\(2x^3+x^2-4x-12\)

\(=2x^3+5x^2+6x-4x^2-10x-12\)

\(=\left(2x^3+5x^2+6x\right)-\left(4x^2+10x+12\right)\)

\(=x\left(2x^2+5x+6\right)-2\left(2x^2+5x+6\right)\)

\(=\left(x-2\right)\left(2x^2+5x+6\right)\)

\(a,2x^3+x^2-4x-12=\left(2x^3-4x^2\right)+\left(5x^2-10x\right)+\left(6x-12\right)=2x^2\left(x-2\right)+5x\left(x-2\right)+6\left(x-2\right)=\left(x-2\right)\left(2x^2+5x+6\right)\)

\(b,x^5-xy^4+x^4y-y^5=x\left(x^4-y^4\right)+y\left(x^4-y^4\right)=\left(x+y\right)\left(x^4-y^4\right)=\left(x+y\right)\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x+y\right)^2\left(x-y\right)\left(x^2+y^2\right)\)

\(c,\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)-9=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]-9=\left(x^2+8x+7\right)\left(x^2+8x+15\right)-9\)

đặt \(x^2+8x+11=y\)

\(\left(x^2+8x+7\right)\left(x^2+8x+15\right)-9=\left(y-4\right)\left(y+4\right)-9=y^2-16-9=y^2-25=\left(y-5\right)\left(y+5\right)=\left(x^2+8x+11-5\right)\left(x^2+8x+11+5\right)=\left(x^2+8x+6\right)\left(x^2+8x+16\right)=\left(x^2+8x+6\right)\left(x+4\right)^2\)

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

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

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

\(\Leftrightarrow-14x+7=0\)

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

a) \(\dfrac{5}{24}+x=\dfrac{7}{12}\)

<=> \(x=\dfrac{7}{12}-\dfrac{5}{24}=\dfrac{14}{24}-\dfrac{5}{24}=\dfrac{9}{24}=\dfrac{3}{8}\)

b) \(x-\dfrac{3}{4}=\dfrac{1}{2}\)

<=> \(x=\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{2}{4}+\dfrac{3}{4}=\dfrac{5}{4}\)

c) bn ghi rõ đề chút

ok bạn

(2\7.x+3\7) :2 1\5-3\7=1

hai phần bảy nhân x cộng ba phần bảy chia hai một phần năm trừ 3 phần bảy bằng một

Giải:

a) \(\dfrac{5}{24}+x=\dfrac{7}{12}\) 

              \(x=\dfrac{7}{12}-\dfrac{5}{24}\) 

              \(x=\dfrac{3}{8}\) 

b) \(x-\dfrac{3}{4}=\dfrac{1}{2}\) 

            \(x=\dfrac{1}{2}+\dfrac{3}{4}\) 

            \(x=\dfrac{5}{4}\) 

c) \(\left(\dfrac{2}{7}.x+\dfrac{3}{7}\right):2\dfrac{1}{5}-\dfrac{3}{7}=1\) 

           \(\left(\dfrac{2}{7}.x+\dfrac{3}{7}\right):\dfrac{11}{5}=1+\dfrac{3}{7}\) 

           \(\left(\dfrac{2}{7}.x+\dfrac{3}{7}\right):\dfrac{11}{5}=\dfrac{10}{7}\) 

                        \(\dfrac{2}{7}.x+\dfrac{3}{7}=\dfrac{10}{7}.\dfrac{11}{5}\) 

                        \(\dfrac{2}{7}.x+\dfrac{3}{7}=\dfrac{22}{7}\)           

                                \(\dfrac{2}{7}.x=\dfrac{22}{7}-\dfrac{3}{7}\) 

                                \(\dfrac{2}{7}.x=\dfrac{19}{7}\) 

                                    \(x=\dfrac{19}{7}:\dfrac{2}{7}\) 

