(8+\(\frac{1}{2}\))3:(1+\(\frac{1}{2}\))2
(44-83):27
Giups mik với!!!!!!!!!!
K
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NG
18 tháng 6 2020
a)
\(\frac{8}{11}.\frac{14}{23}+\frac{9}{23}:\frac{11}{8}-\frac{8}{11} \)
\(=\frac{8}{11}.\frac{14}{23}+\frac{9}{23}.\frac{8}{11}-\frac{8}{11}\)
\(=\frac{8}{23}.(\frac{14}{23}+\frac{9}{23}-1)\)
\(=\frac{8}{23}.0\)
=0
b)
\(1,8.\frac{-20}{27}\)+(75%\(-\frac{5}{16}):3\frac{1}{2}\)
=\(\frac{9}{5}.\frac{-20}{27}+(\frac{3}{4}-\frac{5}{16}):\frac{7}{2}\)
=\(\frac{-4}{3}+\frac{1}{8}\)
=\(\frac{-29}{24}\)
P
20 tháng 8 2019
A=1/(x-2)(x-3) + 1/(x-3)(x-4) + 1/(x-4)(x-5) + 1/(x-5)(x-6)=1/8 (ĐKXĐ: x#2,x#3,x#4,x#5,x#6)
A= 1/x-2 -1/x-3 + 1/x-3 -1/x-4 .....-1/x-6=1/8
=>1/x-2 -1/x-6=1/8
=>8(x-6)-8(x-2)=(x-2)(x-6)
=> 8x-48-8x+16=x^2-8x+12
=> x^2-8x-20=0
=> (x-10)(x+2)=0 => x=10,x=-2 thuộc ĐKXĐ
20 tháng 8 2019
Có cần thế ko ạ ??? Shinichi
Điều kiện xác định \(\hept{\begin{cases}x\ne2\\x\ne\\x\ne4\end{cases}3}\)
\(\hept{\begin{cases}x\ne5\\x\ne6\end{cases}}\)
Ta có : \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
\(x^2-7x+12=\left(x-3\right)\left(x-4\right)\)
\(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)
\(x^2-11+30=\left(x-5\right)\left(x-6\right)\)
Phương trình đã tương đương với
\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}-\frac{1}{x-5}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-2}=\frac{1}{8}\Leftrightarrow\frac{4}{\left(x-6\right)\left(x-2\right)}=\frac{1}{8}\)
\(\Leftrightarrow x^2-8x-20=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\)
\(x-10=0\Leftrightarrow x=10\)
hoặc
\(x+2=0\Leftrightarrow x=-2\)
\(\Leftrightarrow\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)thỏa mãn điều kiện phương trình
Phương trình có nghiệm \(x=10;x=-2\)
H9
HT.Phong (9A5)
CTVHS
18 tháng 7 2023
1) \(3^x=\dfrac{9^8}{27^3\cdot81^2}\)
\(\Rightarrow3^x=\dfrac{\left(3^2\right)^8}{\left(3^3\right)^3\cdot\left(3^4\right)^2}\)
\(\Rightarrow3^x=\dfrac{3^{16}}{3^{15}}\)
\(\Rightarrow3^x=3\)
\(\Rightarrow x=1\)
2) \(\dfrac{2^{4-x}}{16^5}=32^6\)
\(\Rightarrow\dfrac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
\(\Rightarrow\dfrac{2^{4-x}}{2^{20}}=2^{30}\)
\(\Rightarrow2^{4-x}=2^{20}\cdot2^{30}\)
\(\Rightarrow2^{4-x}=2^{50}\)
\(\Rightarrow4-x=50\)
\(\Rightarrow x=-46\)
3) \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)
\(\Rightarrow\dfrac{2^{2x-3}}{\left(2^2\right)^{10}}=\left(2^3\right)^3\cdot\left(2^4\right)^5\)
\(\Rightarrow\dfrac{2^{2x-3}}{2^{20}}=2^{29}\)
\(\Rightarrow2^{2x-3}=2^{49}\)
\(\Rightarrow2x-3=49\)
\(\Rightarrow2x=52\)
\(\Rightarrow x=26\)
28 tháng 6 2016
a) \(\left(\frac{1}{3}-\frac{1}{5}\right)^2:\left(\frac{1}{5}\right)^2=\left[\left(\frac{1}{3}-\frac{1}{5}\right):\frac{1}{5}\right]^2=\left(\frac{2}{15}:\frac{1}{5}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
28 tháng 6 2016
c)\(7\frac{1}{23}+\frac{10}{27}-5\frac{1}{23}+\frac{17}{27}+2^3=\left(7\frac{1}{23}-5\frac{1}{23}\right)+\left(\frac{10}{27}+\frac{17}{27}\right)+2^3=2+1+8=11\)
d)\(5.\left(-\frac{5}{2}\right)^2+\frac{1}{5}.\left(-3\right)^2=5.\frac{25}{4}+\frac{1}{5}.9=\frac{125}{4}+\frac{9}{5}=\frac{661}{20}\)
25 tháng 7 2020
Câu a) số lớn lắm
b) \(3^{-3}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^5\cdot3^x=3^8\)
=> \(\frac{1}{27}\cdot3^x=3^3\)
=> \(3^x=3^3:\frac{1}{27}=3^3:\left(\frac{1}{3}\right)^3=3^3:\frac{1^3}{3^3}=3^3\cdot3^3=3^6\)
=> x = 6
b) \(\left(7x+2\right)^{-1}=3^{-2}\)
=> \(\frac{1}{7x+2}=\frac{1}{9}\)
=> 7x + 2 = 9
=> 7x = 7
=> x = 1
Bài 2:
a) \(3^4\cdot\frac{1}{729}\cdot81^3\cdot\frac{1}{9^2}\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot\left(3^4\right)^3\cdot\left(\frac{1}{3}\right)^4\)
\(=3^4\cdot\left(\frac{1}{3}\right)^6\cdot3^{12}\cdot\left(\frac{1}{3}\right)^4=3^{16}\cdot\left(\frac{1}{3}\right)^{10}=\frac{3^{16}}{3^{10}}=3^6\)
b) \(\left(8\cdot2^5\right):\left(2^4\cdot\frac{1}{32}\right)=\left(2^3\cdot2^5\right):\left(2^4\cdot\left(\frac{1}{2}\right)^5\right)\)
\(=2^8:\left(2^4\cdot\frac{1^5}{2^5}\right)=2^8:\left(\frac{2^4}{2^5}\right)=2^8:2^{-1}=512\)
c) \(12^8\cdot9^{12}=\left(2^2\cdot3\right)^8\cdot\left(3^2\right)^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
d) Tương tự
a) Ta có: \(\left(8+\frac{1}{2}\right)^3:\left(1+\frac{1}{2}\right)^2\)
\(=\left(\frac{17}{2}\right)^3:\left(\frac{3}{2}\right)^2\)
\(=\frac{17^3}{8}\cdot\frac{2^2}{3}\)
\(=\frac{17^3}{2\cdot3}=\frac{4913}{6}\)
b) Ta có: \(\left(4^4-8^3\right):2^7\)
\(=\frac{2^8-2^9}{2^7}\)
\(=\frac{2^8\left(1-2\right)}{2^7}\)
\(=-2\)