ai nhanh k 2 lan

a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)

b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)

\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)

c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)

\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)

e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)

f) \(\left|12-x\right|-7=5\)

th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)

th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)

i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)

k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)

a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)

b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)

Làm nốt nha

câu 1 :

 \(A=\frac{-7}{12}:\frac{49}{11}\cdot\frac{5}{121}-\frac{7}{6}\)                                                \(B=\frac{1}{8}-\frac{8}{7}:8-3:\frac{3}{4}\cdot-2^3\)

          \(A=\frac{-11}{84}\cdot\frac{5}{121}-\frac{7}{6}\)                                                        \(B=\frac{1}{8}-\frac{8}{7}:\frac{8}{1}-\frac{3}{1}:\frac{3}{4}\cdot\left(-2^3\right)\)            

           \(A=\frac{-5}{924}-\frac{7}{6}\)                                                                     \(B=\frac{1}{8}-\frac{1}{7}-\left(-32\right)\)

           \(A=\frac{-361}{308}\)                                                                            \(B=\frac{-1}{56}-\left(-32\right)\)

                                                                                                                 \(B=\frac{1791}{56}\)

Câu 2 :

a)\(\frac{22}{7}:x=\frac{11}{7}\)                                                     b)\(\left(1-3x\right)\cdot\frac{4}{3}=-2^3\)

          \(x=\frac{22}{7}:\frac{11}{7}\)                                                 \(\left(1-3x\right)\cdot\frac{4}{3}=-8\)                                                    

          \(x=2\)                                                                      \(\left(1-3x\right)=-8:\frac{4}{3}\)   

                                                                                              \(\left(1-3x\right)=-6\)

                                                                                       \(3x=-6-1=7\)

                                                                                        \(3x=7:3=\frac{7}{3}\)

c ) bằng \(\frac{27}{5}\)nhé

a) \(\frac{2}{5}x-x=\frac{\left(-2018\right)^0}{5^2}\\ x\left(\frac{2}{5}-1\right)=\frac{1}{25}\\ x\left(\frac{2}{5}-\frac{5}{5}\right)=\frac{1}{25}\\ x\cdot\frac{-3}{5}=\frac{1}{25}\\ x=\frac{1}{25}:\frac{-3}{5}\\ x=\frac{1}{25}\cdot\frac{-5}{3}\\ x=\frac{-1}{15}\)Vậy \(x=\frac{-1}{15}\)

b) \(\left|-1\frac{1}{2}x+2x\right|-\frac{7}{4}=0,5\\ \left|x\left(-1\frac{1}{2}+2\right)\right|-\frac{7}{4}=\frac{1}{2}\\ \left|x\cdot\frac{1}{2}\right|=\frac{1}{2}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{2}{4}+\frac{7}{4}\\ \left|x\cdot\frac{1}{2}\right|=\frac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x\cdot\frac{1}{2}=\frac{9}{4}\\x\cdot\frac{1}{2}=\frac{-9}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}:\frac{1}{2}\\x=\frac{-9}{4}:\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{4}\cdot2\\x=\frac{-9}{4}\cdot2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{9}{2}\\x=\frac{-9}{2}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{9}{2};\frac{-9}{2}\right\}\)

c) \(x+\left(x+\frac{2}{7}\right)+\frac{-5}{11}=\frac{4}{11}\\ x+x+\frac{2}{7}=\frac{4}{11}-\frac{-5}{11}\\ 2x+\frac{2}{7}=\frac{4}{11}+\frac{5}{11}\\ 2x+\frac{2}{7}=\frac{9}{11}\\ 2x=\frac{9}{11}-\frac{2}{7}\\ 2x=\frac{63}{77}-\frac{22}{77}\\ 2x=\frac{41}{77}\\ x=\frac{41}{77}:2\\ x=\frac{41}{77\cdot2}\\ x=\frac{41}{154}\)Vậy \(x=\frac{41}{154}\)

d) \(\left|0,25x-20\%\right|+\frac{3}{8}=1\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\frac{3}{8}-\frac{3}{8}\\ \left|\frac{1}{4}x-\frac{1}{5}\right|=1\\ \Rightarrow\left[{}\begin{matrix}\frac{1}{4}x-\frac{1}{5}=1\\\frac{1}{4}x-\frac{1}{5}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=1+\frac{1}{5}\\\frac{1}{4}x=\left(-1\right)+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{5}{5}+\frac{1}{5}\\\frac{1}{4}x=\frac{-5}{5}+\frac{1}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{1}{4}x=\frac{6}{5}\\\frac{1}{4}x=\frac{-4}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}:\frac{1}{4}\\x=\frac{-4}{5}:\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{6}{5}\cdot4\\x=\frac{-4}{5}\cdot4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{24}{5}\\x=\frac{-16}{5}\end{matrix}\right.\)Vậy \(x\in\left\{\frac{24}{5};\frac{-16}{5}\right\}\)

Câu a:

2x - 3y + xy = 13

(2x + xy) - (3y + 6) = 13 - 6

x(2 + y) - 3(y + 2) = 7

(y + 2)(x - 3) = 7

Ư(7) = {-7; -1; 1; 7}

Lập bảng ta có:

Theo bảng trên ta có:

(x; y) = (-4; -3); (2; -9); (4; 5); (10; -1)

Vậy các giá trị x; y nguyên thỏa mãn đề bài là:

(x; y) = (-4; -3); (2; -9); (4; 5); (10; -1)

Câu e:

x^2 - 3xy = 7

x(x - 3y) = 7

Ư(7) = {-7; -1; 1; 7}

Lập bảng ta có:

Theo bảng trên ta có: (x; y) = (-7; -2); (-1; 2); (1; -2); (7; 2)

Vậy (x; y) = (-7; -2); (-1; 2); (1; -2); (7; 2)

khó quá bn ơi

2b,