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2x\(^2\)+8x+6= 2x\(^2\)+2x+6x+6= 2x(x+1)+6(x+1)=(2x+6)(x+1)

3x\(^2\)-8x+5= 3x\(^2\)-3x-5x+5=3x(x-1)-5(x-1)= (3x-5)(x+1)

2x\(^2\)-30x+28= 2x\(^2\)-2x-28x+28= 2x(x-1)-28(x-1)= 2(x-19)(x-1)

4x\(^2\)+8x+3= 4x\(^2\)+2x+6x+3= 2x(2x+1)+3(2x+1)= (2x+3)(2x+1)

\(2x^3-8x^2+8x=2x\left(x^2-4x+4\right)=2x\left(x-2\right)^2\)

\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\) ( ĐKXĐ: \(x\ne0;x\ne2\) )

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

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

\(\Rightarrow4x-4-7x+14=10-2x-x\)

\(\Leftrightarrow-3x+2x+x=10+4-14\)

\(\Leftrightarrow0=0\)

          Vậy pt đã cho có nghiệm đúng với mọi x

\(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)

\(\Leftrightarrow\dfrac{8x^2}{3\left(1-2x\right)\left(1+2x\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{1+8x}{4\left(1+2x\right)}\)

\(\Leftrightarrow\dfrac{-32x^2}{12\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x.4\left(1+2x\right)-\left(1+8x\right).3\left(2x-1\right)}{12\left(2x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow8x\left(1+2x\right)-\left(1+8x\right).3.\left(2x-1\right)=-32x^2\)

\(\Leftrightarrow8x+16x^2-6x+3-48x^2+24x+32x^2=0\)

\(\Leftrightarrow26x+3=0\)

\(\Leftrightarrow x=-\dfrac{3}{26}\)

Vậy:......

*vn:vô nghiệm.

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

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

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

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

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

\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

\(\dfrac{7}{8x}+\dfrac{5-x}{4x^2-8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8x-16}\)
ĐKXĐ: x ≠ 0; x ≠ 2

\(< =>\dfrac{14x-28+20-4x}{16x\left(x-2\right)}=\dfrac{8x-8+2x}{16x\left(x-2\right)}\)
Suy ra: 14x - 28 + 20 - 4x = 8x - 8 + 2x
<=> 14x - 8x - 2x - 4x = 28 - 20 - 8
<=> 0x = 0
Vậy: S = { x | x ≠ 0;2 }

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 \(x\ne0;x\ne2\)

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

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

\(\Leftrightarrow12x^2+3+8x+2=0\)

\(\Leftrightarrow12x^2+8x+5=0\)

\(\text{Δ}=8^2-4\cdot12\cdot5=64-240< 0\)

DO đo: Phương trình vô nghiệm