\(3^5\)

Hướng dẫn giải:

a) 3 x (20 – 5)

Cách 1:

3 x (20 – 5) = 3 x 15 = 45

Cách 2:

3 x (20 – 5) = 3 x 20 – 3 x 5 = 60 – 15 = 45

b) 20 x (40 – 1)

Cách 1:

20 x (40 – 1) = 20 x 39 = 780

Cách 2:

20 x (40 – 1) = 20 x 40 – 20 x 1 = 800 – 20 = 780

7/12 nha!

K mình đi mà!

\(\frac{10}{20}+\frac{5}{30}:2\)

\(=\frac{1}{2}+\frac{1}{6}:2\)

\(=\frac{1}{2}+\frac{1}{12}\)

\(=\frac{7}{12}\)

10/20+5/30:2

=1/2+1/6:2

=1/2+1/12

=6/12+1/12

=6+1/12

=7/12

a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\) 

\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)

Nên x + 1 = 0

=> x = -1

còn b vs c thì sao ạ

b) Sai đề à bạn đề \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)  hả đề này mk làm đc 

\(\frac{10}{25}=\frac{10:5}{25:5}=\frac{2}{5}\)

\(\frac{5}{20}=\frac{5:5}{20:5}=\frac{1}{4}\)

\(\frac{6}{30}=\frac{6:6}{30:6}=\frac{1}{5}\)

\(\frac{60}{20}=\frac{60:20}{20:20}=3\)

1) Ta có: \(5\left(x-2\right)=3x+10\)

\(\Leftrightarrow5x-10-3x-10=0\)

\(\Leftrightarrow2x-20=0\)

\(\Leftrightarrow2\left(x-10\right)=0\)

Vì 2>0

nên x-10=0

hay x=10

Vậy: x=10

2) Ta có: \(x^2\left(x-5\right)-4x+20=0\)

\(\Leftrightarrow x^2\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)

Vậy: x∈{-2;2;5}

3) Ta có: \(\frac{3x+1}{4}+\frac{8x-21}{20}=\frac{3\left(x+2\right)}{5}-2\)

\(\Leftrightarrow\frac{5\left(3x+1\right)}{20}+\frac{8x-21}{20}-\frac{12\left(x+2\right)}{20}+\frac{40}{20}=0\)

\(\Leftrightarrow15x+5+8x-21-12\left(x+2\right)+40=0\)

\(\Leftrightarrow15x+5-8x-21-12x-24+40=0\)

\(\Leftrightarrow-5x=0\)

hay x=0

Vậy: x=0

4) ĐKXĐ: x≠5; x≠-5

Ta có: \(\frac{3}{4x-20}+\frac{7}{6x+30}=\frac{15}{2x^2-50}\)

\(\Leftrightarrow\frac{3}{4\left(x-5\right)}+\frac{7}{6\left(x+5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}+\frac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}-\frac{180}{12\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow9x+45+14x-70-180=0\)

\(\Leftrightarrow23x-205=0\)

\(\Leftrightarrow23x=205\)

hay \(x=\frac{205}{23}\)(tm)

Vậy: \(x=\frac{205}{23}\)

A=1,319387755 hoặc 1293/980