\(53\cdot\left(51+12\right)+53\left(49+91\right)-3\cdot53\)

\(=53\cdot63+53\cdot140-3\cdot53\)

\(=53\cdot60+53\cdot140=53\cdot200=10600\)

\(b=2022\cdot2026=\left(2024-2\right)\left(2024+2\right)\)

\(=2024\cdot2024-2\cdot2\)

=a-4

=>a-4=b

=>a=b+4

=>a>b

Diện tích xung quanh căn phòng là:

\(\left(8+5\right)\cdot2\cdot4=13\cdot8=104\left(m^2\right)\)

Diện tích không cần quét vôi là:

\(2,2\cdot1,2\cdot2+1,4\cdot0,8\cdot4=9,76\left(m^2\right)\)

Diện tích cần quét vôi là:

\(104+8\cdot5-9,76=134,24\left(m^2\right)\)

Số tiền cần phải trả là:

\(134,24\cdot10000=1342400\left(đồng\right)\)

Cảm ơn bạn rất nhiều!

Bài đúng nhưng mình chỉnh sửa cho bạn phần trình bày nhé (bạn có thể thay đổi hoặc không).

Theo đề bài ta có:
\(\left(a+b\right)\) chia hết cho \(m\) và \(a\) chia hết cho \(m\) 
\(\Rightarrow\left(a+b\right)=x.m;a=y.m\left(x,y\inℕ;x\ge y\right)\)
\(\Rightarrow\left(a+b\right)-a=x.m-y.m\\ \Rightarrow b=m.\left(x-y\right)\)
Vì \(m.\left(x-y\right)⋮m\) nên \(b⋮m\)
Vậy...


 

\(450=2\cdot3^2\cdot5^2;354=2\cdot3\cdot59\)

=>\(ưcln\left(450;354\right)=3\cdot2=6\)

=>\(ƯC\left(450;354\right)=Ư\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

Ta có :

\(450\times2\times3^2\)\(\times5^2\)

\(354=2\times3\times59\)

\(\RightarrowƯCLN\left(450,354\right)\)\(=2\times3\)\(=6\)

\(\RightarrowƯC\left(450,354\right)\)\(=Ư\left(6\right)\)\(\left\{-1;1;2;-2;3;-3;6;-6\right\}\)

\(A=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{97\cdot98}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{98}\)

\(=\dfrac{1}{5}-\dfrac{1}{98}=\dfrac{93}{490}\)

\(A=\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{97.98}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{98}\)

\(=\dfrac{1}{5}-\dfrac{1}{98}\)

\(=\dfrac{93}{490}\)

Câu B đề bài ko đúng, vì 2 số hạng đầu thì dưới mẫu cách nhau 4 đơn vị (7-3=4, 11-7=4, nhưng số hạng cuối lại chỉ cách 3 đơn vị)

a: \(2,3\cdot\left(-6,4\right)+\left(-3,6\right)\cdot\left|-2,3\right|\)

\(=2,3\left(-6,4-3,6\right)\)

\(=2,3\cdot\left(-10\right)=-23\)

b: \(\left|-\dfrac{2}{5}+\dfrac{1}{2}\right|\cdot\left(1\dfrac{3}{4}-0,75\right)\)

\(=\left|-\dfrac{4}{10}+\dfrac{5}{10}\right|\left(1+\dfrac{3}{4}-\dfrac{3}{4}\right)\)

\(=\dfrac{1}{10}\)

c: \(\left(0,75+\left|-\dfrac{1}{2}\right|-\dfrac{1}{3}\right):0,5+1\dfrac{3}{2}\)

\(=\left(\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{1}{3}\right):\dfrac{1}{2}+\dfrac{5}{2}\)

\(=\left(\dfrac{9}{12}+\dfrac{6}{12}-\dfrac{4}{12}\right)\cdot2+\dfrac{5}{2}\)

\(=\dfrac{11}{12}\cdot2+\dfrac{5}{2}=\dfrac{11}{6}+\dfrac{5}{2}=\dfrac{11}{6}+\dfrac{15}{6}=\dfrac{26}{6}=\dfrac{13}{3}\)

d: \(\dfrac{1}{2}-\left|\left(-3\right)\cdot\dfrac{1}{6}\right|+\dfrac{5}{3}-\dfrac{9}{12}\)

\(=\dfrac{1}{2}-\left|-\dfrac{3}{6}\right|+\dfrac{20}{12}-\dfrac{9}{12}\)

\(=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{11}{12}=\dfrac{11}{12}\)

xy+3y-x=7

=>y(x+3)-x-3=4

=>(x+3)(y-1)=4

=>\(\left(x+3;y-1\right)\in\left\{\left(1;4\right);\left(4;1\right);\left(-1;-4\right);\left(-4;-1\right);\left(2;2\right);\left(-2;-2\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-2;5\right);\left(1;2\right);\left(-4;-3\right);\left(-7;0\right);\left(-1;3\right);\left(-5;-1\right)\right\}\)

\(\left(x+2\right)^2-\left(2x+1\right)\left(x+2\right)=0\\ < =>\left(x+2\right)\left[\left(x+2\right)-\left(2x+1\right)\right]=0\\ < =>\left(x+2\right)\left(x+2-2x-1\right)=0\\ < =>\left(x+2\right)\left(1-x\right)=0\\ < =>\left[{}\begin{matrix}x+2=0\\1-x=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Vậy: ... 

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

=>(x+2)(x+2-2x-1)=0

=>(x+2)(-x+1)=0

=>\(\left[{}\begin{matrix}x+2=0\\-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

a: \(\left|-0,25\right|+\left\{\left(4\cdot8\right)\cdot125-\left(-0,5\right)^2\right\}\)

=0,25+4000-0,25

=4000

b: \(\left(2,7+\left|-4.4\right|\right)-\left[\left(-5,6\right)-\left|-7,3\right|\right]\)

=2,7+4,4+5,6+7,3

=10+10=20

c: \(\left(-5,44\right)+4\cdot\left(1,25+0,11\right)\)

\(=-5,44+5+0,44\)

=-5+5=0

d: \(\left[\left|-6,72\right|+\left|-5,27\right|\right]-\left(0,72+1,27\right)\)

=6,72+5,27-0,72-1,27

=6+5=11