3x-21/3+4x-32/2 = 5x-37/4 +9x-33/6

7x+(-21/3-32/2) = 14x+(-37/4 - 33/6 )

7x+(-42/6-96/6) = 14x+ (-111/12 - 66/12)

7x+(-138/6) =14x + (-177/12) 

7x - 14x =138/6 +(-177/12)

-7x = 276/12+ (-177/12)

-7x = 99/12

-x = 23/28

x = -23/28 

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

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

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

a) Kết quả M = (x + l)(2x - 3);

b) Kết quả M = (2x - 1)(x - 2).

a) Ta có : 2 x : 2 2 = 2 5  nên x = 7.

b) Ta có: 3 x : 3 2 = 3 5  nên x = 7.

c) Ta có : 4 4 : 4 x = 4 2  nên x = 2.

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

a,  \(\frac{\left(5-2x\right)}{3}=\frac{\left(4x-1\right)}{-5}\)

\(\Leftrightarrow-5(5-2x)=3\left(4x-1\right)\)

\(\Leftrightarrow10x-25=12x-3\)

\(\Leftrightarrow10x-12x=25-3\)

\(\Leftrightarrow-2x=22\)

\(\Leftrightarrow x=-11\)

b,  \(\frac{\left(12-3x\right)}{32}=\frac{6}{\left(4-x\right)}\)

\(\Leftrightarrow\frac{3\left(4-x\right)}{32}=\frac{6}{\left(4-x\right)}\)

\(\Leftrightarrow3(4-x)\left(4-x\right)=32.6\)

\(\Leftrightarrow(4-x)\left(4-x\right)=32.2\)

\(\Leftrightarrow(4-x)^2=64\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=8\\4-x=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=12\end{cases}}\)

c,  \(\frac{\left(10-2x\right)}{6}=\frac{27}{\left(5-x\right)}\)

\(\Leftrightarrow\frac{2\left(5-x\right)}{6}=\frac{27}{\left(5-x\right)}\)

\(\Leftrightarrow2(5-x)\left(5-x\right)=27.6\)

\(\Leftrightarrow(5-x)\left(5-x\right)=27.3\)

\(\Leftrightarrow(5-x)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}5-x=9\\5-x=-9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=14\end{cases}}\)

a, \(\frac{5-2x}{3}=\frac{4x-1}{-5}\Leftrightarrow-25+10x=12x-3\Leftrightarrow-22-2x=0\Leftrightarrow x=-11\)

b, \(\frac{12-3x}{32}=\frac{6}{4-x}\Leftrightarrow\frac{12-3x}{32}=\frac{18}{12-3x}\)

\(\Leftrightarrow\left(12-3x\right)^2=576\Leftrightarrow12-3x=\pm2\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=\frac{14}{3}\end{cases}}\)

c, \(\frac{10-2x}{6}=\frac{27}{5-x}\Leftrightarrow\frac{10-2x}{6}=\frac{54}{10-2x}\)

\(\Leftrightarrow\left(10-2x\right)^2=324\Leftrightarrow10-2x=\pm18\)\(\Leftrightarrow\orbr{\begin{cases}x=14\\x=-4\end{cases}}\)

\(\frac{5-2x}{3}=\frac{4x-1}{-5}\)

=> -5(5 - 2x) = 3(4x - 1)

=> -25 + 10x = 12x - 3

=> -25 + 10x - 12x +3  = 0

=> (-25 + 3) + (10x - 12x) = 0

=> -22 - 2x = 0

=> 2x = -22

=> x = -11

\(\frac{12-3x}{32}=\frac{6}{4-x}\)

=> (12 - 3x)(4 - x) = 32.6

=> 12(4 - x) - 3x(4 - x) = 192

=> 48 - 12x - 12x + 3x² = 192

=> 48 - 24x + 3x² = 192

=> 3(x² - 8x + 16) = 192

=> x² - 8x + 16 = 64

=>  x² - 4x - 4x + 16 = 64

=> x(x - 4) - 4(x - 4) = 64

=> (x - 4)² = 64

=> (x - 4)² = (\(\pm\)8)²

=> \(\orbr{\begin{cases}x-4=8\\x-4=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=12\\x=-4\end{cases}}\)

\(\frac{10-2x}{6}=\frac{27}{5-x}\)

=> (10 - 2x)(5 - x) = 27.6

=> 10(5 - x) - 2x(5 - x) = 162

=> 50 - 10x - 10x + 2x² = 162

=> 50 - 20x + 2x² = 162

=> 2(x² - 10x + 25) = 162

=> x² - 10x + 25 = 81

=> x² - 5x - 5x + 25 = 81

=> x(x - 5) - 5(x - 5) = 81

=> (x - 5)² = 81

=> (x - 5)² = (\(\pm\)9)²

=> \(\orbr{\begin{cases}x-5=9\\x-5=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=14\\x=-4\end{cases}}\)