a. 2x - 138 = 2³ . 3²

2x              = 72 + 138

x                = 210 : 2

x                = 105

b. 213 - (x - 6) = 1339 : 13

213 - (x - 6)     = 103

x - 6                = 213 - 103

x                     = 110 + 6

x                     = 116

2.x - 138 = 2³ . 3²

2.x - 138 = 8 . 9

2.x - 138 = 72

2.x = 72 + 138

2.x = 210

x = 105

b) 231 - (x - 6) = 1339 : 13

231 - x + 6 = 103

231 - x = 103 - 6

231 - x = 97

x = 231 - 97

x = 134

\(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\Rightarrow\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=k\Rightarrow\hept{\begin{cases}x=2k\\y=\frac{3}{2}k\\z=\frac{4}{3}k\end{cases}}\)

Mà xyz = -108 => \(2k\cdot\frac{3}{2}k\cdot\frac{4}{3}k=-108\Rightarrow4k^3=-108\Rightarrow k^3=-27\Rightarrow k=-3\)

\(\Rightarrow\hept{\begin{cases}x=2.\left(-3\right)=-6\\y=\frac{3}{2}.\left(-3\right)=\frac{-9}{2}\\z=\frac{4}{3}.\left(-3\right)=-4\end{cases}}\)

Vậy x = -7, y = -9/2 , z = -4

\(< =>\left(\left(x+2\right)^3-\left(x-2\right)^3\right)-16-108=0\)

\(< =>\left(x+2-x+2\right)\left(\left(x+2\right)^2+\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\right)-16-108=0\)

\(< =>4\left(x^2+4x+4+x^2-4+x^2-4x+4\right)-16-108=0\)

\(< =>4\left(3x^2+4\right)-16-108=0\)

\(< =>12x^2+16-16-108=0\)

\(< =>12x^2-108=0\)

\(< =>12\left(x^2-9\right)=0\)

\(< =>x^2-9=0\)

\( < =>\left(x-3\right)\left(x+3\right)=0\)

\(< =>\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}< =>\orbr{\begin{cases}x=3\\x=-3\end{cases}}}\)

(x+2)³-(x-2)³-16=108

x³+3x².2+3.x.2²+2³-x³+3x².2-3.x.2²+2³=124

(x³-x³)+(6x²+6x²)+(12x-12x)+(8+8)=124

12x²+16=124

12x²=108

x²=9

x=-3 hoặc x=3

1: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{20}=\frac{y}{9}=\frac{z}{6}=\frac{x-2y+4z}{20-2\cdot9+4\cdot6}=\frac{13}{26}=\frac12\)

=>\(\begin{cases}x=20\cdot\frac12=10\\ y=9\cdot\frac12=\frac92\\ z=6\cdot\frac12=3\end{cases}\)

2: \(\frac{x}{3}=\frac{y}{4}\)

=>\(\frac{x}{15}=\frac{y}{20}\left(1\right)\)

\(\frac{y}{5}=\frac{z}{7}\)

=>\(\frac{y}{20}=\frac{z}{28}\left(2\right)\)

Từ (1),(2) suy ra \(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}\)

mà 2x+3y-z=186

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{2x+3y-z}{2\cdot15+3\cdot20-28}=\frac{186}{62}=3\)

=>\(\begin{cases}x=3\cdot15=45\\ y=3\cdot20=60\\ z=3\cdot28=84\end{cases}\)

3: \(\frac{x}{2}=\frac{2y}{5}=\frac{4z}{7}\)

=>\(\frac{x}{2}=\frac{y}{2,5}=\frac{z}{1,75}\)

mà 3x+5y+7z=123

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{2}=\frac{y}{2,5}=\frac{z}{1,75}=\frac{3x+5y+7z}{3\cdot2+5\cdot2,5+7\cdot1,75}=\frac{123}{30,75}=4\)

=>\(\begin{cases}x=4\cdot2=8\\ y=4\cdot2,5=10\\ z=4\cdot1,75=7\end{cases}\)

4: \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\)

=>\(\frac{x}{2}=\frac{y}{\frac32}=\frac{z}{\frac43}\)

Đặt \(\frac{x}{2}=\frac{y}{\frac32}=\frac{z}{\frac43}=k\)

=>\(x=2k;y=\frac32k;z=\frac43k\)

xyz=-108

=>\(2k\cdot\frac32k\cdot\frac43k=-108\)

=>\(4k^3=-108\)

=>\(k^3=-27\)

=>k=-3

=>\(\begin{cases}x=2\cdot\left(-3\right)=-6\\ y=\frac32\cdot\left(-3\right)=-\frac92\\ z=\frac43\cdot\left(-3\right)=-4\end{cases}\)

Đặt \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}=t\Leftrightarrow\hept{\begin{cases}x=2t\\y=\frac{3}{2}t\\z=\frac{4}{3}t\end{cases}}\)

\(xyz=2t.\frac{3}{2}t.\frac{4}{3}t=4t^3=-108\Leftrightarrow t^3=-27\Leftrightarrow t=-3\)