                                    \(x=\dfrac{19}{2}\)

a)\(\dfrac{2}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}+\dfrac{1}{3}\right).\dfrac{4}{5}=1.\dfrac{4}{5}=\dfrac{4}{5}\)

b)\(\dfrac{2}{3}.\dfrac{4}{5}-\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}-\dfrac{1}{3}\right).\dfrac{4}{5}=\dfrac{1}{3}.\dfrac{4}{5}=\dfrac{4}{15}\)

a) \(\dfrac{2}{3}\times\dfrac{4}{5}+\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{4}{5}\times1=\dfrac{4}{5}\)

b) \(\dfrac{2}{3}\times\dfrac{4}{5}-\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{3}=\dfrac{4}{15}\)

c) \(\dfrac{1}{2}:\dfrac{3}{4}+\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}+\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{2}{3}=\dfrac{8}{9}\)

d) \(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}-\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{1}{3}=\dfrac{4}{9}\)

c)\(\dfrac{1}{2}:\dfrac{3}{4}+\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}.\dfrac{4}{3}+\dfrac{1}{6}.\dfrac{4}{3}=\left(\dfrac{1}{2}+\dfrac{1}{6}\right).\dfrac{4}{3}=\dfrac{2}{3}.\dfrac{4}{3}=\dfrac{8}{9}\)

d)\(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}.\dfrac{4}{3}-\dfrac{1}{6}.\dfrac{4}{3}=\left(\dfrac{1}{2}-\dfrac{1}{6}\right).\dfrac{4}{3}=\dfrac{1}{3}.\dfrac{4}{3}=\dfrac{4}{9}\)

\(F=-3\left(x-8\right)\left(2x+1\right)-\left(x+5\right)\left(2-3x\right)-4x\left(x-6\right)\)

\(=-3\left(-3-8\right)\left(-6+1\right)-\left(5-3\right)\left(2+9\right)+12\left(-9\right)\)

\(=-3\left(-11\right)\left(-5\right)-\left(-2\right)11-12.9\)

\(=-165+22-108=22-273=-251\)

\(G=\left(5x-4\right)\left(5-2x\right)-7x\left(x^2-4x+3\right)+\left(x^2-4x\right)\left(7x-2\right)\)

\(=\left(5-4\right)\left(5-2\right)-7\left(1-4+3\right)+\left(1-4\right)\left(7-2\right)\)

\(=3-7.0+5.\left(-3\right)=3-15=-12\)

\(H=\left(-3x+5\right)\left(x-6\right)-\left(x-1\right)\left(x^2-2x+3\right)+\left(x+2\right)\left(x^2-3\right)\)

\(=\left(3+5\right)\left(-1-6\right)-\left(-1-1\right)\left(1+2+3\right)+\left(-1+2\right)\left(1-3\right)\)

\(=8\left(-7\right)-\left(-2\right)6+1\left(-2\right)=-56+12-2=-46\)

\(L=5x\left(x-1\right)\left(2x+3\right)-10x\left(x^2-4x+5\right)-\left(x-1\right)\left(x-4\right)\)

\(=-\frac{5}{3}\left(-\frac{4}{3}\right)\left(-\frac{2}{3}+3\right)+\frac{10}{3}\left(\frac{1}{9}+\frac{4}{3}+5\right)-\left(-\frac{4}{3}\right)\left(-\frac{1}{3}-4\right)\)

\(=\frac{20}{9}\left(\frac{7}{3}\right)+\frac{10}{3}\left(\frac{13}{9}+5\right)+\frac{4}{3}\left(-\frac{13}{3}\right)\)

\(=\frac{140}{27}+\frac{10}{3}.\frac{58}{9}-\frac{52}{9}\)

\(=\frac{140}{27}+\frac{580}{27}-\frac{156}{27}=\frac{140+580-156}{27}=\frac{720-156}{27}=\frac{564}{27}\)

