Vậy phương trình đã cho có tập nghiệm S = {- 1/2; 3}.

Ta có: 

5 x 2 − 2 x + 10 2 = 3 x 2 + 10 x − 8 2 ⇔ 5 x 2 − 2 x + 10 2 − 3 x 2 + 10 x − 8 2 = 0 ⇔ 5 x 2 − 2 x + 10 + 3 x 2 + 10 x − 8 5 x 2 − 2 x + 10 − 3 x 2 − 10 x + 8 = 0 ⇔ 8 x 2 + 8 x + 2 2 x 2 − 12 x + 18 = 0 ⇔ 2 4 x 2 + 4 x + 1 .2 x 2 − 6 x + 9 = 0 ⇔ 2 x + 1 2 x − 3 2 = 0 ⇔ 2 x + 1 2 = 0 x − 3 2 = 0 ⇔ 2 x + 1 = 0 x − 3 = 0 ⇔ x = − 1 2 x = 3

Vậy tập nghiệm của phương trình là  S   = − 1 2 ;   3

Đáp án cần chọn là: C

\(D=\frac{4^7.2^8}{3.2^{15}.16^2+5.2.\left(2^{10}\right)^2}\)

\(=\frac{2^{14}.2^8}{3.2^{15}.2^8+5.2.2^{20}}\)

\(=\frac{2^{22}}{3.2^{23}+5.2^{21}}\)

\(=\frac{2^{22}}{2^{21}.\left(2^2.3+5\right)}\)

\(=\frac{2}{2^2.3+5}\)

\(=\frac{2}{17}\)

\(D=\frac{\left(2^2\right)^7.2^8}{3.2^{15}.\left(2^4\right)^2+5.2^2.2^{20}}\)

\(=\frac{2^{14}.2^8}{3.2^{15}.2^8+5.2^{22}}\)

\(=\frac{2^{22}}{3.2^{23}+5.2^{22}}\)

\(=\frac{2^{22}}{2^{22}\left(3.2+5\right)}=\frac{1}{11}\)

XIN LỖI VIẾT NHẦM ĐỀ BÀI NÊN SAI RỒI 

Bài 1:

\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)

Bài 2:

\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)

c: \(\dfrac{x^4-x-14}{x-2}\)

\(=\dfrac{x^4-2x^3+2x^3-4x^2+4x^2-8x+7x-14}{x-2}\)

\(=x^3+2x^2+4x+7\)

9: \(\left(-2x\right)\left(3x^2-2x+4\right)=-6x^3+4x^2-8x\)

\(a,\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+1\right)\left(x-3\right)-\left(x-1\right)\left(x+2\right)\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(7-5x\right)=0\)

\(\Leftrightarrow x-1=0;x+2=0\)hoặc \(7-5x=0\)

\(\Leftrightarrow x=1;x=-2\)hoặc \(x=\frac{7}{5}\)

KL....

\(b,\left(5x^2-2x+10\right)^2=\left(x^2+10x-8\right)^2\)

\(\Leftrightarrow\left(5x^2-2x+10\right)^2-\left(x^2+10x-8\right)^2=0\)

\(\Leftrightarrow\left(5x^2-2x+10-x^2-10x+8\right)\left(5x^2-2x+10+x^2+10x-8\right)=0\)

\(\Leftrightarrow\left(4x^2-12x+18\right)\left(6x^2+8x+2\right)=0\)

\(\Leftrightarrow\left(x^2-3x+\frac{9}{2}\right)\left(6x^2+6x+2x+2\right)=0\)

\(\Leftrightarrow\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{9}{4}\right)\left(6x+2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[\left(x-\frac{3}{2}\right)^2+\frac{9}{4}\right]\left(3x+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=-1\end{cases}}\)Vì \(\left(x-\frac{3}{2}\right)^2+\frac{9}{4}>0\forall x\)

Vậy ..

b: \(\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)

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

\(=\left(x-1\right)^2\)

c: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=5x^3+14x^2+12x+8\)