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a: =>1/6x=-49/60
=>x=-49/60:1/6=-49/60*6=-49/10
b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2
=>x=17/15 hoặc x=-13/15
c: =>1,25-4/5x=-5
=>4/5x=1,25+5=6,25
=>x=125/16
d: =>2^x*17=544
=>2^x=32
=>x=5
i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5
=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2
=>x=14,4 hoặc x=9,6
j: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
2) -12:\(\left(-\dfrac{5}{6}\right)^2\)=\(-12:\dfrac{25}{36}=-12\cdot\dfrac{36}{25}=-\dfrac{432}{25}\)
s) \(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
= \(-\dfrac{1}{12}-\dfrac{55}{24}=-\dfrac{2}{24}-\dfrac{55}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)
t) \(-1,75-\left(-\dfrac{1}{9}-2\dfrac{1}{18}\right)=-1,75-\left(-\dfrac{2}{18}-\dfrac{37}{18}\right)\)
= -1,75-(\(-\dfrac{13}{6}\)) = \(-\dfrac{7}{4}+\dfrac{13}{6}=\dfrac{5}{12}\)
c) \(\left(\sqrt{\dfrac{1}{9}}-0,5\right)^3+\dfrac{-1}{3}=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^3-\dfrac{1}{3}\)
= \(\left(-\dfrac{1}{6}\right)^3-\dfrac{1}{3}=\dfrac{-1}{216}-\dfrac{1}{3}=-\dfrac{73}{216}\)
d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{4}{25}}\right)^2-2\dfrac{1}{2}=\left(\dfrac{1}{2}-\dfrac{2}{5}\right)^2-\dfrac{5}{2}\)
= \(\left(\dfrac{1}{10}\right)^2-\dfrac{5}{2}=\dfrac{1}{100}-\dfrac{250}{100}=-\dfrac{249}{100}=-2,49\)
a,\(\dfrac{2}{7}x-\dfrac{1}{2}=\dfrac{3}{4}:\sqrt{\dfrac{49}{64}}\)
\(\Leftrightarrow\dfrac{2}{7}x-\dfrac{1}{2}=\dfrac{6}{7}\)
\(\Leftrightarrow\dfrac{2}{7}x=\dfrac{19}{14}\)
\(\Leftrightarrow x=\dfrac{19}{4}\)
Với mọi \(x\in R\)
\(\left|x+2016\right|+\left|x+2017\right|+\left|x+2018\right|\ge0\Leftrightarrow6x\ge0\Leftrightarrow x\ge0\)
với \(x\ge0\) ta được: \(\left\{{}\begin{matrix}\left|x+2016\right|=x+2016\\\left|x+2017\right|=x+2017\\\left|x+2018\right|=x+2018\end{matrix}\right.\)
\(pt\Leftrightarrow3x+6051=6x\Leftrightarrow3x=6051\Leftrightarrow x=2017\)
a) \(7-\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}=7\)
\(\Rightarrow x=\left(\sqrt{7}\right)^2\)
b) \(5\sqrt{x}+1=40\)
\(\Rightarrow5\sqrt{x}=39\)
\(\Rightarrow\sqrt{x}=7,8\)
\(\Rightarrow x=\left(\sqrt{7,8}\right)^2\)
c) \(\dfrac{5}{12}\sqrt{x}-\dfrac{1}{6}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{5}{12}\sqrt{x}=\dfrac{1}{2}\)
\(\Rightarrow\sqrt{x}=1,2\)
\(\Rightarrow x=\left(\sqrt{1,2}\right)^2\)
d) \(4x^2-1=0\)
\(\Rightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\Rightarrow x=0,5\\2x+1=0\Rightarrow x=-0,5\end{matrix}\right.\)
e) \(\sqrt{x+1}-2=0\)
\(\Rightarrow\sqrt{x+1}=2\)
\(\Rightarrow x+1=1,414\)
\(\Rightarrow x=0,414\)
f) \(2x^2+0,82=1\)
\(\Rightarrow2x^2=0,18\)
\(\Rightarrow x^2=0,09\)
\(\Rightarrow x=\pm0,3\)
g) Không có kết quả
a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
Tính:
C = \(3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right).2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)\)
= \(\dfrac{7}{2}.\dfrac{4}{49}-\left[\left(2+0,\left(1\right).4\right).\dfrac{27}{5}\right].\dfrac{-5}{42}\)
= \(\dfrac{1.2}{1.7}-\left[\left(2+\dfrac{1.4}{9}\right).\dfrac{27}{5}\right].\dfrac{-5}{42}\)