\(M=-7x\left(x-5\right)-\left(x-1\right)\left(x^2-x-2\right)+x^2\left(x-3\right)-5x\left(x-8\right)\)

\(=\frac{-7}{2}\left(\frac{1}{2}-5\right)+\frac{\left(\frac{1}{4}-\frac{1}{2}-2\right)}{2}+\frac{1}{4}\left(\frac{1}{2}-3\right)-\frac{5}{2}\left(\frac{1}{2}-8\right)\)

\(=\frac{7}{2}.\frac{9}{2}-\frac{9}{8}-\frac{1}{4}.\frac{5}{2}+\frac{5}{2}.\frac{15}{2}\)

\(=\frac{63}{4}-\frac{9}{8}-\frac{5}{8}+\frac{75}{4}=\frac{138}{4}-\frac{7}{4}=\frac{131}{4}\)

a: \(\Leftrightarrow\left|x\cdot\dfrac{7}{3}-\dfrac{3}{4}\right|=1+\dfrac{1}{3}+\dfrac{2}{3}=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{7}{3}-\dfrac{3}{4}=2\\x\cdot\dfrac{7}{3}-\dfrac{3}{4}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{33}{28}\\x=-\dfrac{15}{28}\end{matrix}\right.\)

b: \(\Leftrightarrow\left|x\cdot\dfrac{2}{3}-\dfrac{1}{3}\right|=\dfrac{6}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{6}{5}\\x\cdot\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-13}{10}\\x=\dfrac{23}{10}\end{matrix}\right.\)

a) A(x) = 5x⁴ - 5 + 6x³ + x⁴ - 5x - 12

= (5x⁴ + x⁴) + (- 5 - 12) + 6x³ - 5x

= 6x⁴ - 17 + 6x³ - 5x

= 6x⁴ + 6x³ - 5x - 17

B(x) = 8x⁴ + 2x³ - 2x⁴ + 4x³ - 5x - 15 - 2x²

= (8x⁴ - 2x⁴) + (2x³ + 4x³) - 5x - 15 - 2x²

= 4x⁴ + 6x³ - 5x - 15 - 2x²

= 4x⁴ + 6x³ - 2x² - 5x - 15

b) C(x) = A(x) - B(x)

=  6x⁴ + 6x³ - 5x - 17 - (4x⁴ + 6x³ - 2x² - 5x - 15)

= 6x⁴ + 6x³ - 5x - 17 - 4x⁴ - 6x³ + 2x² + 5x + 15

= ( 6x⁴ - 4x⁴) + ( 6x³ - 6x³) + (- 5x + 5x) + (-17 + 15) + 2x²

= 2x⁴ - 2 + 2x² 

= 2x⁴ + 2x² - 2

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

2) \(5x^2-5y^2-10x+10y=5\left(x^2-y^2\right)-10\left(x-y\right)\)

            \(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)=\left(x-y\right)\left(5x+5y-10\right)\)

3) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)

4) \(ax-bx-a^2+2ab-b^2=x\left(a-b\right)-\left(a^2-2ab+b^2\right)\)

                            \(=x\left(a-b\right)-\left(a-b\right)^2=\left(a-b\right)\left(x-a+b\right)\)

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

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

6) \(x^2+4x-y^2+4=x^2+4x+4-y^2=\left(x+2\right)^2-y^2\)

                                   \(=\left(x+2-y\right)\left(x+2+y\right)\)

Phân tích các đa thức sau thành nhân tử :

1) x^3 + x^2y - 4x - 4y

2) x^3 - 3x^2 +1 - 3x

3) 3x^2 - 6xy + 3y^2 - 12z^2

4) x^2 - 2x - 15

5) 2x^2 +3x - 5

6) 2x^2 - 18

7) x^2 - 7xy + 10y^2

8) x^3 - 2x^2 + x - xy^2

Làm nhanh giúp mình với nhé .....mình đang cần gấp[[[[