1: Ta có: \(20-2\left(x+4\right)=4\)

\(\Leftrightarrow2\left(x+4\right)=16\)

\(\Leftrightarrow x+4=8\)

hay x=4

5: Ta có: \(\left(x+1\right)^3=27\)

\(\Leftrightarrow x+1=3\)

hay x=2

1: 2*2*2*2*2*2=2^6

2: x*x*x*x=x^4

3: =7^4*6^4=42^4

4: =5^2*3^2*4^2=60^2

5: =2*3^2*5^5

6: =10^3*10^3=10^6

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)

Áp dụng t/c dtsbn:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+y^2+z^2}{4+9+16}=\dfrac{116}{29}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.4=16\\y^2=4.9=36\\z^2=16.16=16^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=6\\z=16\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\y=-6\\z=-16\end{matrix}\right.\end{matrix}\right.\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)

Ta có: \(x^2+y^2+z^2=116\)

\(\Leftrightarrow4k^2+9k^2+16k^2=116\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=4k=8\end{matrix}\right.\)

Trường hợp 2: k=-2

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=4k=-8\end{matrix}\right.\)

Hhigh             

sửa đề : \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

áp dụng t/c dãy t/s = nhau

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{58}{3+4+5}=\frac{58}{12}=\frac{29}{6}\)

\(\frac{x}{3}=\frac{29}{6}\Rightarrow x=\frac{29}{2}\)

\(\frac{y}{4}=\frac{29}{6}\Rightarrow y=\frac{58}{3}\)

\(\frac{z}{5}=\frac{29}{6}\Rightarrow z=\frac{145}{6}\)

vậy ...

chết rồi mình nhìn nhầm đề nhé : phải là :  3+4-5 = 58/2=29

từ đây bạn tìm x;y;z nhé!

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{x+y-z}{3+4-5}=\frac{58}{2}=29\)

\(\Rightarrow\hept{\begin{cases}x=28.3=54\\y=28.4=112\\z=28.5=140\end{cases}}\)

|2.x+4|=6

TH1: 2.x+4 = 6                

x = 1

TH2: 2.x+4 = - 6                

x = -5

Vậy x thuộc 1 và -5

|2-3.x|=5

Th1: 2-3.x=5

x = -1

Th2: 2-3.x= -5

x = 7/3

Vậy x thuộc -1 và 7/3

|7-x|=9

TH1: 7-x =9

x = -2

TH2: 7-x = -9

x = 16

Vậy.........

a) Ta có: \(\left|2x+4\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+4=6\\2x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6-4=2\\2x=-6-4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{1;-5\right\}\)

b) Ta có: \(\left|2-3x\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}2-3x=5\\2-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=3\\-3x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-1;\dfrac{7}{3}\right\}\)

c) Ta có: \(\left|7-x\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}7-x=9\\7-x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=2\\-x=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=16\end{matrix}\right.\)

Vậy: \(x\in\left\{-2;16\right\}\)

a. |2x + 4| = 6

2x + 4 = 6 hoặc 2x + 4 = -6

2x      = 6 - 4 hoặc 2x   = -6 - 4

2x    = 2 hoặc 2x   = -10 

x     = 2 : 2 hoặc x = -10 : 2

x = 1 hoặc x = -5

Vậy x = 1 hoặc x = -5

|2 - 3x| = 5

2 - 3x = 5 hoặc 2 - 3x = -5

3x  = 2 - 5 hoặc 3x   = -5 - 2

3x = -3 hoặc 3x = -7

x = -3 : 3 hoặc  x = -7 : 3 (ko tính được)

x = -1 

Vậy x = -1

c. |7 - x| = 9

7 - x = 9 hoặc 7 - x = -9

x   = 7 - 9 hoặc x  = 7 - (-9)

x = -2 hoặc x = 16 

Vậy x = -2 hoặc x = 16

a, x = -80

b, x = -11 hoặc -5

c, x =4

d, x  thuộc 1 , 7

 sorry mik ko bik dùng dấu thuộc nhé

a,-80

b,-5

c,4

d,1;7

a=-80

b=-5

c=4

d=1;7

\(\dfrac{x^3+8}{x^2+2x+1}.\dfrac{x^2+3x+2}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{x^3+2^3}{\left(x+1\right)^2}.\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{1^2-x^2}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{\left(x+2\right)\left(x+1\right)}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(1-x\right)\left(x+1\right)^2}\)

a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)

b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)

\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)