b)

ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)

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

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

Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Leftrightarrow2x^2-14=2x^2+x-10\)

\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)

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

\(\Leftrightarrow-x=4\)

hay x=-4(nhận)

Vậy: S={-4}

\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)

\(\Leftrightarrow36x=-3\)

\(\Leftrightarrow x=-\dfrac{1}{12}\)

\(k.x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x-\dfrac{1-2x}{3}}{5}\)

\(\Leftrightarrow\dfrac{15x}{15}+\dfrac{10x+x-1}{15}=\dfrac{15}{15}-\dfrac{9x-1+2x}{15}\)

\(\Leftrightarrow15x+9x-1=14-7x\)

\(\Leftrightarrow31x=15\)

\(\Leftrightarrow x=\dfrac{15}{31}\)

câu h tương tự bài trước câu g mik làm nhé

\(\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\) 

⇔ \(\dfrac{3\left(2x+1\right)^2}{15}-\dfrac{5\left(x-1\right)^2}{15}=\dfrac{7x^2-14x-5}{15}\)

⇔ \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)

⇔ \(3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)

⇔ \(12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)

⇔ \(7x^2+22x-2=7x^2-14x-5\) ⇔ \(36x+3=0\) ⇔ x=\(\dfrac{-1}{12}\)

\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)

\(\Leftrightarrow36x=-3\)

hay x=-1/12

câu 2 đề sai

ok tớ sẽ giải nhunh ! sửa câu 2 đi rồi tớ sẽ làm cho bn !

câu 1 ) thì đúng

câu 2 sai đề

bài 1 chắc bạn sai đề. Mình lười lắm nên link đây nhé https://diendantoanhoc.net/topic/96618-sqrtx8frac3x27x84x2/

Điều kiện: 4

\(x\ge\frac{1}{2}\)

Ta có: 

\(x\left(\sqrt{2x-1}-3\right)=\frac{2\left(2x^2-7x-15\right)}{x^2-6x+13}\)

\(\Leftrightarrow x.\frac{2\left(x-5\right)}{\sqrt{2x-1}+3}=\frac{2\left(x-5\right)\left(2x+3\right)}{x^2-6x+13}\)

\(\Leftrightarrow2\left(x-5\right)\left(\frac{x}{\sqrt{2x-1}+3}-\frac{2x+3}{x^2-6x+13}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\frac{x}{\sqrt{2x-1}+3}-\frac{2x+3}{x^2-6x+13}\left(1\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow\frac{\left(x-3\right)+3}{\sqrt{2x-1}+3}-\frac{\left(2x-1\right)+4}{\left(x-3\right)^2+4}=0\)

Đặt \(\hept{\begin{cases}\left(x-3\right)=a\\\sqrt{2x-1}=b\ge0\end{cases}}\)

\(\Rightarrow\frac{a+3}{b+3}-\frac{b^2+4}{a^2+4}=0\)

Tới đây thì đơn giản rồi nhé