a, \(x\) + 99: 3 = 55

    \(x\) + 33 = 55

     \(x\)        = 55 - 33

      \(x\)       = 22

b, (\(x\) - 25):15 = 20

       \(x\) - 25  = 20 x 15

       \(x\) - 25  = 300

        \(x\)         = 300 + 25

          \(x\)        = 325

c, (3\(x\) - 15).7 = 42

      3\(x\) - 15 = 42:7

       3\(x\) - 15 = 6

        3\(x\)         = 6 + 15

         3\(x\)         = 21

            \(x\)         = 21: 3

             \(x\)         = 7

d, (8\(x\) - 16).(\(x\) -5) = 0

       \(\left[{}\begin{matrix}8x-16=0\\x-5=0\end{matrix}\right.\)

         \(\left[{}\begin{matrix}8x=16\\x=5\end{matrix}\right.\)

          \(\left[{}\begin{matrix}x=16:8\\x=5\end{matrix}\right.\)

           \(\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy \(x\) \(\in\) {2; 5}

1: Ta có: \(x^{10}-4x^8+4x^6\)

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

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

2: Ta có: \(m^3+27\)

\(=\left(m+3\right)\left(m^2-3m+9\right)\)

3: Ta có: \(x^3+8\)

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

4: Ta có: \(\frac{1}{27}+a^3\)

\(=\left(\frac{1}{3}+a\right)\left(\frac{1}{9}-\frac{a}{3}+a^2\right)\)

5: Ta có: \(8x^3+27y^3\)

\(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

6: Ta có: \(\frac{1}{8}x^3+8y^3\)

\(=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)

7: Ta có: \(8x^6-27y^3\)

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

8: Ta có: \(\frac{1}{8}x^3-8\)

\(=\left(\frac{1}{2}x-2\right)\left(\frac{1}{4}x^2+x+4\right)\)

9: Ta có: \(\frac{1}{64}x^6-125y^3\)

\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{4}x^2y+25y^2\right)\)

10: Ta có: \(\left(a+b\right)^3-c^3\)

\(=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)\cdot c+c^2\right]\)

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

11: Ta có: \(x^3-\left(y-1\right)^3\)

\(=\left[x-\left(y-1\right)\right]\cdot\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]\)

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

12: Ta có: \(x^6+1\)

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

1) \(x^{10}-4x^8+4x^6\)

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

2) \(m^3+27=m^3+3^3=\left(m+3\right)\left(m^2-3m+3^2\right)\)

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

4) \(\frac{1}{27}+a^3=\left(\frac{1}{3}\right)^3+a^3=\left(\frac{1}{3}+a\right)\left[\left(\frac{1}{3}\right)^2-\frac{1}{3}a+a^2\right]\)

5) \(8x^3+27y^3=\left(2x\right)^3+\left(3y\right)^3=\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

6) \(\frac{1}{8}x^3+8y^3=\left(\frac{1}{2}x\right)^3+\left(2y\right)^3=\left(\frac{1}{2}x+2y\right)\left[\left(\frac{1}{2}x\right)^2-\frac{1}{2}x.2y+\left(2y\right)^2\right]=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)

8) \(\frac{1}{8}x^3-8=\left(\frac{1}{2}x\right)^3-2^3=\left(\frac{1}{2}x-2\right)\left[\left(\frac{1}{2}x\right)^2+\frac{1}{2}x.2+2^2\right]=\left(\frac{1}{2}x-2\right)\left(\frac{1}{4}x^2+x+4\right)\)

10) \(\left(a+b\right)^3-c^3=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]=\left(a+b-c\right)\left[\left(a^2+2ab+b^2\right)+ac+bc+c^2\right]=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)11) \(x^3-\left(y-1\right)^3=\left(x-y+1\right)\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]=\left(x-y+1\right)\left[x^2+xy-x+\left(y^2-2y+1\right)\right]=\left(x-y+1\right)\left(x^2+xy-x+y^2-2y+1\right)\)

P/s: Đăng ít thôi chớ bạn!

