a) \(x-2=-6\)

\(x=-6+2\)

\(x=-4\)

b) \(15-\left(x-7\right)=-21\)

\(x-7=36\)

\(x=43\)

c) \(4.\left(3x-4\right)-2=18\)

\(4\left(3x-4\right)=20\)

\(3x-4=5\)

\(3x=9\)

\(x=3\)

d) \(\left(3x-6\right)+3=32\)

\(3x-6=29\)

\(3x=29+6\)

\(3x=35\)

\(x=\frac{35}{3}\)

e) \(\left(3x-6\right).3=32\)

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

\(3x=\frac{32}{3}+6\)

\(3x=\frac{50}{3}\)

\(x=\frac{50}{9}\)

f) \(\left(3x-6\right):3=32\)

\(3x-6=96\)

\(3x=102\)

\(x=34\)

g) \(\left(3x-6\right)-3=32\)

\(3x-6=35\)

\(3x=41\)

\(x=\frac{41}{3}\)

h) \(\left(3x-2^4\right).7^3=2.7^4\)

\(\left(3x-2^4\right)=2.7=14\)

\(\left(3x-16\right)=14\)

\(3x=14+16=30\)

\(x=10\)

i) \(\left|x\right|=\left|-7\right|\)

\(\left|x\right|=7\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

k) \(\left|x+1\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

l) \(\left|x-2\right|=3\)

\(\Rightarrow\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

m) \(x+\left|-2\right|=0\)

\(x+2=0\)

\(x=-2\)

o) \(72-3\left|x+1\right|=9\)

\(3\left|x-1\right|=63\)

\(\left|x-1\right|=21\)

\(\Rightarrow\orbr{\begin{cases}x-1=21\\x-1=-21\end{cases}\Rightarrow\orbr{\begin{cases}x=22\\x=-20\end{cases}}}\)

p) Ta có: \(\left|x-1\right|=3\)

\(\Rightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}\)

mà \(x+1< 0\)

\(\Rightarrow x-1=-3\)

\(\Rightarrow x=-2\)

q) \(\left(x-2\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)

hok tốt!!

31x72 - 31x70 - 31 x 2 - 31

=31x(72 -70 -2 -1)

=31 x (-1)

= -31

những câu sau thì lam tương tự nha bạn ^_^

a) -45 : ( 3x - 17 ) = 3²

3x - 17 = -45 : 9

3x - 17 = -5

3x = 12

x = 4

b) \(\left(2x-8\right)\left(-2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-8=0\\-2x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)

Vậy.....

c) \(72:\left|4x-3\right|=2^3\)

\(\left|4x-3\right|=9\)

\(\Rightarrow\orbr{\begin{cases}4x-3=9\\4x-3=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-3}{2}\end{cases}}\)

Vậy.....

a) \(\frac{3}{7}-\frac{1}{7}x=\frac{2}{3}\)

=> \(\frac{1}{7}x=\frac{3}{7}-\frac{2}{3}=-\frac{5}{21}\)

=> \(x=-\frac{5}{21}:\frac{1}{7}=-\frac{5}{21}\cdot7=-\frac{5}{3}\)

b) \(3x^2-2=72\)=> 3x² = 74 => x² = 74/3 => x không thỏa mãn

c) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)

=> \(\left(19x+2\cdot25\right):14=5^2-4^2=9\)

=> \(\left(19x+50\right):14=9\)

=> \(19x+50=126\)

=> \(19x=76\)

=> x = 4

d) \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{16}+x:\frac{1}{32}=343\)

=> \(x\cdot2+x\cdot4+x\cdot8+x\cdot16+x\cdot32=343\)

=> \(x\left(2+4+8+16+32\right)=343\)

=> x . 62 = 343

=> x = 343/62

a)19 - (x + 2³)=2⁴- 6

   19 - (x + 2³) = 16 - 6 

    19 - (x + 2³) = 10

     (x + 2³) = 19 - 10

      x + 2³= 9

      x + 2³ = 3³

      ^{x + 2 = 3}

      ^{x= 3-2}

       ^{x= 1}

sửa lại :

a)19 - (x + 2³)=2⁴- 6

   19 - (x + 2³) = 16 - 6 

    19 - (x + 2³) = 10

     (x + 2³) = 19 - 10

      x + 2³= 9

    =>  x + 8= 9             x= 1

      ^{=> x + 8 =-9                     x= -17}

sửa lại b) 4³ + 3² : (x + 1) = 65

a.\(\left(3x\right)^2-4\left(x-3\right)^2=0\)

<=> \(9x^2-4\left(x^2-6x+9\right)=0\)

<=> \(9x^2-4x^2+24x-36=0\)

<=>\(5x^2+24x-36=0\)

giải pt bậc hai thì pt có hai nghiệm x={1,2;-6}

a) (3x)² - 4(x- 3)² = 0

\(\Leftrightarrow\) (3x - 2x + 6)(3x + 2x - 6) = 0

\(\Leftrightarrow\) (x+ 6)(5x - 6) = 0

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+6=0\\5x-6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\x=\dfrac{6}{5}\end{matrix}\right.\)

Vậy phượng trình có tập nghiệm là: S = {-6;\(\dfrac{6}{5}\)}

b) x³ + x² + 4 = 0

\(\Leftrightarrow\) x³ + 2x² - x² + 4 = 0

\(\Leftrightarrow\) (x³ + 2x²) - (x² - 4) = 0

\(\Leftrightarrow\) x²(x + 2) - (x + 2)(x - 2) = 0

\(\Leftrightarrow\) (x² - x + 2)(x + 2) = 0

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x^2-x+2=0\left(vôli\right)\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\) x = -2

Vậy phương trình có tập nghiệm là: S={-2}

c) (x - 1)²(x - 3) + (1 - x)²(x + 3) = 72

\(\Leftrightarrow\) (x - 1)²(x - 3) + (x - 1)²(x + 3) = 72

\(\Leftrightarrow\) (x - 1)²(x - 3 + x + 3) = 72

\(\Leftrightarrow\) 2x(x² - 2x + 1) = 72

\(\Leftrightarrow\) 2x³ - 4x² + 2x - 72 = 0

\(\Leftrightarrow\) 2(x³ - 2x² + x - 36) = 0

\(\Leftrightarrow\) x³ - 2x² + x - 36 = 0

\(\Leftrightarrow\) x³ - 4x² + 2x² - 8x + 9x - 36 = 0

\(\Leftrightarrow\) (x³ - 4x²) + (2x² - 8x) + (9x - 36) = 0

\(\Leftrightarrow\) x²(x - 4) + 2x(x - 4) + 9(x - 4)= 0

\(\Leftrightarrow\) (x² + 2x + 9)(x - 4) = 0

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x^2+2x+9=0\left(vôli\right)\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\) x = 4

Vậy phương trình có tập nghiệm là: S={4}

\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

2/

a,x=0 or x=-3

b,x=2 or x=5

c,x=1

a) x ∈ { - 5 ; 1 }            b) x ∈ ∅

c) x = 0 .                  d)  x = 1 4