a, \(xy=9\) ( \(x\) < y)

Ư(9) = { -9; -3; -1; 1; 3; 9}

Lập bảng ta có: 

Theo bảng trên ta có: Vì \(x\) < y

nên (\(x;y\)) = (-9; -1); (1; 9)

b, (\(x\) - 6)(y +2) =7

   Ư(7) = {-7; -1; 1; 7}

Lập bảng ta có:

Theo bảng trên ta có:

Các cặp \(x;y\) thỏa mãn đề bài là:

(\(x;y\)) = (-1; -3); (5; -9); (7; 5); (13; -1)

\(a,\left(-8.2+17\right).\left[x+\left(-2\right).3\right]=20\)

\(\Rightarrow\left(-16+17\right).\left(x-6\right)=20\)

\(\Rightarrow1.\left(x-6\right)=20\)

\(\Rightarrow x=20+6\)

\(\Rightarrow x=26\)

\(b,\left(-5\right).x+5=\left(-3\right).\left(-8\right)+6\)

\(\Rightarrow-5x+5=24+6\)

\(\Rightarrow-5x+5=30\)

\(\Rightarrow-5x=25\)

\(\Rightarrow x=25:\left(-5\right)\)

\(\Rightarrow x=-5\)

\(c,-152-\left(3x+1\right)=\left(-2\right).\left(-27\right)\)

\(\Rightarrow-152-3x-1=54\)

\(\Rightarrow-3x=54+152+1\)

\(\Rightarrow-3x=207\)

\(\Rightarrow x=-69\)

\(d,\left(x+6\right)\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=4\end{matrix}\right.\)

a) \(\left(-8,2+17\right).\left(x+\left(-2\right).3\right)=20\)

\(\Rightarrow8,8\cdot\left(x-6\right)=20\)

\(\Rightarrow x-6=\dfrac{25}{11}\)

\(\Rightarrow x=\dfrac{91}{11}\)

b) \(\left(-5\right)x+5=\left(-3\right).\left(-8\right)+6\)

\(\Rightarrow\left(-5\right)x+5=30\)

\(\Rightarrow\left(-5\right)x=25\)

\(\Rightarrow x=-5\)

c) \(-152-\left(3x+1\right)=\left(-2\right).\left(-27\right)\)

\(\Rightarrow-152-\left(3x+1\right)=54\)

\(\Rightarrow-\left(3x+1\right)=206\)

\(\Rightarrow-3x-1=206\)

\(\Rightarrow-3x=207\)

\(\Rightarrow x=-69\)

d) \(\left(x+6\right)\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=4\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`-9/34*17/4`

`=`\(\dfrac{-9}{17\cdot2}\cdot\dfrac{17}{4}\)

`=`\(-\dfrac{9}{2}\cdot\dfrac{1}{4}\)

`=`\(-\dfrac{9}{8}\)

`b)`

\(\dfrac{17}{15}\div\dfrac{4}{3}\)

`=`\(\dfrac{17}{15}\cdot\dfrac{3}{4}\)

`=`\(\dfrac{17}{3\cdot5}\cdot\dfrac{3}{4}\)

`=`\(\dfrac{17}{5}\cdot\dfrac{1}{4}\)

`=`\(\dfrac{17}{20}\)

`c)`

\(4\dfrac{1}{5}\div\left(-2\dfrac{4}{5}\right)\)

`=`\(4\dfrac{1}{5}\cdot\left(-\dfrac{5}{14}\right)\)

`=`\(\dfrac{21}{5}\cdot\left(-\dfrac{5}{14}\right)\)

`=`\(-\dfrac{21}{14}=-\dfrac{3}{2}\)

a) \(\dfrac{-9}{34}\cdot\dfrac{17}{4}\)

\(=\dfrac{-9\cdot17}{34\cdot4}\)

\(=-\dfrac{153}{136}\)

\(=\dfrac{9}{8}\)

b) \(\dfrac{17}{15}:\dfrac{4}{3}\)

\(=\dfrac{17}{15}\cdot\dfrac{3}{4}\)

\(=\dfrac{17\cdot3}{15\cdot4}\)

\(=\dfrac{51}{60}=\dfrac{17}{20}\)

c) \(4\dfrac{1}{5}:\left(-2\dfrac{4}{5}\right)\)

\(=\dfrac{21}{5}:-\dfrac{14}{5}\)

\(=\dfrac{21}{5}\cdot-\dfrac{5}{14}\)

\(=\dfrac{21\cdot-5}{5\cdot14}\)

\(=-\dfrac{105}{70}=\dfrac{3}{2}\)

Đặt \(P\left(n\right)=3.7^{2n+1}+6.2^{2n+2}\)

Ta thấy \(P\left(0\right)=45⋮45\), luôn đúng.

Giả sử khẳng định đúng đến \(n=k\), khi đó \(P\left(k\right)=3.7^{2k+1}+6.2^{2n+2}⋮45\). Ta cần chứng minh khẳng định đúng với \(n=k+1\). Thật vậy:

\(P\left(k+1\right)=3.7^{2\left(k+1\right)+1}+6.2^{2\left(k+1\right)+2}\)

\(=3.7^{2k+3}+6.2^{2k+4}\)

\(=49.3.7^{2k+1}+4.6.2^{2k+2}\)

\(=4\left(3.7^{2k+1}+6.2^{2k+2}\right)+45.3.7^{2k+1}\)

Hiển nhiên \(45.3.7^{2k+1}⋮45\). Lại có \(4\left(3.7^{2k+1}+6.2^{2k+2}\right)\) theo giả thiết quy nạp nên suy ra \(P\left(k+1\right)⋮45\), suy ra khẳng định đúng với mọi \(n\inℕ\). Ta có đpcm

