a)(x/10 -3/2)^2 =1/25 

TH1 : x/10 -3/2 =1/5

=> x/10 =17

=> x=170

TH2 :x/10 -3/2 =-1/5

x/10 =1,3

x=13

Vậy x =170 hoặc 13

B) |x-1|+4=15

=>|x-1|=11

TH1:x-1=11=> x=12

TH2: x-1=-11 => x=-10

Vậy x=12 hoặc -10

\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)

\(10,\left(x+3\right)^2-x^2=45\)

\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)

Vậy \(S=\left\{6\right\}\)

\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)

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

Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)

\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)

a) 

Ta có : vì|1/2-1/3+x| lớn hơn hoặc bằng 0 

Còn -1/4-|y| bé hơn hoặc bằng 0

=> ko tồn tại x

b) 

Ta có: |x-y| lớn hơn hoặc bằng 0 và|y+9/25| lớn hơn hoặc bằng 0 mà:

| x-y|+ |y+9/25| =0 => |x-y| =0 và |y+9/25|=0

 Xét |y+9/25| có:

| y+9/25|=0 => y+9/25=0 => y=-9/25

Thay y = -9/25 vào |x-y| =0 => x=-9/25

 Vậy x=y=-9/25

a) x = 15

b) x = 11

c) x = 68

d) x = 310

(x4)3=x6x19

a) x = 15.              b)      x = 11.        c) x = 68.              d) x = 310

Tìm xx biết:    \left(x^{4}\right)^{3}=\dfrac{x^{19}}{x^{6}}(x4)3=x6x19​

Trả lời: x=x= 

a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)

c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)

d) Giống câu c

⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3

c) ${}^{}-10x=-25\Rightarrow {}^{}-10x+$

\((x-6)(3x-9)>0\)
TH1:
\(\orbr{\begin{cases}x-6< 0\\3x-9< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< 6\\x< 3\end{cases}}\)\(\Rightarrow x< 3\)
TH2:
\(\orbr{\begin{cases}x-6>0\\3x-9>0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>6\\x>3\end{cases}}\)\(\Rightarrow x>6\)
Vậy \(x< 3\) hoặc \(x>6\)thì \((x-6)(3x-9)>0\)
Học tốt!

20.
\((2x-1)(6-x)>0\)
TH1:
\(\orbr{\begin{cases}2x-1>0\\6-x>0\end{cases}\Rightarrow\orbr{\begin{cases}x< \frac{1}{2}\\x< 6\end{cases}}\Rightarrow x< 6}\)
TH2
\(\orbr{\begin{cases}2x-1< 0\\6-x< 0\end{cases}\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x>6\end{cases}}\Rightarrow x>\frac{1}{2}}\)
Vậy \(x< 6\)hoặc \(x>\frac{1}{2}\)thì \((2x-1)(6-x)>0\)
 

21.
\((2-x)(x+7)< 0\)
TH1.
\(\orbr{\begin{cases}2-x>0\\x+7< 0\end{cases}\Rightarrow\orbr{\begin{cases}x< 2\\x>-7\end{cases}}\Rightarrow-7< x< 2}\)
TH2.
\(\orbr{\begin{cases}2-x< 0\\x+7>0\end{cases}\Rightarrow\orbr{\begin{cases}x>2\\x< -7\end{cases}}\Rightarrow2< x< -7}\)(vô lí)
Vậy \(-7< x< 2\) thì \((2-x)(x+7)< 0\)
 

Bài 2:

a, |x-1| -x +1=0

|x-1| = 0-1+x

|x-1| = -1 + x

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

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

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

x = 2-x

2x = 2

x = 2:2

x=1

b, |2-x| -2 = x

|2-x| = x+2

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

2-x = x+2

x+x = 2-2

2x = 0

x = 0

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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}