= \(\dfrac{2}{7}-\left[\dfrac{18+4}{9}.\dfrac{27}{5}\right].\dfrac{-5}{42}\)
= \(\dfrac{2}{7}-\left[\dfrac{22.9}{3.5}\right].\dfrac{-5}{42}\)
= \(\dfrac{2}{7}-\dfrac{198}{15}.\dfrac{-5}{42}=\dfrac{2}{7}-\dfrac{11}{3}.\dfrac{-1}{7}\)
= \(\dfrac{2}{7}+\dfrac{11}{21}\) = \(\dfrac{6+11}{21}\) = \(\dfrac{17}{21}\)
1) Tính
a) 253 : 52 = (52)3 : 52 = 56 : 52 = 54 = 625
\(b)\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^6=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^6=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^9\) d) 9 . 32 . \(\dfrac{1}{81}\) . 32 = 32 . 32 . \(\dfrac{1}{3^4}\) . 32 = 9
2) Tìm x thuộc Q, biết:
a) 3x + 2 = 27
=> 3x + 2 = 33
x + 2 = 3
x = 3 - 2
x = 1
b) \(\left(\dfrac{1}{2}x-3\right)^4=81\)
\(\Rightarrow\left(\dfrac{1}{2}x-3\right)^4=3^4\)
\(\dfrac{1}{2}x-3=3^{ }\)
\(\dfrac{1}{2}x=3+3\)
\(\dfrac{1}{2}x=9\)
\(x=9:\dfrac{1}{2}\)
\(x=18\)
c) \(\left(x-\dfrac{1}{2}\right)^3=-27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(-3\right)^3\)
\(x-\dfrac{1}{2}=-3\)
\(x=-3+\dfrac{1}{2}\)
\(x=\dfrac{-5}{2}\)
d) 5 . 5x + 1 = 125
5x + 1 = 125 : 5
5x + 1 = 25
5x + 1 = 52
x + 1 = 2
x = 2 - 1
x = 1.
a: \(\Leftrightarrow11x^3+11x^2-6x^2-6x+10x+10=0\)
\(\Leftrightarrow\left(x+1\right)\left(11x^2-6x+10\right)=0\)
=>x=-1
c: \(\Leftrightarrow x^2\left(\sqrt{5}-1\right)-x\sqrt{5}+1=0\)
\(a=\sqrt{5}-1;b=-\sqrt{5};c=1\)
Vì a+b+c=0 nên pt có hai nghiệm là:
\(x_1=1;x_2=\dfrac{c}{a}=\dfrac{1}{\sqrt{5}-1}=\dfrac{\sqrt{5}+1}{4}\)
d: Ta có: \(x^2\left(1+\sqrt{3}\right)+x-\sqrt{3}=0\)
\(a=1+\sqrt{3};b=1;c=-\sqrt{3}\)
Vì a-b+c=0 nên phương trình có hai nghiệm là:
\(x_1=-1;x_2=\dfrac{\sqrt{3}}{\sqrt{3}+1}\)
BÀi 2:
Cả 4 câu áp dụng tính chất này: \(\sqrt{a^2}=a\)
a)\(\sqrt{\frac{3^2}{7^2}}=\frac{3}{7}\)
b)\(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{92^2}}=\frac{3+39}{7+92}=\frac{42}{99}=\frac{14}{33}\)
c)\(\frac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{7^2}-\sqrt{91^2}}=\frac{3-39}{7-91}=\frac{-36}{-84}=\frac{3}{7}\)
d)\(\sqrt{\frac{39^2}{91^2}}=\frac{39}{91}=\frac{3}{7}\)
b)Vì BCNN(3;5) = 15
\(\Rightarrow\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x}{2.5}=\frac{y}{3.5}=\frac{x}{10}=\frac{y}{15};\frac{y}{5}=\frac{z}{7}\Leftrightarrow\frac{y}{5.3}=\frac{z}{7.3}=\frac{y}{15}=\frac{z}{21}\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.10=20\\y=2.15=30\\z=2.21=42\end{matrix}\right.\)
Vậy...
c)Vì BCNN(2;3;5) = 30
\(\Rightarrow2x=3y=5z\Leftrightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}=\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
WTFFFFFF>>>
d)dễ... áp dụng tính chất DTBN là ra 1/2 rồi tính
e)Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(x=\frac{y}{2}=\frac{z}{4}=\frac{4x}{4}=\frac{3y}{6}=\frac{2x}{8}=\frac{4x-3y+2x}{4-6+8}=\frac{36}{6}=6\)
\(\Rightarrow\left\{{}\begin{matrix}x=6.1=6\\y=6.2=12\\z=6.4=24\end{matrix}\right.\)
Vậy...
a, x=2=35/2
x=log(35/2)
x=log(35)-log(20)
x=log(35)-1
b) \(\left(\sqrt{9}+\sqrt{4}\right).\sqrt{x}=10\)
\(\left(3+2\right).\sqrt{x}=10\)
\(5.\sqrt{x}=10\)
\(\sqrt{x}=2\)
\(x=\sqrt{2}\)