a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

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

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

câu 1:

a)x-1=5-x\(\Leftrightarrow\)x+x=5+1\(\Leftrightarrow\)2x=6\(\Leftrightarrow\)x=3

Vậy tập nghiệm của PT (a) là S={3}

b)3+x=2-x\(\Leftrightarrow\)x+x=2-3\(\Leftrightarrow\)2x=-1\(\Leftrightarrow\)x=-0,5

Vậy tập nghiệm của PT (b) là:S={-0,5}

câu 2:

a) 3x+7=2x-3\(\Leftrightarrow\)3x-2x=-3-7\(\Leftrightarrow\)x=-10

Vậy tập nghiệm của PT (a) là:S={-10}

b)4-(x-2)=(3-2x)\(\Leftrightarrow\)4-x+2=3-2x\(\Leftrightarrow\)-x+2x=-4+3-2\(\Leftrightarrow\)x=-3

Vậy tập nghiệm của PT (b) là:S={-3}

Câu 3:

a)\(\dfrac{5x-4}{2}=\dfrac{16x+1}{7}\Leftrightarrow\dfrac{7\left(5x-4\right)}{14}=\dfrac{2\left(16x+1\right)}{14}\)

\(\Leftrightarrow\)35x-28=32x+2\(\Leftrightarrow\)35x-32x=2+28\(\Leftrightarrow\)3x=30\(\Leftrightarrow\)x=10

Vậy tập nghiệm của PT (a) là :S={10}

b)\(\dfrac{12x+5}{3}=\dfrac{2x-7}{4}\Leftrightarrow\dfrac{4\left(12x+5\right)}{12}=\dfrac{3\left(2x-7\right)}{12}\)

\(\Leftrightarrow\)48x+20=6x-21\(\Leftrightarrow\)48x-6x=-20-21\(\Leftrightarrow\)42x=-41\(\Leftrightarrow\)x=\(-\dfrac{41}{42}\)

Vậy tập nghiệm của PT (b) là:S={\(-\dfrac{41}{42}\)}

ĐKXĐ: \(x\ne\left\{2;4\right\}\)

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

\(\Rightarrow\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right)\left(x-4\right)\)

\(\Leftrightarrow x^2-7x+12+x^2-4x+4=-x^2+6x-8\)

\(\Leftrightarrow3x^2-17x+24=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\) (thỏa mãn)

\(\Leftrightarrow\left(\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}+\sqrt{a}\right).\dfrac{\sqrt{a}-2}{\sqrt{a}-1}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{a}-2}+\sqrt{a}.\dfrac{\sqrt{a}-2}{\sqrt{a}-1}\)

\(\Leftrightarrow\dfrac{1+\sqrt{a}\sqrt{a}-2}{\sqrt{a}-2}.\dfrac{\sqrt{a}-2}{\sqrt{a}-1}\)

\(\Leftrightarrow\dfrac{1+a-2a}{\sqrt{a}-1}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}\Leftrightarrow\sqrt{a}-1\)

a. Với a = -3 ta được:

\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}+\dfrac{27-3}{x^2-9}=0\)

\(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{24}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow x^2+6x+9-x^2+6x-9+24=0\)

\(\Leftrightarrow12x+24=0\)

\(\Leftrightarrow x=-2\)

Giải phương trình :

\(\dfrac{x-a}{x+a}-\dfrac{x+a}{x-a}+\dfrac{3a^2+a}{x^2-a^2}=0\)

a) Với a = -3

\(\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}+\dfrac{27+3}{x^2-3^2}=0\)

ĐKXĐ : \(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne3\end{matrix}\right.\)

Ta có : \(\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}+\dfrac{27+3}{x^2-3^2}\)

\(\Leftrightarrow\) \(\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{27+3}{\left(x+3\right)\left(x-3\right)}=0\)

Khử mẫu ta có : \(\left(x-3\right)^2-\left(x+3\right)^2+27+3=0\)

⇔ \(x^2+6x+9-x^2+6x-9+30=0\)

\(\Leftrightarrow12x+30=0\)

\(\Leftrightarrow12x=-30\)

\(\Leftrightarrow x=-\dfrac{5}{2}\)

Tập nghiệm của pt là: \(S=\left\{-\dfrac{5}{2}\right\}\)

b) Với a = 1

\(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{3+3}{x^2-1}=0\)

ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)

Ta có : \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{3+3}{x^2-1}=0\)

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

Khử mẫu ta có : \(\left(x-1\right)^2-\left(x+1\right)^2+6=0\)

\(\Leftrightarrow x^2+x-1-x^2+x+1+6=0\)

\(\Leftrightarrow2x+6=0\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

Tập nghiệm của pt là : \(S=\left\{-3\right\}\)

ĐKXĐ: x khác a

b. Với a = 1 ta có pt:

\(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{3+1}{x^2-1}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

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

\(\Leftrightarrow-4x+4=0\)

\(\Leftrightarrow-4x=-4\)

\(\Leftrightarrow x=1\)( loại)

Vậy pt vô nghiệm với a = 1

a.

\(\Leftrightarrow4x^2-6x+1+\dfrac{1}{\sqrt{3}}\sqrt{\left(4x^2-2x+1\right)\left(4x^2+2x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2-2x+1}=a>0\\\sqrt{4x^2+2x+1}=b>0\end{matrix}\right.\) ta được:

\(2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

\(\Leftrightarrow\left(a-\dfrac{b}{\sqrt{3}}\right)\left(2a+\sqrt{3}b\right)=0\)

\(\Leftrightarrow a=\dfrac{b}{\sqrt{3}}\)

\(\Leftrightarrow3a^2=b^2\)

\(\Leftrightarrow3\left(4x^2-2x+1\right)=4x^2+2x+1\)

\(\Leftrightarrow...\)

b.

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow2a^2-b^2+\dfrac{1}{\sqrt{3}}ab=0\)

Lặp lại cách làm câu a

a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)

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

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

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

Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1

b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)

\(\Leftrightarrow x< 5\)

c) \(\dfrac{-1}{2x+3}< 0\)

dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp

a) Thay a = -1 vào phương trình

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

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

\(\Rightarrow2x^2-10=2\left(x+3\right)\left(x+1\right)=2x^2+8x+6\)

\(\Rightarrow2x^2+8x+6-2x^{10}+10=0\)

\(\Rightarrow8x+16=0\Rightarrow x=-2\)

b, c Làm tương tự như câu a

d)

Phương trình nhận x = 1 làm nghiệm

=> \(\dfrac{1+a}{1+3}+\dfrac{1-3}{1-a}=2\)

\(\Rightarrow\dfrac{a+1}{4}+\dfrac{2}{a-1}=2\)

\(\Rightarrow\dfrac{a^2-1+8}{4\left(a-1\right)}=2\)

\(\Rightarrow a^2+7=2\left(4a-1\right)=8a-2\)

\(\Rightarrow a^2-8x+9=0\)

\(\Rightarrow\left[{}\begin{matrix}a=4+\sqrt{7}\\a=4-\sqrt{7}\end{matrix}\right.\)