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a) \(a^2\cdot a^3\cdot a^7\cdot b^2\cdot b\)
\(=\left(a^2\cdot a^3\cdot a^7\right)\cdot\left(b^2\cdot b\right)\)
\(=a^{12}\cdot b^3\)
b) \(b^6\cdot b\cdot c^7\cdot c^8\)
\(=\left(b^6\cdot b\right)\cdot\left(c^7\cdot c^8\right)\)
\(=b^7\cdot c^{15}\)
c) \(a^8\cdot a^9\cdot a\cdot c\cdot c^{20}\)
\(=\left(a^8\cdot a^9\cdot a\right)\cdot\left(c\cdot c^{20}\right)\)
\(=a^{18}\cdot c^{21}\)
d) \(a^2\cdot a^3\cdot b^4\cdot c\cdot c^3\)
\(=\left(a^2\cdot a^3\right)\cdot b^4\cdot\left(c\cdot c^3\right)\)
\(=a^5\cdot b^4\cdot c^4\)
a) Kiểm tra lại nhé
b) \(b^6.b^7.c^8\)
\(=b^{6+7}.c^8=b^{13}.c^8\)
c) \(a^8.a^9.a.c.c^{20}\)
\(=a^{8+9+1}.c^{1+20}\)
\(=a^{18}.c^{21}\)
d) \(a^2.a^3.b^4.c.c^3\)
\(=a^{2+3}.b^4.c^{1+3}\)
\(=a^5.b^4.c^4\)
\(#WendyDang\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`4x = 320 \div 16`
`=> 4x = 20`
`=> x = 20 \div 4`
`=> x=5`
Vậy, `x=5`
`b)`
`(x-10) \div 5 = 2`
`=> x - 10 = 2*5`
`=> x - 10 = 10`
`=> x = 10 + 10`
`=> x=20`
Vậy, `x=20`
`c)`
`(20+x) \div 5 = 25`
`=> 20+x = 25*5`
`=> 20+x = 125`
`=> x = 125 - 20`
`=> x = 105`
Vậy, `x=105.`
`@` `\text {Kaizuu lv uuu}`
\(a.\left[\left(6.x-39\right).7\right].4=12\)
\(\left(6.x-39\right).7=12:4\)
\(6.x-39=3:7\)
\(6.x=\frac{3}{7}+39\)
\(x=\frac{276}{7}:6\)
\(x=\frac{46}{7}\)
( câu a )
\(b.200-\left(2.x+6\right)=4^3\)
\(200-\left(2.x+6\right)=64\)
\(2.x+6=200-64\)
\(2.x=136-6\)
\(x=130:2\)
\(x=65\)
( câu b )
a, Ta có :
xy=6
yz=-14
xz=-21
=>(xyz)2=1764=>xzy=42 hoặc -42
+)xyz=42
=>z=42:6=7
=>x=-3
=>y=-2
+)xyz=-42
=>z=-7
=>y=2
=>x=3
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
a) \(\frac{3^{20}\cdot4+3^{20}\cdot5}{3^{21}}=\frac{3^{20}\cdot3^2}{3^{21}}=3\)
b)\(5^{20}\cdot4+5^{20}\cdot7-5^{17}=5^{20}\cdot11-5^{17}=5^{17}\left(5^3\cdot11-1\right)\)
a) 320.(4+5) : 321=3