\(\Leftrightarrow\hept{\begin{cases}x=2.\left(-3\right)=-6\\y=\frac{3}{2}.\left(-3\right)=-\frac{9}{2}\\z=\frac{4}{3}.\left(-3\right)=-4\end{cases}}\)

\(\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+2}{4}\)   => \(\frac{3x+3}{6}=\frac{2y+4}{6}=\frac{z+2}{4}\)(1)

Áp dụng tính chất dãy tỉ số bằng nhau ta có 

TỪ(1) => \(\frac{3x+3+2y+4+z+2}{6+6+4}=\frac{\left(3x+2y+z\right)+\left(3+4+2\right)}{16}\)

=\(\frac{105+9}{16}=\frac{57}{8}\)

b)tương tự câu a

a) Ta có :\(\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+2}{4}\)

=> \(\frac{3x+3}{6}=\frac{2y+4}{6}=\frac{z+2}{4}\)

Lại có 3x - 2y + z = 105

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{3x+3}{6}=\frac{2y+4}{6}=\frac{z+2}{4}=\frac{3x+3-2y-4+z+2}{6-6+4}=\frac{\left(3x-2y+z\right)+3-4+2}{4}\) 

                                                                                                                      \(=\frac{105+1}{4}=\frac{106}{4}=26,5\)

=> x = 52 ; y = 77,5 ; z = 104

b) Ta có : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}\)

Đặt \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}=k\Rightarrow\hept{\begin{cases}x^2=4k\\y^2=9k\\z^2=16k\end{cases}}\)

Lại có x² - y² + 2z² = 108

=> 4k - 9k + 2.16k = 108

=> -5k + 32k = 108

=> 27k = 108

=> k = 4

=> x = \(\pm\)4 ; y = \(\pm\)6 ; z = \(\pm\)8

Vì \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)=> x ; y ; z cùng dấu

=> các cặp số (x;y;z) thỏa mãn bài toán là (-4;-6;-8) ; (4;6;8)

Xin lỗi viết sai đề ! 

              (x+3)⁴-(x-3)⁴-24.x².(x-1)=108

<=> [(x+3)²-(x-3)² ]-24.x².(x-1)=108

<=> (x+3-x+3).(x+3+x-3)-24.x².(x-1)=108

<=> 6.2x-24.x².(x-1)=108

<=> 12x.[1-2x.(x-1)]=108

<=> x.[1-2x.(x-1)]=9

Ta có : 9=9.1=3.3=(-9).(-1)=(-3).(-3)

Thay \(\hept{\begin{cases}x=3\\1-2x.\left(x-1\right)=3\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=ko\left(tm\right)\end{cases}}\left(kotm\right)}\)

Thay \(\hept{\begin{cases}x=-3\\1-2x.\left(x-1\right)=-3\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\x=2\end{cases}}\left(tm\right)\)

Thay \(\hept{\begin{cases}x=1\\1-2x.\left(x-1\right)=9\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\x=ko\left(tm\right)\end{cases}\left(kotm\right)}\)

Thay \(\hept{\begin{cases}x=9\\1-2x.\left(x-1\right)=1\end{cases}\Rightarrow\hept{\begin{cases}x=9\\x=0\end{cases}\left(tm\right)}}\)

Thay \(\hept{\begin{cases}x=-1\\1-2x.\left(x-1\right)=-9\end{cases}\Rightarrow}\hept{\begin{cases}x=-1\\x=....\end{cases}\left(tm\right)}\)

Thay \(\hept{\begin{cases}x=-9\\1-2x.\left(x-1\right)=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-9\\x=...\end{cases}\left(tm\right)}}\)

           \(\left(x+3\right)^4-\left(x-3\right)^4-24.x^2.\left(x-1\right)=108.\)

\(\Leftrightarrow\)\(\left[\left(x+3\right)^2-\left(x-3\right)^2\right].\left[\left(x+3\right)^2+\left(x-3\right)^2\right].24x^2.\left(x-1\right)=108\)

\(\Leftrightarrow\)\(\left(x+3-x+3\right).\left(x+3+x-3\right).\left[\left(x^2+6x+9\right)+\left(x^2-6x+9\right)\right].24x^2.\left(x-1\right)=108\)

\(\Leftrightarrow\)\(6.2x.\left(x^2+6x+9+x^2-6x+9\right).24x^2.\left(x-1\right)=108\)

\(\Leftrightarrow\)\(12x.\left(2x^2+18\right).24x^2.\left(x-1\right)=108\)

\(\Leftrightarrow\)\(288x^3.\left[2.\left(x^2+9\right)\right].\left(x-1\right)=108\)

\(\Leftrightarrow\)\(\left(x^4-x^3\right).\left(x^2+9\right).2=\frac{3}{8}\)

\(\Leftrightarrow\)\(x^8+9x^4-x^6-9x^3=\frac{3}{16}\)

\(\Leftrightarrow\)\(x^2.\left(x^4-x^3\right)+9.\left(x^4-x^3\right)=\frac{3}{16}\)

\(\Leftrightarrow\)\(\left(x^2+9\right).\left(x^4-x^3\right)=\frac{3}{16}\)

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