Bài 3:

a: Ta có: \(23\left(42-x\right)=23\)

\(\Leftrightarrow42-x=1\)

hay x=41

b: Ta có: 15(x-3)=30

nên x-3=2

hay x=5

Bài 1: 

a: 32+89+68=100+89=189

b: 64+112+236=300+112=412

c: \(1350+360+650+40=2000+400=2400\)

a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)

=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)

=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)

b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)

c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)

=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)

=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)

=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)

d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)

=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)

=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)

e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)

=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)

=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)

f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)

=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)

=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)

=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)

a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)

   \(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)

    \(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)

     \(\dfrac{72}{17x}\)        = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)

      \(\dfrac{72}{17x}\)      = - \(\dfrac{1}{14}\)

      17\(x\)       = 72.(-14)

       17\(x\)     = - 1008

         \(x\)       = - 1008 : 17

         \(x\)       = - \(\dfrac{1008}{17}\)

Vậy \(x\) \(=-\dfrac{1008}{17}\)

b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))³ = - \(\dfrac{24}{27}\)

          - \(\dfrac{32}{27}\)  + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))³ 

          (3\(x-\dfrac{7}{9}\))³ = - \(\dfrac{8}{27}\)

         (3\(x-\dfrac{7}{9}\))³ = (- \(\dfrac{2}{3}\))³

           3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)

           3\(x\)        = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
           3\(x\)        = \(\dfrac{1}{9}\)

             \(x\)        = \(\dfrac{1}{9}\) : 3

             \(x\)       = \(\dfrac{1}{27}\)

Vậy \(x=\dfrac{1}{27}\)

c) 2x - 49 = 5 . 3²

2x - 49 = 5 . 9

2x - 49 = 45

2x = 45 + 49

2x = 94

x = 94 : 2

x = 47

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) x = 8.

b) x = -5.

c) x = 5.

d) x = -3.

Bài 1: 

a) =25,97+(6,54+103,46)

=25,97+110

=135,97

b)136x75+75x64

=75x(136+64)

=75x200

=15 000

c) (21/8+1/2):5/16

=(21/8+4/8)x16/5

=25/8x16/5

=10

d)3/17-4/5+14/17

=(3/17+14/17)-4/5

=1-4/5

=1/5

Bài 2:

a)720:\([41-(2x-5)]\)=120

              41 - (2x-5)                 =720:120

              41 - (2x-5)                 =6

                      2x-5                   =41-6

                      2x-5                   =35

                      2x                      =35+5

                      2x                      =40

                      x                        =40:2

                      x                        =20

b)2/3 x X +3/4=3

   2/3 x X        =3-3/4

   2/3 x X        =12/4-3/4

   2/3 x X        =9/4

             x        =9/4:2/3

             x        =9/4x3/2

             x        =27/8

c) x+0,34=1,19x1,02

    x+0,34=1,2138

    x         =1,2138-0,34

    x         =0,8738

Bài 3:

Tổng số tiền An dùng mua đồ:

125 000 + 85 000 + 60 000 + 65 000 = 335 000 (đồng)

Số tiền An còn lại sau khi mua đồ:

350 000 - 335 000 = 15 000 (đồng)

Đ.số: 15 000 đồng

Bài 2:

a, IV: Bốn(4); XXVII: Hai mươi bảy (27), XXX: ba mươi (30), M:một nghìn (1000)

b, 7: VII; 15: XV; 29: XXIX

Con bai tim x thi sao ban?

2x + 32 = 32

2x = 0

x = 0

Chọn A

A nha