Bài 2:

a, (-2)3.34 = (-8) . 81 = 648

b,54.(-3)2 =  625 . 9 = 5625

Bài 3:

a, 2x-25=45  <=> 2x = 70  <=> x= 35

Vậy  x= 35

b,3x+17=2  <=> 3x = -15   <=> x = -5

Vậy  x= -5

c,/x/ ≤ 8       <=>    x ≤ 8     hoặc     x ≤ -8

Vậy x ≤ 8  hoặc  x ≤ -8

d,/ x-1/=0     <=>    x - 1 = 0    <=> x = 1

Vậy  x= 1

2x + 1/2 = -5/3

2x = -13/6

x = -13/12

2x + ½ = -5/3

       2x = -5/3 - 1/2

       2x = -13/6

         x = -13/6 : 2

         x = -13/12

1/7 - 3/5 x = 3/5

       -3/5 x = 3/5 - 1/7

       -3/5 x = 16/35

              x = 16/35 : -3/5

              x = -16/21

-3x - ¾ = 5/6

      -3x = 5/6 + 3/4

      -3x = 19/12

         x = 19/12 : (-3)

         x = -19/36

3/7 - 1/2 x = 5/3

       -1/2 x = 5/3 - 3/7

       -1/2 x = 26/21

              x = 26/21 : (-1/2)

              x = -52/21

2x - ¾ = 5/8

      2x = 5/8 + 3/4

      2x = 11/8

        x = 11/8:2

        x = 11/16

¼ x - |-7/5| = -5/3

   ¼ x - 7/5 = -5/3

           ¼ x = -5/7 + 7/5

           ¼ x = 24/35

               x = 24/35 : 1/4

              x = 96/35

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

2: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

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

Đặt \(x^2+x+1=a\)ta có

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

Thay \(a=x^2+x+1\)ta được

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

\(=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)Kl...

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

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

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

Đặt \(x^2+8x+7=a\) Ta có

\(a\left(a+8\right)+15=a^2+8a+15=a^2+5a+3a+15=a\left(a+5\right)+3\left(a+5\right)=\left(a+5\right)\left(a+3\right)\)

Thay \(a=x^2+8x+15\)ta được

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

\(=\left(x^2+6x+2x+12\right)\left(x^2+8x+10\right)\)

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

em nham cau dau tien la (4/5+1/2) : -13/5 + 3/2

|1/2x| = 3 - 2x

ĐKXĐ : 3 - 2x \(\ge\)0 => 2x \(\ge\) 3 => x \(\ge\)3/2

Ta có: |1/2x| = 3 - 2x

=> \(\orbr{\begin{cases}\frac{1}{2}x=3-2x\\\frac{1}{2}x=-3+2x\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{1}{2}x+2x=3\\\frac{1}{2}x-2x=-3\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{2}x=3\\-\frac{3}{2}x=-3\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{6}{5}\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)

=> x = 2 

|5x| = x - 12

ĐKXĐ : x - 12 \(\ge\)0 => x \(\ge\)12

Ta có: |5x| = x - 12

=> \(\orbr{\begin{cases}5x=x-12\\5x=-x+12\end{cases}}\)

=> \(\orbr{\begin{cases}5x-x=-12\\5x+x=12\end{cases}}\)

=> \(\orbr{\begin{cases}4x=-12\\6x=12\end{cases}}\)

=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)(ktm)

=> pt vô nghiệm

|2x - 5| = x + 1

ĐKXĐ: x + 1 \(\ge\)0 => x \(\ge\)-1

Ta có: |2x - 5| = x + 1

=> \(\orbr{\begin{cases}2x-5=x+1\\2x-5=-x-1\end{cases}}\)

=> \(\orbr{\begin{cases}2x-x=1+5\\2x+x=-1+5\end{cases}}\)

=> \(\orbr{\begin{cases}x=6\\3x=4\end{cases}}\)

=> \(\orbr{\begin{cases}x=6\\x=\frac{4}{3}\end{cases}}\)(tm)

Vậy ...

|7 - 2x| + 7 = 2x

=> |7 - 2x| = 2x - 7

ĐKXĐ: 2x - 7 \(\ge\)0 => 2x \(\ge\) 7 => x \(\ge\) 7/2

Ta có: |7 - 2x| = 2x - 7

=> \(\orbr{\begin{cases}7-2x=2x-7\\7-2x=7-2x\end{cases}}\)

=> 7 + 7 = 2x + 2x

hoặc x tùy ý (TMĐK)

=> 4x = 14 => x = 7/2

hoặc x tùy ý (Tm ĐK)

Vậy ...

a: \(=\dfrac{4x-2+6x^2-6x+2x^2+1}{2x\left(2x-1\right)}=\dfrac{8x^2-2x-1}{2x\left(2x-1\right)}\)

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

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=-1\\x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}\)

\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)

\(=\frac{6.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}{7.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}\)

\(=\frac{6}{7}\)

Tìm x

\(a,3x(2x+1)=0\)

\(\Rightarrow\hept{\begin{cases}3x=0\\2x+1=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)

Vậy \(x=0\)hoặc \(x=\frac{-1}{2}\)

\(b.\frac{2}{3}-\frac{1}{3}(x-\frac{3}{2})-\frac{1}{2}(2x+1)=5\)

\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

\(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-x(\frac{1}{3}+1)=5\)

\(\frac{4}{3}x=\frac{2}{3}-5\)

\(\frac{4}{3}x=\frac{-13}{3}\)

\(x=\frac{-13}{3}\div\frac{4}{3}\)

\(x=\frac{-13}{4}\)

Chúc ban học tốt

\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)

\(=\frac{6.\left(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011}\right)}{7.\left(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011}\right)}=\frac{6}{7}\)

Tìm x

a)bạn tự làm vì nó rất dễ nha

b)\(\frac{2}{3}-\frac{1}{3}.\left(x-\frac{3}{2}\right)-\frac{1}{2}.\left(2x+1\right)=5\)

\(\frac{2}{3}-\frac{1}{3}x-\frac{1}{2}-\frac{1}{2}.2x-\frac{1}{2}=5\)

\(\frac{2}{3}-\frac{1}{3}x-\frac{1}{2}-\frac{2}{2}x-\frac{1}{2}=5\)

\(\frac{2}{3}-\frac{1}{3}x-\frac{1}{2}-x-\frac{1}{2}=5\)

\(\frac{1}{3}x-x=5+\frac{1}{2}+\frac{1}{2}+\frac{2}{3}\)

\(\left(\frac{1}{3}-1\right)x=5+1+\frac{2}{3}\)

\(\frac{-2}{3}x=6+\frac{2}{3}\)

\(\frac{-2}{3}x=\frac{20}{3}\)

\(x=\frac{20}{3}:\left(\frac{-2}{3}\right)\)

\(x=\frac{20}{3}.\left(\frac{3}{-2}\right)\)

x=-10

Vậy x=-10

Chúc bạn học tốt