*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 \(S=\left\{-2;\pm5\right\}\)

a: \(=\dfrac{2x^4+x^3-5x^2-3x-3}{x^2-3}\)

\(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)

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

b: \(=\dfrac{x^5+x^2+x^3+1}{x^3+1}=x^2+1\)

c: \(=\dfrac{2x^3-x^2-x+6x^2-3x-3+2x+6}{2x^2-x-1}\)

\(=x+3+\dfrac{2x+6}{2x^2-x-1}\)

d: \(=\dfrac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)

\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)

\(=x^2-2x-5\)

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: Ta có: \(8x+11-3=5x+x-3\)

\(\Leftrightarrow8x+8=6x-3\)

\(\Leftrightarrow2x=-11\)

hay \(x=-\dfrac{11}{2}\)

b: Ta có: \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

\(\Leftrightarrow2x\left(x^3+6x^2+12x+8\right)-8x^2=2\left(x^3-8\right)\)

\(\Leftrightarrow2x^4+12x^3+24x^2+16x-8x^2-2x^3+16=0\)

\(\Leftrightarrow2x^4+10x^3+16x^2+16x+16=0\)

\(\Leftrightarrow2x^4+4x^3+6x^3+12x^2+4x^2+8x+8x+16=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x^3+6x^2+4x+8\right)=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

c: Ta có: \(\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\)

\(\Leftrightarrow2x^2-3x+2x-3-2x^2-10x+x+5=0\)

\(\Leftrightarrow-10x+2=0\)

\(\Leftrightarrow-10x=-2\)

hay \(x=\dfrac{1}{5}\)

d: Ta có: \(\dfrac{1}{10}-2\cdot\left(\dfrac{1}{2}t-\dfrac{1}{10}\right)=2\left(t-\dfrac{5}{2}\right)-\dfrac{7}{10}\)

\(\Leftrightarrow\dfrac{1}{10}-t+\dfrac{1}{5}=2t-5-\dfrac{7}{10}\)

\(\Leftrightarrow-t-2t=-\dfrac{57}{10}-\dfrac{3}{10}=-6\)

hay t=2

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Đề bài mình viết thiếu là CM biểu thức sau không phụ thuộc vào x ( nghĩa là kết quả phải ra số tự nhiên không có x ) 

\(A=\left(2x+1\right)\left(x-1\right)-2x\left(x+2\right)-5\left(-x+3\right)+4\)

\(=2x^2-2x+x-1-2x^2-4x+5x-15+4\)

\(=-12\left(đpcm\right)\)

\(B=\left(4x+3\right)\left(2x-5\right)-\left(8x+1\right)\left(x+3\right)+13\left(3x+1\right)+2\)

\(=8x^2-20x+6x-15-\left(8x^2+24x+x+3\right)+39x+13+2\)

\(=-3\left(đpcm\right)\)

\(a)\left(2x+5\right)\left(2x-7\right)-\left(-4x-3\right)^2=16\\ \Leftrightarrow4x^2-14x+10x-35-\left(16x^2+24x-9\right)=16\\ \Leftrightarrow-12x^2-28x-44=16\\ \Leftrightarrow-12x^2-28x-60=0\\ \Leftrightarrow3x^2+7x+15=0\\ \Delta=b^2-4ac=7^2-4.3.15=-131< 0\)

Vậy phương trình vô nghiệm

\( b)(8x^2 + 3)(8x^2 - 3) - (8x^2 - 1)^2 = 22\)

\(\Leftrightarrow64x^4-9-\left(64x^4-16x^2+1\right)=22\\ \Leftrightarrow-10+16x^2=22\\ \Leftrightarrow16x^2=32\\ \Leftrightarrow x^2=2\\ \Leftrightarrow x=\pm\sqrt{2}\)

Vậy \(x=\sqrt{2},x=-\sqrt{2}\)

\(c)49x^2+14x+1=0\\ \Leftrightarrow\left(7x+1\right)^2=0\\ \Leftrightarrow7x+1=0\\ \Leftrightarrow7x=-1\)

\(\Leftrightarrow\)\(x=-\dfrac{1}{7}\)

Vậy \(x=-\dfrac{1}{7}\)

\(\Leftrightarrow\)\(x=-\dfrac{1}{7}\)

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)