-2\le x\le3\)

Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

`=>(\sqrt{3x^2})^2=(x+2)^2`

`=>3x^2=x^2+4x+4`

`=>3x^2-x^2-4x-4=0`

`=>2x^2-4x-4=0`

`=>x^2-2x-2=0`

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

`=>(x-1)^2=3`

`=>(x-1)^2=(\pm \sqrt{3})^2`

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

Bài 1 

a, `3x-7\sqrt{x}+4=0`            ĐKXĐ : `x>=0`

`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`

`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`

`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`

TH1 :

`3\sqrt{x}-4=0`

`<=>\sqrt{x}=4/3`

`<=>x=16/9` ( tm )

TH2

`\sqrt{x}-1=0`

`<=>\sqrt{x}=1` (tm)

Vậy `S={16/9;1}`

b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17`     ĐKXĐ : `x>=1`

`<=>(1/2-9/2+3)\sqrt{x-1}=-17`

`<=>-\sqrt{x-1}=-17`

`<=>\sqrt{x-1}=17`

`<=>x-1=289`

`<=>x=290` ( tm )

Vậy `S={290}`

Bài 1: 

a) Ta có: \(3x-7\sqrt{x}+4=0\)

\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

b) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\cdot\left(-1\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

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

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

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

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

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

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

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

Mình đề câu a phải như vậy nè:

\(a,\hept{\begin{cases}\frac{1}{x-2}+\frac{1}{y-1}=1\\\frac{2}{x-2}-\frac{3}{y-1}=1\end{cases}}\)\(Đkxđ:\hept{\begin{cases}x\ne2\\y\ne1\end{cases}}\)

Đặt: \(X=\frac{1}{x-2};Y=\frac{1}{y-1}\)

Ta có hệ sau:

 \(\hept{\begin{cases}X+Y=1\\2X-3Y=1\end{cases}\Leftrightarrow\hept{\begin{cases}X=1-Y\\2\left(1-Y\right)-3Y=1\end{cases}}}\Leftrightarrow\hept{\begin{cases}X=1-Y\\2-5Y=1\end{cases}\Leftrightarrow\hept{\begin{cases}X=\frac{4}{5}\\Y=\frac{1}{5}\end{cases}}}\)

Với \(X=\frac{4}{5}\Rightarrow\frac{1}{x-2}=\frac{4}{5}\Leftrightarrow4\left(x-2\right)=5\Leftrightarrow x=\frac{13}{4}\)

Với \(Y=\frac{1}{5}\Rightarrow\frac{1}{y-1}=\frac{1}{5}\Leftrightarrow y-1=5\Leftrightarrow y=6\)

Vậy nghiệm của hệ pt là: \(\left(x;y\right)=\left(\frac{13}{4};6\right)\)

Câu b e nghĩ đề như vậy nè:

\(b,\hept{\begin{cases}\frac{7}{\sqrt{x-7}}-\frac{4}{\sqrt{y+6}}=\frac{5}{3}\\\frac{5}{\sqrt{x-7}}+\frac{3}{\sqrt{y+6}}=\frac{3}{6}\end{cases}}\) \(Đkxđ:\hept{\begin{cases}x>7\\x>-6\end{cases}}\)

Đặt \(\frac{1}{\sqrt{x-7}}=a\left(a>0\right);\frac{1}{\sqrt{y+6}}=b\left(b>0\right)\)

Ta có hệ pt mới: \(\hept{\begin{cases}7a-4b=\frac{5}{3}\\5a+3b=\frac{13}{6}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{1}{6}\end{cases}}\left(tmđk\right)\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{\sqrt{x-7}}=\frac{1}{3}\\\frac{1}{\sqrt{y+6}}=\frac{1}{6}\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x-7}=3\\\sqrt{y+6}=6\end{cases}}\Leftrightarrow\hept{\begin{cases}x-7=9\\x+6=36\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16\\y=30\end{cases}\left(tmđk\right)}\)

Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(16;30\right)\)

\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}\)

a: \(f\left(-x\right)=-2\cdot\left(-x\right)^3+3\cdot\left(-x\right)\)

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

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

=-f(x)

Vậy: f(x) là hàm số lẻ

c: TXĐ: D=[-2;2]

Nếu \(x\in D\Leftrightarrow-x\in D\)

\(f\left(-x\right)=\sqrt{6-3\cdot\left(-x\right)}-\sqrt{6+3\cdot\left(-x\right)}\)

\(=\sqrt{6+3x}-\sqrt{6-3x}\)

\(=-f\left(x\right)\)

Vậy: f(x) là hàm số lẻ

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

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

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

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

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

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)\)

\(=3\sqrt{x}-6\)

b) \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)

\(\Leftrightarrow3\sqrt{x}-6=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)   (1)

ĐKXĐ: \(x>0\)

\(\left(1\right)\Leftrightarrow3x-6\sqrt{x}=4\sqrt{x}-1\)

\(\Leftrightarrow3x-6\sqrt{x}-4\sqrt{x}+1=0\)

\(\Leftrightarrow3x-10\sqrt{x}+1=0\)   (2)

Đặt \(t=\sqrt{x}\ge0\)

\(\left(2\right)\Leftrightarrow3t^2-10t+1=0\)

\(\Delta'=25-4=22\)

Phương trình có hai nghiệm phân biệt:

\(t_1=\dfrac{5+\sqrt{22}}{3}\) (nhận)

\(t_2=\dfrac{5-\sqrt{22}}{3}\) (nhận)

Với \(t=\dfrac{5+\sqrt{22}}{3}\) \(\Leftrightarrow\sqrt{x}=\dfrac{5+\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47+10\sqrt{22}}{9}\) (nhận)

Với \(t=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow\sqrt{x}=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47-10\sqrt{22}}{9}\) (nhận)

Vậy \(x=\dfrac{47+10\sqrt{22}}{9};x=\dfrac{47-10\sqrt{22}}{9}\) thì \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)

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

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

b: P=(4căn x-1)/căn x

=>3x-6căn x-4căn x+1=0

=>3x-10căn x+1=0

=>x=(47+10căn 22)/9 hoặc x=(47-10căn 22)/9