\(\dfrac{6^5.\left(-12\right)^6}{\left(-4\right)^9.\left(-3\right)^{10}}\)
Mọi người làm chi tiết từng bước một giùm em nhé và em cần gấp
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\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}\)
\(=\left(\frac{3}{7}\right)^9\)
a) \(5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^{6-5}+1=5+1=6\)
b) \(\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6\)
\(=\left(\dfrac{3}{7}\right)^{21-6}=\left(\dfrac{3}{7}\right)^{15}\)
c) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8-27+12}{27}=-\dfrac{7}{27}\)
\(a)5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^1+1=6\)
\(b,\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{49-40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3=\left(\dfrac{3}{7}\right)^{21}:[\left(\dfrac{3}{7}\right)^2]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6=\left(\dfrac{3}{7}\right)^{21-6}\)
\(=\left(\dfrac{3}{7}\right)^{15}\)
\(c,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=3.\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)
\(=\dfrac{8-9+4}{9}=\dfrac{1}{3}\)
Bài 2:
1: \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\)
=>\(\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{5}{6}=\dfrac{1}{12}+\dfrac{10}{12}=\dfrac{11}{12}\)
=>x=11
2: \(\dfrac{2}{3}-1\dfrac{4}{15}x=-\dfrac{3}{5}\)
=>\(\dfrac{2}{3}-\dfrac{19}{15}x=-\dfrac{3}{5}\)
=>\(\dfrac{19}{15}x=\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10+9}{15}=\dfrac{19}{15}\)
=>\(x=\dfrac{19}{15}:\dfrac{19}{15}=1\)
3: \(\dfrac{\left(-3\right)^x}{81}=-27\)
=>\(\left(-3\right)^x=\left(-3\right)^3\cdot\left(-3\right)^4=\left(-3\right)^7\)
=>x=7
4: \(\left|x+0,237\right|=0\)
=>x+0,237=0
=>x=-0,237
5: \(\left(x-1\right)^2=25\)
=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
6: \(\left|2x-1\right|=5\)
=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
7: \(\left(x-1\right)^3=-\dfrac{8}{27}\)
=>\(\left(x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)
=>\(x-1=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}+1=\dfrac{1}{3}\)
8: \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)
=>\(\dfrac{5}{3}:\dfrac{x}{4}=20\)
=>\(\dfrac{20}{3x}=20\)
=>3x=20/20=1
=>\(x=\dfrac{1}{3}\)
9: \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:2\dfrac{2}{3}\)
=>\(\dfrac{\dfrac{8}{3}}{x}=\dfrac{\dfrac{16}{9}}{\dfrac{8}{3}}\)
=>\(\dfrac{16}{9}\cdot x=\dfrac{8}{3}\cdot\dfrac{8}{3}=\dfrac{64}{9}\)
=>16x=64
=>x=64/16=4
Bài 3:
1: Ta có: x-24=y
=>x-y=24
mà \(\dfrac{x}{7}=\dfrac{y}{3}\)
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{24}{4}=6\)
=>\(x=6\cdot7=42;y=6\cdot3=18\)
2: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}\)
mà x-y=48
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}=\dfrac{x-y}{5-7}=\dfrac{48}{-2}=-24\)
=>\(x=-24\cdot5=-120;y=-24\cdot7=-168;z=-24\cdot2=-48\)
3: \(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}\)
mà x-y=4009
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}=\dfrac{x-1+3-y}{2005+2006}=\dfrac{4009+2}{4011}=1\)
=>\(x-1=2005;3-y=2006\)
=>x=2005+1=2006; y=3-2006=-2003
5: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)
mà 2x+3y-z=-14
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{2\cdot3+3\cdot5-7}=\dfrac{-14}{14}=-1\)
=>\(x=-3;y=-5;z=-7\)
Bạn tách ra từng CH khác nhau đi nhé. Gộp 1 trong tất cả rất khó nhìn và lâu.
\(\dfrac{6^5.\left(-12\right)^6}{\left(-4\right)^9\left(-3\right)^{10}}\)
\(=\dfrac{6^5.12^6}{\left(-4\right)^9.3^{10}}\)
\(=\dfrac{2^5.3^5.2^{12}.3^6}{\left(-1\right).2^{18}.3^{10}}\)
\(=\dfrac{2^{17}.3^{11}}{\left(-1\right).2^{18}.3^{10}}\)
\(=\dfrac{3}{\left(-1\right).2}\)
\(=\dfrac{-3}{2}\)
~ Chúc bạn học tốt ~!
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
`a, (2 sqrt 3 + sqrt 5)sqrt 3 - sqrt 60`
`= 2 sqrt 3 . sqrt 3 + sqrt 5 . sqrt 3 - sqrt(4 . 15)`
`= 2 . 3 + sqrt 15 - 2 sqrt 15`.
`= 6 - sqrt 15`.
`b, (5 sqrt 2 + 2 sqrt 5)sqrt 5 - sqrt250`
`= 5 sqrt 2 . sqrt 5 + 2 sqrt 5 . sqrt 5 - sqrt(25.10)`
`= 5 sqrt 10 + 10 - 5 sqrt 10`
`= 10`.
5) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)
\(=2\sqrt{3}\cdot\sqrt{3}+\sqrt{5}\cdot\sqrt{3}-\sqrt{2^2\cdot15}\)
\(=2\cdot3+\sqrt{15}-2\sqrt{15}\)
\(=6+\left(1-2\right)\sqrt{15}\)
\(=6-\sqrt{15}\)
6) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{2}\cdot\sqrt{5}+2\sqrt{5}\cdot\sqrt{5}-\sqrt{5^2\cdot10}\)
\(=5\sqrt{10}+2\cdot5-5\sqrt{10}\)
\(=\left(5-5\right)\sqrt{10}+10\)
\(=0+10\)
\(=10\)
Bài 1:
a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)
c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
\(E=\dfrac{98:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right)\cdot\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}\\ E=\dfrac{98}{\dfrac{3}{5}}+\dfrac{\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\\ E=\dfrac{490}{3}+\dfrac{\dfrac{7}{4}}{7}=\dfrac{490}{3}+\dfrac{1}{4}=\dfrac{1963}{12}\)
bạn ơi chỗ kia mik nhìn hơi loạn tí bạn giải thích giúp mik với
Ta có: \(\dfrac{6^5\cdot\left(-12\right)^6}{\left(-4\right)^9\cdot\left(-3\right)^{10}}\)
\(=-\dfrac{3^5\cdot2^5\cdot12^6}{4^9\cdot3^{10}}\)
\(=-\dfrac{2^5\cdot3^6\cdot4^6}{4^9\cdot3^5}\)
\(=-\dfrac{2^5\cdot3}{4^3}\)
\(=-\dfrac{2^5}{2^6}\cdot3=-\dfrac{3}{2}\)
ta được \(\dfrac{6^5.12^6}{4^8.\left(-4\right).3^{10}}\) \(=\dfrac{2^5.3^5.2^{12}.3^6}{2^{16}.\left(-4\right).3^{10}}\) \(=\dfrac{2^{17}.3^{11}}{2^{16}.\left(-4\right).3^{10}}=\dfrac{-6}{4}=\dfrac{-3}{2}\)