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`@` `\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 \(a< \dfrac{3}{4}\)
\(\dfrac{1}{50}-\dfrac{1}{50.49}-\dfrac{1}{49.48}-...-\dfrac{1}{2.1}\\ =-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{48.49}+\dfrac{1}{49.50}-\dfrac{1}{50}\right)\\ =-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{48}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{50}-\dfrac{1}{50}\right)\\ =-\left(1-\dfrac{1}{50}-\dfrac{1}{50}\right)\\ =-\dfrac{24}{25}\)
\(\left(-\dfrac{10}{3}\right)^5.\left(-\dfrac{6}{5}\right)^4\\ =\left(-\dfrac{10}{3}\right)^5.\left(\dfrac{6}{5}\right)^4\\ =-\left(\dfrac{10}{3}\right)^5.\left(\dfrac{6}{5}\right)^4\\ =-\dfrac{10^5.6^4}{3^5.5^4}=-\dfrac{2^5.5^5.3^4.2^4}{3^5.5^4}\\ =-\dfrac{2^9.3^4.5^4.5}{3.3^4.5^4}\\ =-\dfrac{2^9.5}{3}=-\dfrac{2560}{3}\)
`@` `\text {Ans}`
`\downarrow`
\(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2\)
`=`\(\left(\dfrac{6}{14}+\dfrac{7}{14}\right)^2\)
`=`\(\left(\dfrac{13}{14}\right)^2=\dfrac{169}{196}\)
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 \(a=2,b=7,c=-4\)
a) -3 < x < 3
=> \(x=-2;-1;0;1;2\)
Tổng x là \(-2+\left(-1\right)+0+1+2=0\)
b) \(-5\le x< 4\)
=> \(x=-5;-4;-3;-2;-1;0;1;2;3\)
Tổng x là \(-5+\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3=-9\)
c) -7 < x < 11
=> \(x=-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10\)
Tổng x là
\(-6+\left(-5\right)+\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+4+5+6+7+8+9+10=34\)
d) \(15\le x\le13\)
Không có x nào thoả mãn đề bài
a) -2;-1;0;1;2
b)-5;-4;-3;-2;-1;0;1;2;3
c)-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10
d) mình chịu nhé
\(a,\left(-8.2+17\right).\left[x+\left(-2\right).3\right]=20\)
\(\Rightarrow\left(-16+17\right).\left(x-6\right)=20\)
\(\Rightarrow1.\left(x-6\right)=20\)
\(\Rightarrow x=20+6\)
\(\Rightarrow x=26\)
\(b,\left(-5\right).x+5=\left(-3\right).\left(-8\right)+6\)
\(\Rightarrow-5x+5=24+6\)
\(\Rightarrow-5x+5=30\)
\(\Rightarrow-5x=25\)
\(\Rightarrow x=25:\left(-5\right)\)
\(\Rightarrow x=-5\)
\(c,-152-\left(3x+1\right)=\left(-2\right).\left(-27\right)\)
\(\Rightarrow-152-3x-1=54\)
\(\Rightarrow-3x=54+152+1\)
\(\Rightarrow-3x=207\)
\(\Rightarrow x=-69\)
\(d,\left(x+6\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=4\end{matrix}\right.\)
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) \(\left(x+6\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=4\end{matrix}\right.\)