Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

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

\(=2x^3+6x\)

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

\(=27x-55\)

\(\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}\)

\(=\dfrac{4\left(x-3\right)+x+7}{x^2-9}\)

\(=\dfrac{4x-12+x+7}{x^2-9}\)

\(=\dfrac{5x-5}{x^2-9}\)

Bài 1:
\(a,\left(x+4\right)\left(x+3\right)-7x=x^2+4x+3x+12-7x=x^2+12\\ b,\left(x+4\right)^2+x-16=x^2+8x+16+x-16=x^2+9x\\ c,\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}=\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{4x-12+x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{5x-5}{\left(x+3\right)\left(x-3\right)}\)

Bài 2:
\(7a-7b=7\left(a-b\right)\\ b,x^2-8x+16=\left(x-4\right)^2\\ c,ax-ay+3x-3y=a\left(x-y\right)+3\left(x-y\right)=\left(a+3\right)\left(x-y\right)\\ d,x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)

1: =>3^x=81

=>x=4

2: =>2^x=8

=>x=3

3: =>x^3=2^3

=>x=2

4: =>x^20-x=0

=>x(x^19-1)=0

=>x=0 hoặc x=1

5: =>2^x=32

=>x=5

6: =>(2x+1)^3=9^3

=>2x+1=9

=>2x=8

=>x=4

7: =>x^3=115

=>\(x=\sqrt[3]{115}\)

8: =>(2x-15)^5-(2x-15)^3=0

=>(2x-15)^3*[(2x-15)^2-1]=0

=>2x-15=0 hoặc (2x-15)^2-1=0

=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1

=>x=15/2 hoặc x=8 hoặc x=7

1. Tìm số tự nhiên x biết:

1) \(3^x.3=243\)

\(3^x=243:3\)

\(3^x=81\)

\(3^x=3^4\)

\(\Rightarrow x=4\)

_____

2) \(7.2^x=56\)

\(2^x=56:7\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

_____

3) \(x^3=8\)

\(x^3=2^3\)

\(\Rightarrow x=3\)

_____

4) \(x^{20}=x\)

\(x^{20}-x=0\)

\(x\left(x^{19}-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x=1\)

5) \(2^x-15=17\)

\(2^x=17+15\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

_____

6) \(\left(2x+1\right)^3=9.81\)

\(\left(2x+1\right)^3=729=9^3\)

\(\rightarrow2x+1=9\)

\(2x=9-1\)

\(2x=8\)

\(x=8:2\)

\(\Rightarrow x=4\)

_____

7) \(x^6:x^3=125\)

\(x^3=125\)

\(x^3=5^3\)

\(\Rightarrow x=5\)

_____

8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)

_____

9) \(3^{x+2}-5.3^x=36\)

\(3^x.\left(3^2-5\right)=36\)

\(3^x.\left(9-5\right)=36\)

\(3^x.4=36\)

\(3^x=36:4\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

_____

10) \(7.4^{x-1}+4^{x+1}=23\)

\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)

\(4^{x-1}.\left(7+4^2\right)=23\)

\(4^{x-1}.\left(7+16\right)=23\)

\(4^{x-1}.23=23\)

\(4^{x-1}=23:23\)

\(4^{x-1}=1\)

\(4^{x-1}=4^1\)

\(\rightarrow x-1=0\)

\(x=0+1\)

\(\Rightarrow x=1\)

Chúc bạn học tốt

`Answer:`

1) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(=[x\left(x+3\right)][\left(x+1\right)\left(x+2\right)]+1\)

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

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

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

2) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)

\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)

\(=[\left(12x^2+11x+0,5\right)+1,5][\left(12x^2+11x+0,5\right)-1,5]-4\)

\(=\left(12x^2+11x+0,5\right)^2-\left(1,5\right)^2-4\)

\(=\left(12x^2+11x+0,5\right)^2-\left(2,5\right)^2\)

\(=\left(12x^2+11x+0,5-2,5\right)\left(12x^2+11x+0,5+2,5\right)\)

\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)

3) \(\left(x^2+6x+5\right)\left(x^2+10x+21\right)+15\)

\(=\left(x^2+x+5x+5\right)\left(x^2+3x+7x+21\right)+15\)

\(=\left(x+1\right)\left(x+5\right)\left(x+3\right)\left(x+7\right)+15\)

\(=[\left(x+1\right)\left(x+7\right)][\left(x+5\right)\left(x+3\right)]+15\)

\(=\left(x^2+x+7x+7\right)\left(x^2+3x+5x+15\right)+15\)

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

Đặt \(v=x^2+=8x+11\)

Đa thức có dạng sau: \(\left(v-4\right)\left(v+4\right)+15\)

\(=v^2-4^2+15\)

\(=v^2-1\)

\(=\left(v+1\right)\left(v-1\right)\)

\(=\left(x^2+8x+11+1\right)\left(x^2+8x+11-1\right)\)

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

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

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

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

\(=x^4-ax^2-a.\left(x^2-a\right)-6x^2+4x+2a\)

\(=x^4-ax^2-\left(ax^2-aa\right)-6x^2+4x+2a\)

\(=x^4-2ax^2+a^2-6x^2+2a+4x\)

6) \(a^2-b^2-c^2+2bc-2a+1\)

\(=\left(a^2-2a+1\right)-\left(b^2-2bc+c^2\right)\)

\(=\left(a-1\right)^2-\left(b-c\right)^2\)

\(=\left(a-b+c-1\right)\left(a+b-c-1\right)\)

7) \(4a^2-4b^2+16bc-16c^2\)

\(=4a^2-\left(4b^2-16bc+16c^2\right)\)

\(=\left(2a\right)^2-\left(2b-4c\right)^2\)

\(=\left(2a-2b+4c\right)\left(2a+2b-4c\right)\)

\(=2.\left(a-b-2c\right).2\left(a+b-2c\right)\)

\(=4\left(a-b-2c\right)\left(a+b-2c\right)\)

\(4,7\div0,25+5,3\times4\)

\(=18,8+21,2\)

\(=40\)

\(3\times\left(a-2\right)+150=240\)

\(3\times\left(a-2\right)=90\)

\(a-2=30\)

\(a=32\)

\(\dfrac{1}{9}+a+\dfrac{7}{12}=\dfrac{17}{18}\)

\(\dfrac{1}{9}+a=\dfrac{13}{36}\)

\(a=\dfrac{1}{4}\)

\(\left(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}+\dfrac{1}{6}\times\dfrac{1}{7}+\dfrac{1}{7}\times\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)

\(\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\right)\times a=\dfrac{9}{16}\)

\(\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)

\(\dfrac{3}{8}\times a=\dfrac{9}{16}\)

\(a=\dfrac{3}{2}\)

c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x=0\)

\(\Leftrightarrow x\left(3x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)

\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)

a.\(\Leftrightarrow\left(x-1\right)^3+8-x^3+3x\left(x+2\right)=17\)

    \(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)

   \(\Leftrightarrow9x+7=17\)

   \(\Leftrightarrow9x=10\Leftrightarrow x=\frac{10}{9}\)