-2/x=-x/0,32

ta có : -2.0,32=-x.x

=> 2.0,32=x.x

=> 0,64=x²

=> 0,8²=x²

=> x=0,8

Vậy x=0,8

8-x/3=2x-3/5

ta có : (8-x).5=(2x-3).3

=>     40-5x=6x-9

=>     40+9=6x+5x

=>      49 =11x

=>  x= 49:11

=>   x=49/11

Vậy x=49/11

\(3.\left(X-\frac{1}{2}\right)-3.\left(X-\frac{1}{3}\right)=X\)

\(\Leftrightarrow3.\left[\left(X-\frac{1}{2}\right)-\left(X-\frac{1}{3}\right)\right]=X\)

\(\Leftrightarrow3.\left[X-\frac{1}{2}-X+\frac{1}{3}\right]=X\)

\(\Leftrightarrow3.\left[\left(X-X\right)+\left(-\frac{1}{2}+\frac{1}{3}\right)\right]=X\)

\(\Leftrightarrow3.\left[0+\left(-\frac{1}{6}\right)\right]=X\)

\(\Leftrightarrow3.\left(-\frac{1}{6}\right)=X\)

\(\Leftrightarrow X=-\frac{1}{2}\)

\(3\left(X-\frac{1}{2}\right)-3\left(X-\frac{1}{3}\right)=X\)

\(\Leftrightarrow3X-\frac{3}{2}-3X+1=X\)

\(\Leftrightarrow3X-3X-X=\frac{3}{2}-1\)

\(\Leftrightarrow-X=\frac{1}{2}\)

\(\Leftrightarrow X=-\frac{1}{2}\)

Ta có\(\frac{a}{b}=\frac{b}{c}\)

Theo tính chất thì \(a\cdot c=b\cdot b\)

=>\(a\cdot c=b^2\)

=>Hoặc a=b=c hoặc a=c=-b.

Với a=c=-b=>a/b ko thể = c/a.

Vậy bài toán đc cm.

ta có : a/b=b/c=c/a

=> a/b=b/c=c/a=a+b+c/b+c+a=1

=> a=b=c=1

Vậy a=b=c

Ta có: \(\frac{32-x}{3}=\frac{x-18}{5}=\frac{32-x+x-18}{3+5}=\frac{14}{8}=\frac{7}{4}\)

\(\Rightarrow\frac{32-x}{3}=\frac{7}{4}\Leftrightarrow4\left(32-x\right)=7.3\)

\(\Leftrightarrow128-4x=21\Rightarrow4x=128-21=107\)

\(\Leftrightarrow x=\frac{107}{4}\)

Đáp số: \(x=\frac{107}{4}\)

32-x/3=x-18/5

=> (32-x).5=(x-18).3

=> 160-5x=3x-54

=> 160+54=3x+5x

=> 214=8x

=> x=214:8

=> x=26,75

Vậy x=26,75.

\(\sqrt{\left(3x-12\right)^2}=\sqrt{16x^2}\)

=>\(\sqrt{\left(3x-12\right)^2}=\sqrt{4x^2}\)

=>|3x-12|=|4x|

=>3x-12=4x hoặc 3x-12=-4x

=>3x-4x=12 hoặc 3x+4x=12

=>x=-12 hoặc x=\(\frac{12}{7}\)

Ta thấy A gồm có 99 số hạng nên ta nhóm mỗi nhóm 3 số hạng.

Ta có: A = 1 + 5 + 5² + 5³ + 5⁴ + 5⁵ +...+ 5⁹⁷ + 5⁹⁸ + 5⁹⁹

             = (1 + 5 + 5² )+ (5³ + 5⁴ + 5⁵ )+...+( 5⁹⁷ + 5⁹⁸ + 5⁹⁹ )

             =(1 + 5 + 5² )+ 5³(1 + 5 + 5² ) +...+ 5⁹⁷(1 + 5 + 5² )

             = ( 1 + 5 + 5²)(1 + 5³+....+5⁹⁷)

             = 31(1 + 5³+....+5⁹⁷)

Vì có một thừa số là 31 nên A chia hết cho 31.

 P/s Đừng để ý câu trả lời của mình

\(x:\frac{1}{3}=\frac{12}{99}:\frac{15}{90}\)

\(x:\frac{1}{3}=\frac{12}{99}:\frac{1}{6}\)

\(x:\frac{1}{3}=\frac{8}{11}\)

\(x=\frac{8}{11}X\frac{1}{3}\)

\(x=\frac{8}{33}\)

tk mk nhe

Ta có: x:1/3=12/99:15/90

x:1/3=8/11

x=8/33

Vậy x=8/33

\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Rightarrow x\left(x^2+5x-5x-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=3\)

\(\Rightarrow x^3+5x^2-5x^2-25x-x^3+2x^2-4x-2x^2+4x-8=3\)

\(\Rightarrow\left(x^3-x^3\right)+\left(5x^2-5x^2+2x^2-2x^2\right)+\left(-25x-4x+4x\right)-8=3\)

\(\Rightarrow-25x=3+8=11\)

\(\Rightarrow x=\frac{11}{-25}=\frac{-11}{25}\)

Vay x = \(\frac{-11}{25}\)

\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x^3-25x-\left(x^3-2x^2+4x+2x^2-4x+8\right)=3\)

\(\Leftrightarrow x^3-25x-x^3+2x^2-4x-2x^2+4x-8-3=0\)

\(\Leftrightarrow-25x-11=0\)

\(\Leftrightarrow-25x=11\)

\(\Leftrightarrow x=-\frac{11}{25}\)

a) Ta có: B=34.10\(^{-9}\)=34.\(\frac{1}{10^9}\)=34.\(\frac{1}{10}\).\(\frac{1}{10^8}\)=3,4.10\(^{-8}\)

Vậy A=B

b) Ta có: A=10\(^{-4}\)+10\(^{-3}\)+10\(^{-2}\)=0,0111

B=10\(^{-9}\)=\(\frac{1}{10^9}\)=\(\frac{1}{1000000000}\)=0,000000001

\(\Rightarrow\)A= 11100000B

x=1;2

sai roi