ko

nói đùa thôi =)

\(\left\{{}\begin{matrix}A\subset X\\X\subset B\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}X=\left\{1;2;3;4\right\}\\X=\left\{1;2;3;4;5\right\}\\X=\left\{1;2;3;4;5;6\right\}\\X=\left\{1;2;3;4;5;6;7\right\}\end{matrix}\right.\)

x ϵ {1;2;3;4}

x ϵ {1;2;3;4;5}

x ϵ {1;2;3;4;5;6}

x ϵ {1;2;3;4;5;6;7}

-2³+0,5.x=1,5

⇒ -8+0,5.x=1,5

⇒ 0,5.x=1,5-(-8)

⇒ 0,5.x=9,5

⇒ x=9,5:0,5

⇒ x=19

-2^{3 _{+ 0,5x = 1,5}}

^{_{-8 + 0,5x=1,5}}

^{_{0,5x=1,5- -8}}

^_0,5x=9,5

^{_{x= 9,5: 0,5}}

^_x=19

\(A=\left\{1;2;3;4\right\}\)

\(B=\left\{2;3;4;5;6\right\}\)

 mà \(X\subset\left(A\cap B\right)\)

\(\Rightarrow\left\{{}\begin{matrix}X=\left\{2;3;4\right\}\\X=\left\{2;3\right\}\\X=\left\{2\right\}vàX=\left\{3\right\}vàX=\left\{4\right\}\end{matrix}\right.\)

\(0,5.\sqrt[]{100}-\sqrt[]{\dfrac{1}{2}+6,5}\)

\(=0,5.10-\sqrt[]{\dfrac{1}{2}+\dfrac{13}{2}}\)

\(=5-\sqrt[]{7}\)

21/2

Chiều dài hơn chiều rộng là: 10 cm 

Ta có sơ đồ: 

Theo sơ đồ ta có:

Chiều rộng là: 10: (3-2) \(\times\)  = 5 (cm)

Chiều dài là: 5 + 10 = 15  (cm)

Diện tích hình chữ nhật ban đầu là: 15 \(\times\) 5 = 75 (cm²)

Đs...

\(\left(2x-14\right):12=26\)

\(2x-14=26.12\)

\(2x-14=312\)

\(2x=312+14\)

\(2x=326\)

\(x=326:2\)

\(x=163\)

2x-14=26*12

2x-14=312 

2x=312-14

2x=298

x= 298:2

x=149

\(3^{30}:3^{27}=3^{30-27}=3^3=27\)

Vậy x=27

27

\(\left(x:32+15\right):4=24\)

\(x:32+15=24.4\)

\(x:32+15=96\)

\(x:32=96-15\)

\(x:32=81\)

\(x=81.32\)

\(x=2592\)

Sửa bài

\(\left(x:32+15\right):4=24\)

\(x:32+15=24.4\)

\(x:32+15=6\)

\(x:32=6-15\)

\(x:32=-9\)

\(x=\left(-9\right).32\)

\(x=-288\)

\(x+2y+xy=2\)

\(\Rightarrow x\left(1+y\right)+2y+2-2=2\)

\(\Rightarrow x\left(1+y\right)+2\left(y+1\right)=4\)

\(\Rightarrow\left(y+1\right)\left(x+2\right)=4\)

\(\Rightarrow\left(x+2\right);\left(y+1\right)\in\left\{-1;1;-2;2;-4;4\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-3;-5\right);\left(-1;3\right);\left(-4;-3\right);\left(0;1\right);\left(-6;-2\right);\left(2;0\right)\right\}\)