Trả lời:

\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)\(\left(đkxđ:x\ne0;x\ne2\right)\)

\(\Leftrightarrow\frac{x-1}{2x\left(x-2\right)}-\frac{7}{8x}=\frac{5-x}{4x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}\)

\(\Leftrightarrow\frac{4\left(x-1\right)}{8x\left(x-2\right)}-\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{2\left(5-x\right)}{8x\left(x-2\right)}-\frac{x}{8x\left(x-2\right)}\)

\(\Rightarrow4\left(x-1\right)-7\left(x-2\right)=2\left(5-x\right)-x\)

\(\Leftrightarrow4x-4-7x+14=10-2x-x\)

\(\Leftrightarrow10-3x=10-3x\)

\(\Leftrightarrow-3x+3x=10-10\)

\(\Leftrightarrow0x=0\)( luôn thỏa mãn )

Vậy S = R với \(x\ne0;x\ne2\)

a) \(\left(8x+5\right)^2\left(4x+3\right)\left(2x+1\right)=9\)

\(\Leftrightarrow\left(64x^2+8x+25\right)\left(8x^2+10x+3\right)-9=0\)

Đặt a = \(8x^2+10x+3\)

\(\left(8a+1\right)a-9=0\)

\(\Leftrightarrow8a^2+a-9=0\)

\(\Leftrightarrow\left(a-1\right)\left(8a+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=-\frac{9}{8}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}8x^2+10x+3=1\\8x^2+10x+3=-\frac{9}{8}\end{cases}}\)

mà \(8x^2+10x+3=1\Rightarrow8x^2+10x+2=0\)

\(\Rightarrow2\left(x+1\right)\left(4x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-0,25\end{cases}}\)

cảm ơn bạn còn mấy phần còn lại ạ

a.

\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1\le x\le3\)

b.

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)

Câu b còn 1 cách giải nữa:

Với \(x=0\) không phải nghiệm

Với \(x>0\) , chia 2 vế cho \(\sqrt{x}\) ta được:

\(\sqrt{2x+8+\dfrac{5}{x}}+\sqrt{2x-4+\dfrac{5}{x}}=6\)

Đặt \(\sqrt{2x-4+\dfrac{5}{x}}=t>0\Leftrightarrow2x+8+\dfrac{5}{x}=t^2+12\)

Phương trình trở thành:

\(\sqrt{t^2+12}+t=6\)

\(\Leftrightarrow\sqrt{t^2+12}=6-t\)

\(\Leftrightarrow\left\{{}\begin{matrix}6-t\ge0\\t^2+12=\left(6-t\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\le6\\12t=24\end{matrix}\right.\)

\(\Rightarrow t=2\)

\(\Rightarrow\sqrt{2x-4+\dfrac{5}{x}}=2\)

\(\Leftrightarrow2x-4+\dfrac{5}{x}=4\)

\(\Rightarrow2x^2-8x+5=0\)

\(\Leftrightarrow...\)

x = -0.25 và hình như còn 1 giá trị nữa hay sao ý ạ

bạn gõ công thức toán đi ! như này khó nhìn quá :((

Bài làm

3 - 4x( 25 - 2x ) = 8x² - x - 300

<=> 3 - 100x + 8x² - 8x² + x + 300 = 0

<=> 303 - 99x = 0

<=> 3( 101 - 33x ) = 0

<=> 101 - 33x = 0

<=> x = 101/33

Vậy x = 101/33

3-4x(25-2x)=8x²-x-300

<=>3-100x+8x²=8x²-x-300

<=>3-100x+8x²-8x²+x+300=0

<=>303-99x=0

<=>99x=303

<=>x=101/33

Vậy x=101/33

1: \(\Leftrightarrow\left(x-4\right)^2+14=-9\left(x-4\right)\)

\(\Leftrightarrow x^2-8x+16+14+9x-36=0\)

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

=>(x+3)(x-2)=0

=>x=-3(nhận) hoặc x=2(nhận)

2: \(\Leftrightarrow\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9=0\)

\(\Leftrightarrow16x^2-8x+2x-1-4x^2-2x-12x^2+9=0\)

=>-8x+8=0

hay x=1(nhận)

c: \(\dfrac{1}{2\left(x-3\right)}-\dfrac{3x-5}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{2}\)

\(\Leftrightarrow x-1-2\left(3x-5\right)=\left(x-3\right)\left(x-1\right)\)

\(\Leftrightarrow x^2-4x+3=x-1-6x+10=-5x+9\)

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

=>(x+3)(x-2)=0

=>x=-3(nhận) hoặc x=2(nhận)