Tìm số còn thiếu trong hình dưới đây :
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![](https://rs.olm.vn/images/avt/0.png?1311)
\((28\times15-34\times12)\times a:6+8=24\)
\(\Rightarrow12\times a:6=24-8\)
\(\Rightarrow2\times a=16\)
\(\Rightarrow a=16:2=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời :
13+23+33+...+103
=(1+2+3+4+5+6+7+8+9+10)3
=553
Hok_Tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(x=\frac{a+1}{a+9}=\frac{a+9-8}{a+9}=\frac{a+9}{a+9}-\frac{8}{a+9}=1-\frac{8}{a+9}\)
Để \(x\in Z\)thì \(a+9\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
Vậy \(a\in\left\{-17;-13;-11;-10;-8;-7;-5;-1\right\}\)
b) \(x=\frac{a-1}{a+4}=\frac{a+4-5}{a+4}=\frac{a+4}{a+4}-\frac{5}{a+4}=1-\frac{5}{a+4}\)
Để \(x\in Z\)thì \(a+4\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Vậy \(a\in\left\{-9;-5;-3;1\right\}\)
Bài 2:
a) \(t=\frac{3x-8}{x-5}=\frac{3x-15}{x-5}+\frac{7}{x-5}=\frac{3\left(x-5\right)}{x-5}+\frac{7}{x-5}=3+\frac{7}{x-5}\)
Để \(t\in Z\)thì \(x-5\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Vậy \(x\in\left\{-2;4;6;12\right\}\)
b)\(q=\frac{2x+1}{x-3}=\frac{2x-6}{x-3}+\frac{7}{x-3}=\frac{2\left(x-3\right)}{x-3}+\frac{7}{\left(x-3\right)}=2+\frac{7}{x-3}\)
Để \(q\in Z\)thì \(x-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
Vậy \(x\in\left\{-4;2;4;10\right\}\)
c)\(p=\frac{3x-2}{x+3}=\frac{3x+9}{x+3}-\frac{11}{x+3}=\frac{3\left(x+3\right)}{x+3}-\frac{11}{x+3}=3-\frac{11}{x+3}\)
Để \(p\in Z\)thì \(x+3\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)
Vậy \(x\in\left\{-14;-4;-2;8\right\}\)
Bài 3:
Gọi \(d\inƯC\left(2m+9;14m+62\right)\)
\(\Rightarrow\hept{\begin{cases}\left(2m+9\right)⋮d\\\left(14m+62\right)⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}7\left(2m+9\right)⋮d\\\left(14m+62\right)⋮d\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(14m+63\right)⋮d\\\left(14m+62\right)⋮d\end{cases}}\)
\(\Rightarrow\left[\left(14m+63\right)-\left(14m+62\right)\right]⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
\(\RightarrowƯC\left(2m+9;14m+62\right)=1\)
Vậy \(x=\frac{2m+9}{14m+62}\)là p/s tối giản
![](https://rs.olm.vn/images/avt/0.png?1311)
Nếu cạnh của hình lập phương giảm đi 20% thì diện tích hình lập phương đó sẽ giảm đi 125%
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left(a^4-b^4\right)-\left(2a^3b-2ab^3\right)\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(a^2+b^2-2ab\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a-b\right)^2\)
\(=\left(a-b\right)^3\left(a+b\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(3x^2\left(x-1\right)=0\)
\(\Rightarrow3x^2=0\)hoặc \(x-1=0\)
Vậy \(x=0\)hoặc \(x=1\)
b)\(\left(x+7\right)\left(2+3x\right)=0\)
\(\Rightarrow x+7=0\)hoặc \(2+3x=0\)
Vậy \(x=-7\)hoặc \(x=\frac{-2}{3}\)
#)Giải :
\(\left(3x^2\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x^2=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(\left(x+7\right)\left(2+3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\2+3x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-7\\x=-\frac{2}{3}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(5^4.3^4-\left(15^2-1\right)\left(15^2+1\right)\)
\(=15^4-\left(15^4-1^2\right)\)
\(=1\)
\(\left(18^4+1\right)\left(18-1\right)-9^8.2^8\) câu này bn xem lại đề đi nha, chắc bn chép sai đề rồi
b/ \(\frac{77^2+17^2-34.77}{77^2-17^2}\) \(=\frac{77^2-2.17.77+17^2}{\left(77-17\right)\left(77+17\right)}\)
= \(\frac{\left(77-17\right)^2}{\left(77-17\right)\left(77+17\right)}\)
= \(\frac{77-17}{77+17}\)
\(=\frac{60}{94}=\frac{30}{47}\)
\(\frac{135^2+130.135+65^2}{135^2-65^2}=\frac{135^2+2.65.135+65^2}{\left(135-65\right)\left(135+65\right)}\)
\(=\frac{\left(135+65\right)^2}{\left(135-65\right)\left(135+65\right)}\)
\(=\frac{135+65}{135-65}=\frac{200}{70}=\frac{20}{7}\)
chúc bn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : A = a000 + 400 + 50 +c +2000+ b00+ 40 + 1 + 137
A = ( a000 + b00 + c ) + ( 400 + 50 + 2000+ 40 + 1 +137 )
A = ab0c + 2628
B = ab9c + 2588
B = a000 + b00+ c +90 +2588
B = ab0c + 2678
Vì 2678 > 2588 suy ra B > A
Nhớ k chúc hk tốt
😿 😿