\(\frac{2x}{3}-\frac{2}{y}=\frac{1}{3}\)

\(\Rightarrow\frac{2}{y}=\frac{2x-1}{3}\)

\(\Leftrightarrow y.\left(2x-1\right)=6\)

Tới đây tự lập bảng ra nhé!! hok tốt!!

\(\left(2x-3\right)^2=16\)

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

\(\Rightarrow2x-3=4\)

\(\Rightarrow2x=4+3\)

\(\Rightarrow2x=7\)

\(\Rightarrow x=\frac{7}{2}\)

2x-3)^2=4^2

2x-3=4

2x=7

x=7/2

\(\left(2x-3\right)^2=16\)

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

\(\Leftrightarrow2x-3=4\)

\(\Leftrightarrow2x=4+3=7\)

\(\Leftrightarrow x=\frac{7}{2}\)

Vậy \(x=\frac{7}{2}\)

5x+6⋮x+2

=>5(x+2)-4⋮x+2

Mà x+2⋮x+2 =>5(x+2)⋮x+2

=>4⋮x+2

=>x+2∈Ư(4)={-4;-2;-1;1;2;4}

=>x∈{-6;-4;-3;-1;0;2}

Vì x+2 ⋮ x+2; 5 ∈ N

=> 5(x+2) ⋮ x+2

=> 5x +10 ⋮ x+2

Mà 5x + 6 ⋮ x+2

=> (5x+10)-(5x+6) ⋮ x+2

=> 4 ⋮ x+2

=> x+2 thuộc tập ước của 4

Mà ước của 4 = {1;-1;2;-2;4;-4}

=> x+2 ∈ {1;-1;2;-2;4;-4}

=> x ∈ {-1;-3;0;-4;2;-6}

Vậy x ∈ {-1;-3;0;-4;2;-6}

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

/x+1/=6+3+2x=9+2x

=> \(\left[{}\begin{matrix}x+1=9+2x\\x+1=-\left(9+2x\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x+8=0\\3x=-10\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-8\\x=-\dfrac{10}{3}\end{matrix}\right.\)

/x-1/=6+3+2x=9+2x

=> \(\left[{}\begin{matrix}x-1=9+2x\\x-1=-\left(9+2x\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x+10=0\\3x=-8\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-10\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+...+\frac{3}{10300}\)

\(=\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{100\times103}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)

\(=1-\frac{1}{103}=\frac{102}{103}\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-.....+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{2}-\frac{1}{50}\)

\(=\frac{24}{50}=\frac{12}{25}\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{2}-\frac{1}{50}\)

\(=\frac{12}{25}\)

      \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{2}-\frac{1}{50}\)

\(=\frac{25}{50}-\frac{1}{50}\)

\(=\frac{24}{50}=\frac{12}{25}\)

a, | x - 3/4 | = 1/2

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

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

=>\(\orbr{\begin{cases}x=\frac{2}{4}+\frac{3}{4}\\x=-\frac{2}{4}+\frac{3}{4}\end{cases}}\)

=>\(\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{4}\end{cases}}\)

Vậy....

a) \(|x-\frac{3}{4}|=\frac{1}{2}\)

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

\(< =>\orbr{\begin{cases}x=\frac{1}{2}+\frac{3}{4}\\x=-\frac{1}{2}+\frac{3}{4}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{4}\end{cases}}\)

Vay : x = 5/4 hoặc x =  1/4

b)\(saide\)

Sai de ban oi

a: =16-2+91=14+91=105

b: =9*5+8*10-27=45+53=98

c: =32+65-3*8=8+65=73

d; \(=5^3-10^2=125-100=25\)

e: \(=4^2-3^2+1=8\)

f: =9*16-16*8-8+16*4

=16(9-8+4)-8

=16*5-8

=72

a) \(2^4-50:25+13\cdot7\)

\(=2^4-2+91\)

\(=16-2+91\)

\(=14+91\)

\(=105\)

b) \(3^2\cdot5+2^3\cdot10-3^4:3\)

\(=9\cdot5+8\cdot10-3^3\)

\(=45+80-27\)

\(=98\)

c) \(2^5+5\cdot13-3\cdot2^3\)

\(=32+65-3\cdot8\)

\(=32+65-24\)

\(=73\)

d) \(5^{13}:5^{10}-5^2\cdot2^2\)

\(=5^{13-10}-\left(5\cdot2\right)^2\)

\(=5^3-10^2\)

\(=125-100\)

\(=25\)

e) \(4^5:4^3-3^9:3^7+5^0\)

\(=4^{5-3}-3^{9-7}+1\)

\(=4^2-3^2+1\)

\(=16-9+1\)

\(=8\)

f) \(3^2\cdot2^4-2^3\cdot4^2-2^3\cdot5^0+4^2\cdot2^2\)

\(=3^2\cdot4^2-2^3\cdot4^2-2^3\cdot1+4^2\cdot2^2\)

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

\(=4^2\cdot\left(9-8+4\right)-8\)

\(=16\cdot5-8\)

\(=72\)