a) (5x - 1) : 3 + 1 = 4

=> (5x - 1) : 3 = 3

=> (5x - 1) = 9

=> 5x - 1 = 9

=> 5x = 10

=> x = 2

b) 54 : (16 - x) - 1=5

=> 54:(16-x) = 6

=> 16-x = 9

=> x = 7

A=7 B=1 OR =0 C=2 D=4.5.6

a: \(2^{x}\cdot4=128\)

=>\(2^{x}=\frac{128}{4}=32=2^5\)

=>x=5

b: \(x^{15}=x\)

=>\(x^{15}-x=0\)

=>\(x\left(x^{14}-1\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^{14}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{14}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\end{array}\right.\)

c: \(\left(2x+1\right)^3=125\)

=>\(\left(2x+1\right)^3=5^3\)

=>2x+1=5

=>2x=5-1=4

=>\(x=\frac42=2\)

d: \(\left(x-5\right)^4=\left(x-5\right)^6\)

=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)

=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)

=>\(\left(x-5\right)^4\cdot\left(x-5-1\right)\left(x-5+1\right)=0\)

=>\(\left(x-5\right)^4\cdot\left(x-6\right)\left(x-4\right)=0\)

=>\(\left[\begin{array}{l}x-5=0\\ x-6=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=6\\ x=4\end{array}\right.\)

\(8-12x+6x^2-x^3\)

\(=\left(2-x\right)^3\)

\(125x^3-75x^2+15x-1\)

\(=\left(5x-1\right)^3\)

\(x^2-xz-9y^2+3yz\)

\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x+3y-z\right)\)

\(x^3-x^2-5x+125\)

\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

\(x^3+2x^2-6x-27\)

\(=x^3+5x^2+9x-3x^2-15x-27\)

\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

\(12x^3+4x^2-27x-9\)

\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(4x^2-9\right)\)

\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)

\(4x^4+4x^3-x^2-x\)

\(=4x^3\left(x+1\right)-x\left(x+1\right)\)

\(=x\left(x+1\right)\left(4x^2-1\right)\)

\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)

Là 00000000000000

Câu a:

1 + {-2 - [-3 + (-4 + |x|)]} = 1 - 2 + [(-3 - 4)]

1 + {-2 -[-3 - 4 + |x|]} = 1 - 2 + [-7]

1 + {-2 - [- 7 + |x|]} = 1 - 2 - 7

1 + {- 2 + 7 - |x|} = - 1 - 7

1 + {5 - |x|} = - 8

5 - |x| = - 8 - 1

5 - |x| = - 9

|x| = 5 + 9

|x| = 14

x = - 14 hoặc x = 14

Vậy x ∈ {-14; 14}

Câu b:

34 + (9 - 21) = 3417 - (x + 3417)

34 + (-12) = 3417 - x - 3417

34 - 12 = 3417 - 3417 - x

22 = - x

x = 22 : (-1)

x = - 22

Vậy x = - 22

a) \(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\)

\(\dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\)

\(\left(\dfrac{1}{2}-\dfrac{2}{3}\right)x=\dfrac{7}{12}\)

\(\dfrac{-1}{6}x=\dfrac{7}{12}\)

\(x=\dfrac{7}{12}:\dfrac{-1}{6}\)

\(x=\dfrac{-7}{2}\)

b) \(x:4\dfrac{1}{3}=-2,5\)

\(x:\dfrac{13}{3}=\dfrac{-5}{2}\)

\(x=\dfrac{13}{3}.\dfrac{-5}{2}\)

\(x=\dfrac{-65}{6}\)

c) \(5,5x=\dfrac{13}{15}\)

\(\dfrac{11}{2}x=\dfrac{13}{15}\)

\(x=\dfrac{13}{15}:\dfrac{11}{2}\)

\(x=\dfrac{26}{165}\)

d) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\left(\dfrac{3x}{7}+1\right)=\dfrac{-1}{28}.\left(-4\right)\)

\(\dfrac{3x}{7}+1=\dfrac{1}{7}\)

\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)

\(\dfrac{3x}{7}=\dfrac{-6}{7}\)

\(\Rightarrow3x=-6\\ x=\left(-6\right):3\\ x=-2\)

câu d mình nhầm bạn nha

d)\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\dfrac{3x}{7}+1=\dfrac{-1}{28}.\left(-4\right)\)

\(\left(\dfrac{3x}{7}+1\right)=\dfrac{1}{7}\)

\(\dfrac{3x}{7}=\dfrac{1}{7}-1\)

\(\dfrac{3x}{7}=\dfrac{-6}{7}\)

Ta có : 3x =-6

x = -6:3

x=-2

a, \(x:3\frac{1}{15}=1\frac{1}{2}\)

\(\Rightarrow x=1\frac{1}{2}\cdot3\frac{1}{15}\)

\(\Rightarrow x=\frac{3}{2}\cdot\frac{46}{15}=\frac{3\cdot46}{2\cdot15}=\frac{1\cdot23}{1\cdot5}=\frac{23}{5}=4\frac{3}{5}\)

\(b)x\cdot\frac{15}{28}=\frac{3}{20}\)

\(\Rightarrow x=\frac{3}{20}:\frac{15}{28}=\frac{3}{20}\cdot\frac{28}{15}=\frac{1}{5}\cdot\frac{7}{5}=\frac{7}{25}\)

Tự làm nốt câu cuối :>

Câu a:

18 - |x - 1| = 2^2

18 - |x - 1| = 4

|x - 1| = 18 - 4

|x - 1| = 14

x - 1 = -14 hoặc x - 1 = 14

Th1: x - 1 = -14

x = - 14 + 1

x = -13

Th2: x - 1 = 14

x = 14 + 1

x = 15

Vậy x ∈ {-13; 15}

Câu b:

x - {55 - [49 + (-28 - x)]} = 13 - {47 + [25 - (32 - x)]}

x - {55 - [49 - 28 - x]} = 13 - {47 + [25 - 32 + x]}

x - {55 - 49 + 28 + x} = 13 - {47 + 25 - 32 + x}

x - {6 + 28 + x} = 13 - {72 - 32 + x}

x - {34 + x} = 13 - {40 + x}

x - 34 - x = 13 - 40 - x

(x - x) + x = 13 - 40 + 34

0 + x = - 27 + 34

x = 7

Vậy x = 7

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}