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Ta có:

- 4 x 2 ( 6 x 3   +   5 x 2   –   3 x   +   1 )     =   ( - 4 x 2 ) . 6 x 3   +   ( - 4 x 2 ) . 5 x 2   +   ( - 4 x 2 ) . ( - 3 x )   +   ( - 4 x 2 ) . 1     =   - 24 x 5   –   20 x 4   +   12 x 3   –   4 x 2

Đáp án cần chọn là: C

a: \(=\dfrac{2x\left(3x^2+2\right)+3x^2+2}{3x^2+2}=2x+1\)

b: 

Sửa đề: 6x^4-4x^3+3x-2/3x-2

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

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

where is đề

...

Câu 4:

D=(x+1)(x+3)(x+5)(x+7)+15

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+105+15\)

\(=\left(x^2+8x\right)^2+22\left(x^2+8x\right)+120\)

\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(=\left(x+2\right)\left(x+6\right)x^2+8x+10\)

Câu 2:

b: \(4x^2-12x+9\)

\(=\left(2x\right)^2-2\cdot2x\cdot3+3^2\)

\(=\left(2x-3\right)^2\)

Câu 1:

a: \(4x^2-9y^2=\left(2x\right)^2-\left(3y\right)^2=\left(2x-3y\right)\left(2x+3y\right)\)

b: \(\left(3x+y\right)^3=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot y+3\cdot3x\cdot y^2+y^3\)

\(=27x^3+27x^2y+9xy^2+y^3\)

\(\dfrac{3}{4}\times\dfrac{5}{7}=\dfrac{15}{28};\dfrac{5}{8}\times\dfrac{4}{15}=\dfrac{20}{120}=\dfrac{1}{6};\dfrac{7}{12}\times\dfrac{4}{9}=\dfrac{28}{108}=\dfrac{7}{27};\)

\(\dfrac{1}{6}\times\dfrac{3}{5}=\dfrac{3}{30}=\dfrac{1}{10};\dfrac{12}{21}\times\dfrac{23}{8}=\dfrac{276}{168}=\dfrac{23}{14};\dfrac{13}{4}\times\dfrac{5}{39}=\dfrac{65}{156}=\dfrac{5}{12}\)

\(\dfrac{7}{42}\times\dfrac{13}{21}=\dfrac{91}{882}=\dfrac{13}{126};\dfrac{3}{16}\times\dfrac{4}{15}=\dfrac{12}{240}=\dfrac{1}{20}\)