1) 14x-8x=10+5

x(14-8)=15

x6=15

x=15/6

2)5x-3x=30-15

2x=15

x=15/2

3)làm tương tự

1) x=2,5

2) x=7,5

3) x=4

4) x=7/3

5) x=8,25

11: Ta có: \(\left(x+3\right)^3=125\)

\(\Leftrightarrow x+3=5\)

hay x=2

12: Ta có: \(\left(2x\right)^4=16\)

\(\Leftrightarrow x^4=1\)

hay \(x\in\left\{1;-1\right\}\)

11: Ta có: \left(x+3\right)^3=125(x+3)3=125

$\iff x+3=5$

hay x=2

12: Ta có: \left(2x\right)^4=16(2x)4=16

\Leftrightarrow x^4=1(2x)   = 2   =2  4

           4      

⇔2x=2⇔ x=1

13) 32:(3x-2)=2^3

32:(3x-2)=8

3x-2=32:8=4

3x=4+2=6

x=6:3=2

⇔x=2

5x-16=40+x

=> 5x-16-x = 40

=> 5x-x -16=40

4x-16=40

4x= 40+16

4x=56

x= 56:4

x=14

Vậy...

4x-10=15-x

=> 4x-10+x= 15

4x+x -10=15

5x= 15+10

5x= 25

x= 25:5

x=5

Vậy....

5x -16=40+x

=> 5x-x=40+16

=>4x=56

=>x=56:4

x=14

1)ĐK:`4x^2-12x+9>0`

`<=>(2n-3)^2>0`

`<=>2n-3 ne 0`

`<=>n ne 3/2`

`d)x^2-x+1`

`=(x-1/2)^2+3/4>0AAx`

`=>` bt xd `AAx in RR`

e)ĐK:`x^2-8x+15>0`

`<=>x^2-3x-5x+15>0`

`<=>x(x-3)-5(x-3)>0`

`<=>(x-3)(x-5)>0`

`TH1:` \(\begin{cases}x-3>0\\x-5>0\\\end{cases}\)

`<=>` \(\begin{cases}x>3\\x>5\\\end{cases}\)

`<=>x>5`

`TH2:` \(\begin{cases}x-3<0\\x-5<0\\\end{cases}\)

`<=>` \(\begin{cases}x<3\\x<5\\\end{cases}\)

`<=>x<3`

f)ĐK:`3x^2-7x+20>0`

`<=>x^2-2x+1+2x^2-5x+19>0`

`<=>(x-1)^2+2(x-5/2)^2+13/2>0` luôn đúng

c) Để biểu thức \(\dfrac{1}{\sqrt{4x^2-12x+9}}\) có nghĩa thì \(4x^2-12x+9>0\)

\(\Leftrightarrow\left(2x-3\right)^2>0\)

\(\Leftrightarrow2x-3\ne0\)

\(\Leftrightarrow2x\ne3\)

hay \(x\ne\dfrac{3}{2}\)

d) Để biểu thức \(\dfrac{1}{\sqrt{x^2-x+1}}\) có nghĩa thì \(x^2-x+1>0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}>0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)(luôn đúng)

e) Để biểu thức \(\dfrac{1}{\sqrt{x^2-8x+15}}\) có nghĩa thì \(x^2-8x+15>0\)

\(\Leftrightarrow\left(x-4\right)^2>1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4>1\\x-4< -1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>5\\x< 3\end{matrix}\right.\)

f) Để biểu thức \(\dfrac{1}{\sqrt{3x^2-7x+20}}\) có nghĩa thì \(3x^2-7x+20>0\)

\(\Leftrightarrow x^2-\dfrac{7}{3}x+\dfrac{20}{3}>0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{191}{36}>0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2+\dfrac{191}{36}>0\)(luôn đúng)