\(M=\dfrac{4a-3}{a+2}\left(a\in Z,a\ne-2\right)\)

`M` có gt âm hay `M<0`

TH1 : \(a>-2=>a+2>0\)

\(M=\dfrac{4a-3}{a+2}< 0\\ =>4a-3< 0\) ( Nhân 2 vế BPT cho `a+2>0` )

\(=>a< \dfrac{3}{4}\)

Kết hợp ĐK \(=>-2< a< \dfrac{3}{4}\)

TH2 : \(a< -2=>a+2< 0\)

\(M=\dfrac{4a-3}{a+2}< 0\\ =>4a-3>0\) ( Nhân 2 vế cho `a+2<0` )

\(=>a>\dfrac{3}{4}\) (KTMDK)

Vậy : \(-2< a< \dfrac{3}{4}\) . Mà a là số nguyên nên \(a\in\left\{-1;0\right\}\)

Vì M phải có giá trị âm thì \(M< 0\)

\(M=\dfrac{4a-3}{a+2}\left(a\in Z,a\ne2\right)\)

\(\Rightarrow\dfrac{4a-3}{a+2}< 0\)

\(\Rightarrow4a-3< 0\)

\(\Rightarrow4a< 3\)

\(\Rightarrow a< \dfrac{3}{4}\)

vậy \(a< \dfrac{3}{4}\)

\(\dfrac{1}{50}-\dfrac{1}{50.49}-\dfrac{1}{49.48}-...-\dfrac{1}{2.1}\\ =-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{48.49}+\dfrac{1}{49.50}-\dfrac{1}{50}\right)\\ =-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{48}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{50}-\dfrac{1}{50}\right)\\ =-\left(1-\dfrac{1}{50}-\dfrac{1}{50}\right)\\ =-\dfrac{24}{25}\)

\(\left(-\dfrac{10}{3}\right)^5.\left(-\dfrac{6}{5}\right)^4\\ =\left(-\dfrac{10}{3}\right)^5.\left(\dfrac{6}{5}\right)^4\\ =-\left(\dfrac{10}{3}\right)^5.\left(\dfrac{6}{5}\right)^4\\ =-\dfrac{10^5.6^4}{3^5.5^4}=-\dfrac{2^5.5^5.3^4.2^4}{3^5.5^4}\\ =-\dfrac{2^9.3^4.5^4.5}{3.3^4.5^4}\\ =-\dfrac{2^9.5}{3}=-\dfrac{2560}{3}\)

\(=\dfrac{-100000}{243}\times\dfrac{1296}{625}=\dfrac{\left(-100000\right).1296}{243.625}=\dfrac{\left(-160\right).625.1296}{625.243}=\dfrac{-160.1296}{243}=\dfrac{81.16.\left(-160\right)}{81.3}=\dfrac{16.\left(-160\right)}{3}=\dfrac{-2560}{3}\)

`@` `\text {Ans}`

`\downarrow`

\(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2\)

`=`\(\left(\dfrac{6}{14}+\dfrac{7}{14}\right)^2\)

`=`\(\left(\dfrac{13}{14}\right)^2=\dfrac{169}{196}\)

\(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2=\left(\dfrac{6}{14}+\dfrac{7}{14}\right)^2\\ =\left(\dfrac{13}{14}\right)^2=\dfrac{169}{196}\)

a<b

Hôm nay olm.vn sẽ hướng dẫn các em so sánh lũy thừa bằng lũy thừa trung gian. 

a = 15⁸ và b = 8¹¹

15⁸ < 16⁸ = (2⁴)⁸ = 2³²

8¹¹ = (2³)¹¹ = 2³³

Vì 2³² < 2³³ nên 15⁸ < 8¹¹

Vậy  a <  b 

\(a\left(a+b+c\right)=10,b\left(a+b+c\right)=35,c\left(a+b+c\right)=-20\\ =>a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=10+35+\left(-20\right)\\ =>\left(a+b+c\right)^2=25\\ =>\left[{}\begin{matrix}a+b+c=5\\a+b+c=-5\end{matrix}\right.\)

TH1 : `a+b+c=5`

\(=>\left\{{}\begin{matrix}a=10:5=2\\b=35:5=7\\c=-20:5=-4\end{matrix}\right.\)

TH2 : `a+b+c=-5`

\(=>\left\{{}\begin{matrix}a=10:\left(-5\right)=-2\\b=35:\left(-5\right)=-7\\c=\left(-20\right):\left(-5\right)=4\end{matrix}\right.\)

Vậy : \(\left(a;b;c\right)=\left(2;7;-4\right);\left(-2;-7;4\right)\)

Ta có:

\(a\times\left(a+b+c\right)+b\times\left(a+b+c\right)+c\times\left(a+b+c\right)=10+35+\left(-20\right)\)

\(\left(a+b+c\right)\times\left(a+b+c\right)=25\)

\(\left(a+b+c\right)^2=5^2\)

\(a+b+c=5\)

Thay \(a+b+c=5\) ta có: \(a\times5=10\) 

                                               \(a=10\div5\) 

                                                \(a=2\) 

Thay \(a+b+c=5\) ta có: \(b\times5=35\) 

                                                \(b=35\div5\) 

                                                 \(b=7\) 

Thay \(a+b+c=5\) ta có: \(c\times5=-20\) 

                                                \(c=-20\div5\) 

                                                \(c=-4\) 

Vậy chữ số \(a,b,c\) cần tìm là \(a=2,b=7,c=-4\)