4(x-12)-2=50

4(x-12)=50+2

4(x-12)=52

x-12=52÷4

x-12=13

x=25

4(x-12)-2=50

4(x-12)=50+2

4(x-12)=52

x-12=52÷4

x-12=13

x=25

*4/7x=9/8.0,125 

4/7x=9/8.1/8

4/7x=9/64

x    =9/64:4/7

x    =9/64.7/4

x    =63/256

*x-25%=1/2

x-1/4.x  =1/2

x(1-1/4) =1/2

3/4.x     =1/2

x           =1/2:3/4

x           =1/2.4/3

x           =2/3

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

\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)

\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)

(3x+5)(4-3x)=0

3x+5 =0 hoặc 4-3x=0

3x=-5 hoặc 3x=-4

x=-5/3 hoặc x=-4/3

9(3x-2)=x(2-3x)

9(3x-2)-x(3x-2)=0

(3x-2)(9-x)=0

3x-2=0 hoặc 9-x=0

3x=2 hoặc x= -9

x =2/3 hoặc x=-9 

vậy x =2/3 ; x= -9

25x^2 - 2=0

(5x)^2 -√2^2=0

(5x-√2)(5x+√2)=0

5x=√2 hoặc 5x = -√2

x=√2/5 hoặc x= -√2/5

vậy x=√2/5 ; x=-√2/5

a)7x+2x=918;           

9x=918

x=102.

Vay x=102.

\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\Leftrightarrow\left(4-x\right)\left(4-x\right)=-5\times-5\)
\(\Rightarrow\left(4-x\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)

\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)

\(\Rightarrow\left(4-x\right).\left(4-x\right)=\left(-5\right).\left(-5\right)\)

\(\Rightarrow\left(4-x\right)^2=25\)

\(\Rightarrow\left(4-x\right)^2=5^2\)

\(\Rightarrow4-x=\pm5\)

\(\Rightarrow\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4-5\\x=4-\left(-5\right)\end{matrix}\right.\)

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

=>(4-x)^2=25

=>(x-4)^2=25

=>x-4=5 hoặc x-4=-5

=>x=-1 hoặc x=9

\(4^x+4^{x+3}=4160\)

\(\Leftrightarrow4^x.\left(1+4^3\right)=4160\)

\(\Leftrightarrow4^x.65=4160\)

\(\Leftrightarrow4^x=4160:65=64\)

\(\Rightarrow x=3\)

x=9,16515139?

4 /x=y/21=28/49

vì 21.4/12=7 nên x=7

    28.21/49 =12 nên y=12 (công thức nhân chéo)

vậy ,x=7 ;y=12