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1: \(=\left|\dfrac{-21+5}{35}\right|+\dfrac{5}{7}\cdot\dfrac{-2}{5}\)
\(=\dfrac{16}{35}+\dfrac{-2}{7}=\dfrac{16}{35}-\dfrac{10}{35}=\dfrac{6}{35}\)
2: =>4^x+1=16
=>x+1=2
=>x=1
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
1.
\(\left(\dfrac{-2}{3}\right).0,75+1\dfrac{2}{3}:\left(\dfrac{-4}{9}\right)+\left(\dfrac{-1}{2}\right)^2\)
\(=\left(\dfrac{-2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}.\left(\dfrac{9}{-4}\right)+\dfrac{1}{4}\)
\(=-\dfrac{1}{2}+\dfrac{45}{-12}+\dfrac{1}{4}\)
\(=-\dfrac{6}{12}+\dfrac{-45}{12}+\dfrac{3}{4}\)
\(=\dfrac{-48}{12}\)
\(=-4\)
2.
a) \(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{10}{20}\)
\(\Leftrightarrow x=\dfrac{-11}{20}\)
b) \(\left|x-\dfrac{2}{5}\right|+\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{11}{4}-\dfrac{3}{4}\)
\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=-2\Rightarrow x=-2+\dfrac{2}{5}=\dfrac{-8}{5}\\x-\dfrac{2}{5}=2\Rightarrow x=2+\dfrac{2}{5}=\dfrac{12}{5}\end{matrix}\right.\)
3.
a) \(\dfrac{16}{2^n}=2\)
\(\Leftrightarrow2^n=16:2\)
\(\Leftrightarrow2^n=8\)
\(\Leftrightarrow2^n=2^3\)
\(\Leftrightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{81}=-27\)
\(\Leftrightarrow\left(-3\right)^n=\left(-27\right).81\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^3.\left(-3\right)^4\)
\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\)
\(\Leftrightarrow n=7\)
4. Ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)
\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\) (2)
Từ (1) và (2) suy ra \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Vì \(x-y+x=-49\) ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)
Vậy \(\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)
=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)
=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)
d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)
=> \(\left(2x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
=> 2x-1 = \(\dfrac{-2}{3}\)
=> x= \(\dfrac{1}{6}\)
a) \(\frac{x-1}{-15}=\frac{-60}{x-1}\)
\(\left(x-1\right)^2=\left(-15\right).\left(-60\right)=900\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=300^2\\\left(x-1\right)^2=\left(-300\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=300\\x-1=-300\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=301\\x=-299\end{cases}}\)
b) \(\left|x+\frac{4}{5}\right|+\frac{3}{5}=\frac{2}{5}\)
\(\left|x+\frac{4}{5}\right|=\frac{2}{5}-\frac{3}{5}\)
\(\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)
vì \(\left|x+\frac{4}{5}\right|\ge0\forall x\)mà \(\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)
\(\Rightarrow\)không có giá trị x nào thỏa mãn đề bài trên
c) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
a) \(\Leftrightarrow\left(x-1\right)\left(x-1\right)=\left(-60\right).\left(-15\right)\)
\(\Leftrightarrow\left(x-1\right)^2=900=30^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=30\\x-1=-30\end{cases}\Leftrightarrow\orbr{\begin{cases}x=30+1\\x=-30+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=31\\x=-29\end{cases}}}\)
Vậy x = 31 hoặc x = - 29
b) \(\left|x+\frac{4}{5}\right|+\frac{3}{5}=\frac{2}{5}\)
\(\Leftrightarrow\left|x+\frac{4}{5}\right|=\frac{2}{5}-\frac{3}{5}\)
\(\Leftrightarrow\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)vô lý không có giá trị tuyệt đối của số nào mà nhận giá trị âm
Vậy ko có giá trị nào của x thỏa mãn
c) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
a, Theo đề ta có:
\(2.3^x-405=3^{x-1}\)
=> \(2.3^x-405=3^x:3\)
=> \(405=(2.3^x)-(3^x:3)\)
=>\(405=(2.3^x)-(3^x.\dfrac{1}{3})\)
=> \(405=3^x(2-\dfrac{1}{3})\)
=>\(405=3^x(\dfrac{6}{3}-\dfrac{1}{3})\)
=> \(405=3^x.\dfrac{5}{3}\)
=> \(3^x=405:\dfrac{5}{3}\)
=>\(3^x=405.\dfrac{3}{5}\)
=> \(3^x=81.3\)
=> \(3^x=243\)
=> \(3^x=3^5\)
=> x=5
Vậy:..............................
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{3^3}\)
\(\left(x-\dfrac{1}{2}\right)=\dfrac{1}{3}\)
\(x=\dfrac{5}{6}\)