3\3*5+3\5*7+.....+3/161*163= ...../......
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\(a.83-\left(-756-17\right)+\left(50-756\right)=83+756+17+50-756=\left(83+17\right)+50+\left(756-756\right)=100+50=150\)
\(b.-105+\left(-34-95\right)-166=-105+\left(-34\right)-95-166=\left(-105-95\right)-\left(34+166\right)=\left(-200\right)-200=-400\)
\(c,-\left(39+228-407\right)+\left(118-161\right)=-39-228+407+118-161=\left(-39-161\right)-\left(228-118\right)+407=\left(-200\right)-110=-310\)
\(d,5^{10}:5^8+60:12+\left(-10\right)=5^2+5+\left(-10\right)=5\left(5+1-2\right)=5.4=20\)
\(e,-342-\left(161-342\right)-39=-342-161+342-39=\left(-342+342\right)-\left(161+39\right)=-200\)\(g,7^5:7^3+\left(-187-149\right)-213=7^2+\left(-187\right)-149-213=\left(49-149\right)-\left(187+213\right)=\left(-100\right)-400=-500\)
Bai 2 :
\(\hept{\begin{cases}x+y=7\\x-7=13\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=7\\x=20\end{cases}}}\)
Thay x vào phương trình đầu ta có :
\(20+y=7\Leftrightarrow y=-13\)
Vậy \(\left\{x;y\right\}=\left\{20;-13\right\}\)
Thử \(20-13=7\); \(20-7=13\)( thỏa mãn )
Theo mình thấy thì như thế này
Ta có \(\left(3\sqrt{5}\right)^2=9.5=45\)
Mà \(45>36\)
Nên \(\sqrt{45}>\sqrt{36}\Leftrightarrow3\sqrt{5}>6\)
\(\Leftrightarrow163+3\sqrt{5}>163+6=169\)
\(\Leftrightarrow\sqrt{163+3\sqrt{5}}>\sqrt{169}=13\)
Do đó \(13-\sqrt{163+3\sqrt{5}}<0\)
Nên \(\sqrt{13-\sqrt{163+3\sqrt{5}}}<0\) (không có nghĩa )
Do đó A không thể rút gọn
a: \(7\cdot\left(-2\right)^3-12\cdot\left(-5\right)+\left(-17\right)\)
\(=7\cdot\left(-8\right)+60-17\)
=-56+43
=-13
b: \(1632-37-\left(-157\right)-163-1532\)
\(=\left(1632-1532\right)-37-163+157\)
=100-200+157
=57
c: \(47\cdot\left(-918\right)+\left(-53\right)\cdot918\)
\(=918\left(-47\right)+\left(-53\right)\cdot918\)
\(=918\cdot\left(-47-53\right)\)
\(=918\left(-100\right)=-91800\)
d: \(\left(-52\right)\cdot\left(-281\right)+\left(-52\right)\cdot181\)
\(=\left(-52\right)\left(-281+181\right)\)
\(=\left(-52\right)\cdot\left(-100\right)=5200\)
a: 7⋅(−2)3−12⋅(−5)+(−17)7⋅(−2)3−12⋅(−5)+(−17)
=7⋅(−8)+60−17=7⋅(−8)+60−17
=-56+43
=-13
b: 1632−37−(−157)−163−15321632−37−(−157)−163−1532
=(1632−1532)−37−163+157=(1632−1532)−37−163+157
=100-200+157
=57
c: 47⋅(−918)+(−53)⋅91847⋅(−918)+(−53)⋅918
=918(−47)+(−53)⋅918=918(−47)+(−53)⋅918
=918⋅(−47−53)=918⋅(−47−53)
=918(−100)=−91800=918(−100)=−91800
d: (−52)⋅(−281)+(−52)⋅181(−52)⋅(−281)+(−52)⋅181
=(−52)(−281+181)=(−52)(−281+181)
=(−52)⋅(−100)=5200=(−52)⋅(−100)=5200
\(1,\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=15\end{matrix}\right.\\ 2,7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-12\\y=-28\end{matrix}\right.\\ 3,\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=-\dfrac{9}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{45}{2}\\y=-27\\z=-\dfrac{63}{2}\end{matrix}\right.\\ 4,x:y:z=3:5:7\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
3. Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=\dfrac{-9}{2}\)
\(x=\dfrac{-45}{2}\)
\(y=-27\)
\(z=\dfrac{-63}{2}\)
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3/(3 × 5) + 3/(5 × 7) + ... + 3/(161 × 163)
= 3 × 2 × (1/3 - 1/5 + 1/5 - 1/7 + ... + 1/161 - 1/163)
= 6 × (1/3 - 1/163)
= 6 × 160/489
= 320/163
bhbkjb ụ.