\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)

=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)

=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)

Vậy \(x\in\left\{\frac{9}{20}\right\}\)

\(b,x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

Vậy \(x\in\left\{\frac{13}{12}\right\}\)

\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)

=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)

Vậy \(x\in\left\{\frac{25}{42}\right\}\)

\(d,\left|x+5\right|-6=9\)

=> \(\left|x+5\right|=9+6=15\)

=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)

Vậy \(x\in\left\{10;-20\right\}\)

\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)

\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{6}\)

=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)

\(g,x^2=16\)

=> \(\left|x\right|=\sqrt{16}=4\)

=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

vậy \(x\in\left\{4;-4\right\}\)

\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)

Vậy \(x\in\left\{\frac{5}{6}\right\}\)

\(i,3^3.x=3^6\)

\(x=3^6:3^3=3^3=27\)

Vậy \(x\in\left\{27\right\}\)

\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)

=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)

Vậy \(x\in\left\{\frac{5}{27}\right\}\)

\(k,1\frac{2}{3}:x=6:0,3\)

=> \(\frac{5}{3}:x=20\)

=> \(x=\frac{5}{3}:20=\frac{1}{12}\)

Vậy \(x\in\left\{\frac{1}{12}\right\}\)

\(a.\)

\(x+5=-10\)

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

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

LẠM DỤNG QUÁ NHIỀU

Giải:

4.Theo đề bài ta có:

\(A=7.a+4 \)

\(=17.b+3 \)

\(=23.c+11 (a,b,c ∈ N)\)

Nếu ta thêm 150 vào số đã cho thì ta lần lượt có:

\(A+150=7.a+4+150=7.a+7.22=7.(a+22)\)

\(=17.b+3+150=17.b+17.9=17.(b+9)\)

\(=23.c+11+150=23.c+23.7=23.(c+7) \)

\(\Rightarrow A+150⋮7;17;23\).Nhưng 7, 17 và 23 là ba số đôi một nguyên tố cùng nhau, suy ra \(A+150⋮7.17.13=2737\)

Vậy \(A+150=2737k\left(k=1;2;3;4;...\right)\)

Suy ra: \(A=2737k-150=2737k-2737+2587=2737(k-1)+2587=2737k+2587\)

Do \(2587<2737\)

\(\Rightarrow A\div2737\) dư \(2587\)

Bạn ơi, A=23c+7 chứ. Sao lại= 23c+11?

a,\(\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)

\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x};Đkxđ:x\ne1\)

\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}\left(\frac{-9}{20}\right)=\frac{5}{2-2x}\)

\(\Rightarrow\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2-2x}\)

\(\Rightarrow\frac{1}{x-1}-\frac{5}{2-2x}=\frac{-3}{10}\)

\(\Rightarrow\frac{1}{x-1}-\frac{5}{-2\left(x-1\right)}=\frac{-3}{10}\)

\(\Rightarrow\frac{1}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{3}{10}\)

\(\Rightarrow\frac{7}{2\left(x-1\right)}=\frac{-3}{10}\)

\(\Rightarrow70=-6\left(x-1\right)\)

\(\Rightarrow6x=6-70\)

\(\Rightarrow6x=-64\)

\(\Rightarrow x=\frac{-32}{3}x\ne1\)

b) <=> 4(10+x) = 3(17+x) 

40+4x=51+3x 

4x-3x=51-40 

x=11 

c) <=> 7.(40+x) = 6 .(17+x) 

280+7x =102 + 6x

7x-6x=102-280 

x=-178 

a)  7/x=x/28

=)7*28=x*x

=)196=x^2

=)14^2=x^2

= )x=14

k cho mih di roi mih giai tiep cho nha

Ta có: 3636:(12x−91)=36

12x–91=3636:36

12x–91=101

12x=101+91

12x=192

x=192:12

x=16

Vậy x=16

440 + 2. (125 – x) = 546

2. (125 - x)=546 - 440

2. (125 - x)=106

125 - x=106 : 2

125 - x=53

x=125-53

x=72