\(\frac{3}{5}.x-\frac{1}{2}.x=-\frac{8}{20}\)

\(\left(\frac{3}{5}-\frac{1}{2}\right).x=-\frac{2}{5}\)

               \(\frac{1}{10}.x=-\frac{2}{5}\)

                      \(x=\left(-\frac{2}{5}\right):\frac{1}{10}\)

                      \(x=-4\)

x=-4 bạn nha

CHÚC BẠN HỌC GIỎI

1.

\(\left(2-\frac{1}{5}\right)\cdot\left(\frac{-3}{15}+\frac{1}{3}\right)\)

\(=\left(\frac{10}{5}-\frac{1}{5}\right)\cdot\left(\frac{-3}{15}+\frac{5}{15}\right)\)

\(=\frac{9}{5}\cdot\frac{2}{15}=\frac{18}{75}\)

2.

\(\frac{x}{20}-\frac{2}{5}=\frac{4}{15}\cdot\frac{-9}{8}\)

\(\frac{x}{20}-\frac{2}{5}=\frac{-9}{30}\)

\(\frac{x}{20}=\frac{-9}{30}+\frac{2}{5}\)

\(\frac{x}{20}=\frac{-9}{30}+\frac{12}{30}\)

\(\frac{x}{20}=\frac{1}{10}\)

\(\frac{x}{20}=\frac{2}{20}\)

\(\Rightarrow x=2\)

Vậy x = 2

a) X = {x ∈ N | 8 < x < 15}
→ X = {9, 10, 11, 12, 13, 14}

b) Y = {x ∈ N | x < 7}
→ Y = {0, 1, 2, 3, 4, 5, 6}

c) Z = {x ∈ N | 13 ≤ x ≤ 20}
→ Z = {13, 14, 15, 16, 17, 18, 19, 20}

d) T = {x ∈ N | 4 ≤ x < 9}
→ T = {4, 5, 6, 7, 8}

a) X={9;10;11;12;13;14}

b) Y={0;1;2;3;4;5;6}

c) Z={13;14;15;16;17;18;19;20}

d) T={4;5;6;7;8}

\(a.\)

\(x+5=-10\)

\(\Rightarrow x=-10-5=-15\)

a ) x +5 = -10

x = -10 -5

x = - 15

b) x - ( - 10 ) = 5

x = 5+(-10)

x = -5

c) \(\left|x\right|\) -5 = 3

\(\left|x\right|=8\)

x ϵ { -8 ; 8 }

d) 15 - ( - x ) = 20

Không có số tự nhiên x nào mà 15 ( - x ) = 20

e ) \(\left|x-4\right|=3-\left(-7\right)\\ \left|x-4\right|=10\\ \left|x\right|=14\\ x\in\left\{\pm14\right\}\)

f ) \(\left|x+5\right|=10-\left(-20\right)\\ \left|x+5\right|=30\\ \left|x\right|=25\\ x\in\left\{\pm25\right\}\)

a)

\(9^{8} = \left(\right. 3^{2} \left.\right)^{8} = 3^{16}\)

\(8^{9} = \left(\right. 2^{3} \left.\right)^{9} = 2^{27}\)

ta có:
\(3^{16} = 3. 3^{15} = 3 \left(\right. 3^{3} \left.\right)^{5} = 3.2 7^{5}\)

\(2^{27} = 2^{2} . 2^{25} = 4. \left(\right. 2^{5} \left.\right)^{5} = 4.3 2^{5}\)

⇒ \(3.2 7^{5} < 4.3 2^{5}\) nên \(3^{16} < 2^{27}\)

\(Vậy9^8<8^9\)

\(9^8<8^9\)

\(3^{39}<11^{20}\)

a) \(x +(x + 1) + (x + 2) + ... + (x +30) = 620\)

\(=\left(x+x+...+x+x\right)+\left(1+2+...+30\right)\)

\(=31x+465=620\)

\(=31x=620-465\)

\(=31x=155\)

\(=x=155\div31\)

\(x=5\)

b) \(2+4+6+8+....+2x = 210\)

\(\Rightarrow2.1+2.2+2.3+2.4+...+2.x\)

\(\Rightarrow2.\left(2+4+6+8+...+x\right)=210\)

\(\Rightarrow2+4+6+8+x=210\div2\)

\(\Rightarrow2+4+6+8+...+x=105\)

\(\Rightarrow x=14\)

sao đến 2+4+6+8+...+x = 105 lại => x=14 được bạn

\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)

\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)

\(< =>\frac{-15+12-24}{63}\)

\(< =>\frac{-3}{7}\)

\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{7}\)

\(< =>\frac{29}{35}\)

\(bai2:\)

\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)

\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)

\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)

\(< =>x=\frac{3}{5}:\frac{-3}{4}\)

\(< =>x=\frac{-4}{5}\)

\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)

\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)

Bài 1:

a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\)                                 b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

 \(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\)                                       \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\) 

  \(=\frac{3}{7}.\frac{-9}{9}\)                                                                  \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

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

                                                                                               \(=\frac{29}{35}\)

Bài 2:

a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\)                                               b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

  \(\frac{-3}{4}x\)           \(=\frac{1}{5}+\frac{4}{10}\)                                     \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(\frac{-3}{4}x\)             \(=\frac{3}{5}\)                                            \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)     

         \(x\)              \(=\frac{3}{5}:\frac{-3}{4}\)                                        \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)                                         

         \(x\)              \(=\frac{4}{-5}\)                                                   \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)

                                                                                                             \(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\) 

                                                                                                                          \(x.\frac{10}{3}=\frac{13}{12}\) 

                                                                                                                                    \(x=\frac{13}{12}:\frac{10}{3}\) 

                                                                                                                                     \(x=\frac{13}{40}\)                             

e) \(\left(x-3\right)\left(x^2+1\right)=0\)

\(\Rightarrow\left(x-3\right)=0\)  ( \(x^2+1>0\forall x\))

\(\Rightarrow x=3\)

đ) \(4.8^2=2^x\)

\(2^2.\left(2^3\right)^2=2^x\)

\(2^2.2^6=2^x\)

\(2^8=2^x\)

\(\Rightarrow x=8\)

d) \(\left|x+3\right|=8\)

\(\Rightarrow\orbr{\begin{cases}x+3=8\\x+3=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-11\end{cases}}\)

mấy câu trên dễ rồi tự làm em nhé 

a, 2x + 5 = 7

=> 2x = 2

=> x = 1

vậy____

b: \(\Leftrightarrow x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)

\(\Leftrightarrow x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)

\(\Leftrightarrow x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)

=>x=3/11+20/55=3/11+4/11=7/11

c: \(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-2}{98}-1\right)+\left(\dfrac{x-5}{95}-1\right)=\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{95}\)

\(\Leftrightarrow x-100=1\)

hay x=101