\((3\frac{4}{5}-2x).1\frac{1}{3}=5\frac{5}{7}\)

\(3\frac{4}{5}-2x\)=\(4\frac{2}{7}\)

2x=\(\frac{-17}{35}\)

x=\(\frac{-17}{70}\)

Vậy x=\(\frac{-17}{70}\)

2

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

=>25-5x+3=64

=>25+3-64=5x

=>-36=5x

=>x=-36:5

=>x=-7,2

b,3x-17=x+3

=>3x-x=17+3

=>2x=20

=>x=20:2

=>x=10

c,\(5^x+5^{x+2}=650\)

=>\(5^x+5^x.5^2=650\)

=>\(5^x+5^x.25=650\)

=>\(5^x.\left(1+25\right)=650\)

=>\(5^x.26=650\)

=>\(5^x=650:26\)

=>\(5^x\)=25

=>\(5^x=5^2\)

=>x=2

\(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{6}{7}+\dfrac{7}{8}+\dfrac{6}{7}-\dfrac{5}{6}+\dfrac{4}{5}-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{2}\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(-\dfrac{2}{3}+\dfrac{2}{3}\right)+....+\left(-\dfrac{6}{7}+\dfrac{6}{7}\right)+\dfrac{7}{8}\)

\(=\dfrac{7}{8}\)

Chúc bạn học tốt!!!

\(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{6}{7}+\dfrac{7}{8}+\dfrac{6}{7}-\dfrac{5}{6}+\dfrac{4}{5}-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{2}\)

= \((\dfrac{1}{2}-\dfrac{1}{2})+(-\dfrac{2}{3}+\dfrac{2}{3})+...+(-\dfrac{6}{7}+\dfrac{6}{7})+\dfrac{7}{8}\)

=\(\dfrac{7}{8}\)

1/2-2/3+3/4-4/5+5/6-6/7+7/8+6/7-5/6+4/5-3/4+2/3-1/2

= \(\dfrac{7}{8}\)

a) \(15\cdot5+75\cdot5+5\cdot3+5\cdot7\)

\(=5\left(15+75\right)+5\left(3+7\right)\)

\(=5\cdot90+5\cdot10=5\left(90+10\right)=5\cdot100=500\)

b) \(2\cdot200+4\cdot200+4\cdot50+125\cdot8\)

\(=2.2.100+4\cdot200+4\cdot50+125\cdot2\cdot4\)

\(=4\left(100+200+50+250\right)\)

\(=4\cdot600=2400\)

c) \(125\cdot20\cdot80\cdot2\cdot100\cdot5\)

\(=\left(125\cdot8\right)\cdot\left(2\cdot5\right)\cdot100\cdot20\)

\(=1000\cdot10\cdot100\cdot20\)

\(=20000000\)

\(\left(x-2\right)\left(2x+1\right)-5\left(x+3\right)=2x\left(x-3\right)+4\left(1+2x\right)-2\left(1+x\right)\)

\(2x^2+x-4x-2-5x-15=2x^2-6x+4+8x-2-2x\)

\(x-4x-2-5x-15=-6x+4+8x-2-2x\)

\(\Rightarrow-8x-17=2\)

\(-8x=19\Rightarrow x=-\dfrac{19}{8}\)

Vậy \(x=-\dfrac{19}{8}\)

\(\left(x-2\right)\left(2x+1\right)-5\left(x+3\right)=2x\left(x-3\right)+4\left(1+2x\right)-2\left(1+x\right)\)

\(\Rightarrow2x^2+x-4x-2-5x-15-2x^2+6x-4-8x+2+2x=0\\ \Leftrightarrow-8x-19=0\\ \Leftrightarrow-8x=19\\ \Leftrightarrow x=-\dfrac{19}{8}\)

dễ lắm bạn ơi bọn mình học rùi

bn lấy tử số của từng mỗi rồi chia cho mẫu của từng cái rồi rút gọn lafd ra

\(\dfrac{\dfrac{5}{7}+\dfrac{5}{9}+\dfrac{5}{11}}{\dfrac{25}{7}+\dfrac{25}{9}+\dfrac{25}{11}}\cdot\dfrac{4+\dfrac{4}{73}-\dfrac{4}{115}}{5+\dfrac{5}{73}-\dfrac{1}{23}}\)

=\(\dfrac{5\cdot\left(\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}\right)}{25\left(\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}\right)}\cdot\dfrac{4\left(1+\dfrac{1}{73}+\dfrac{1}{115}\right)}{5+\dfrac{5}{73}-\dfrac{5}{115}}\)

=\(\dfrac{5}{25}\cdot\dfrac{4\left(1+\dfrac{1}{73}-\dfrac{1}{115}\right)}{5\left(1+\dfrac{1}{73}-\dfrac{1}{115}\right)}\)

=\(\dfrac{1}{5}\cdot\dfrac{4}{5}\)=\(\dfrac{4}{25}\)

\(\dfrac{\dfrac{5}{7}+\dfrac{5}{9}+\dfrac{5}{11}}{\dfrac{25}{7}+\dfrac{25}{9}+\dfrac{25}{11}}\cdot\dfrac{4+\dfrac{4}{73}-\dfrac{4}{115}}{5+\dfrac{5}{73}-\dfrac{1}{23}}\)

\(=\dfrac{\dfrac{5}{7}+\dfrac{5}{9}+\dfrac{5}{11}}{5\left(\dfrac{5}{7}+\dfrac{5}{9}+\dfrac{5}{11}\right)}\cdot\dfrac{4+\dfrac{4}{73}-\dfrac{4}{115}}{5+\dfrac{5}{73}-\dfrac{5}{115}}\)

\(=\dfrac{1}{5}\cdot\dfrac{4\left(1+\dfrac{1}{73}-\dfrac{1}{115}\right)}{5\left(1+\dfrac{1}{73}-\dfrac{1}{115}\right)}\)

\(=\dfrac{1}{5}\cdot\dfrac{4}{5}=\dfrac{4}{25}\)

6a + 5 ⋮ 3a-1

=> 6a - 2 + 7 ⋮ 3a-1

=> 2. (3a - 1) + 7 ⋮ 3a-1

Có: 2. (3a - 1) ⋮ 3a - 1 => Để 6a + 5 ⋮ 3a-1 thì:

7 ⋮ 3a - 1

=> 3a - 1 ∈ Ư(7)

=> 3a - 1 ∈ {1; -1; 7; -7}

=> 3a ∈ {2; 0; 8; -6}

Mà: 3a ⋮ 3 => 3a ∈ {0; -6}

=> a ∈ {0; -2}

Vậy: ................

6a+5⋮3a-1

=>6a-3+8⋮3a-1

=>2(3a-1)+8⋮3a-1

=>8⋮3a-1[vì 3a-1⋮3a-1 nên 2(3a-1)⋮3a-1]

=>3a-1∈Ư(8)={1;2;4;8;-1;-2;-4;-8}

Đến đây làm tiếp đc hơm ??

Làm đc thì tự làm nốt nhé bae