\(a)\)

\(\frac{1}{x+1}-\frac{x-1}{x}=\frac{3x+1}{x\left(x+1\right)}\)

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

\(\Leftrightarrow x^2-2x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b)\)

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

\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)

\(\Leftrightarrow x^2+2x+1=x^2+10\)

\(\Leftrightarrow2x-9=0\)

\(\Leftrightarrow2x=9\)

\(\Leftrightarrow x=\frac{2}{9}\)

Rồi sao? đề bài?

\(4(x+1)^2-(2x-1)^2-8(x-1)(x+1)=11\)

\(\Leftrightarrow4\left(x^2+2x+1\right)-\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)

\(\Leftrightarrow4x^2+8x+4-4x^2+4x-1-8x^2+8=11\)

\(\Leftrightarrow-8x^2+12x+11=11\)

\(\Leftrightarrow-4x\left(2x-3\right)=0\)

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

Ta có:

\(4\left(x+1\right)^2-\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\\ \Leftrightarrow4x^2+8x+4-4x^2+4x-1-8x^2+8=11\\ \Leftrightarrow-8x^2+12x=0\\ \Leftrightarrow-4x\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

Bài 2:

a: \(\Leftrightarrow x-1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

=>\(x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)

b: \(\Leftrightarrow x+3\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

=>\(x\in\left\{-2;-4;0;-6;2;-8;12;-18\right\}\)

c: \(\Leftrightarrow x+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(x\in\left\{-2;-4;-1;-5;0;-6;1;-7;3;-9;9;-15\right\}\)

d: =>x+1+15 chia hết cho x+1

=>\(x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

=>\(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

Bài 9 : Tìm x, biết :

a, (x - 2)(x - 3) + (x - 2) - 1 = 0

\(\Leftrightarrow\left(x-2\right)\left(x-3+1\right)-1=0\)

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2+1\right)\left(x-2-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy x ={1; 3}

b, (x + 2)² - 2x(2x + 3) = (x + 1)²

\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)

\(\Leftrightarrow\left(x+2+x+1\right)\left(x+2-x-1\right)-2x\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(1-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x³ + x² = 2x

\(\Leftrightarrow6x^3+x^2-2x=0\)

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

\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)

\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)

\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

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

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

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}=2x+1\Rightarrow\left|x-3\right|=2x+1\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)

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

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

a)Ta có: \(\dfrac{x+3}{x+1}+\dfrac{1}{3}\ge0\)

\(\Leftrightarrow\dfrac{3x+9+x+1}{3\left(x+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{4x+10}{3x+3}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1>0\\4x+10\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-1\\x\le-\dfrac{5}{2}\end{matrix}\right.\)

b) Ta có: \(\dfrac{x+2}{x+3}+\dfrac{1}{3}\le0\)

\(\Leftrightarrow\dfrac{3x+6+x+3}{3\left(x+3\right)}\le0\)

\(\Leftrightarrow\dfrac{4x+9}{3x+9}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+9>0\\4x+9\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow-3< x\le-\dfrac{9}{4}\)

a)\(\dfrac{x+3}{x+1}\ge-\dfrac{1}{3}\left(x\ne-1\right)\)

\(\Leftrightarrow\dfrac{x+3}{x+1}+\dfrac{1}{3}\ge0\)

\(\Leftrightarrow\dfrac{3x+9+x+1}{3x+3}\ge0\)

\(\Leftrightarrow\dfrac{4x+10}{3x+3}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+10\ge0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+10\le0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{5}{2}\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{-5}{2}\\x< -1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>-1\\x\le\dfrac{-5}{2}\end{matrix}\right.\)

 b) \(\dfrac{x+2}{x+3}\le-\dfrac{1}{3}\left(x\ne-3\right)\)

\(\dfrac{x+2}{x+3}+\dfrac{1}{3}\le0\)

\(\Leftrightarrow\dfrac{3x+6+x+3}{3x+9}\le0\)

\(\Leftrightarrow\dfrac{4x+9}{3x+9}\le0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+9\ge0\\3x+9< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+9\le0\\3x+9>0\end{matrix}\right.\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{9}{4}\\x< -3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-\dfrac{9}{4}\\x>-3\end{matrix}\right.\end{matrix}\right.\)    

TH1: loại

TH2: TM

Vậy no của BPT là :\(-\dfrac{9}{4}\ge x>-3\)

chúc bạn học tốt

b: \(8x-\dfrac{1}{\sqrt{2+x}}+5\)

a. -> \(3x^2-\dfrac{1}{\sqrt{5}}.\left(x-m\right)-15=3\)