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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
\(1,\)
\(a,\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)\)
\(=\dfrac{11}{125}+\left(\dfrac{-1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}\)
\(b,-1\dfrac{5}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=\dfrac{-12}{7}.15+\dfrac{2}{7}.\left(-15\right)+\left(105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
\(=-15.\left[\dfrac{12}{7}+\dfrac{2}{7}+\left(-5\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\right]\)
\(=-15.\left[2+\left(-5\right).\dfrac{1}{105}\right]\)
\(=-15.\left(2-\dfrac{1}{21}\right)\)
\(=-15.\dfrac{41}{21}=\dfrac{-615}{21}\)
\(2,\)
\(a,\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{13}\right)\)
\(\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=\dfrac{-15}{28}+\dfrac{11}{13}\)
\(\Leftrightarrow x=\dfrac{-15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}\)
\(\Leftrightarrow x=\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\left(\dfrac{5}{42}+\dfrac{-15}{28}\right)\)
\(\Leftrightarrow x=\dfrac{5}{12}\)
Vậy \(x=\dfrac{5}{12}\)
\(b,\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|-3,75=-2,15\)
\(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6=\dfrac{16}{10}=\dfrac{8}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=\dfrac{8}{5}\\x+\dfrac{4}{15}=\dfrac{-8}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{5}-\dfrac{4}{15}=\dfrac{4}{3}\\x=\dfrac{-8}{5}-\dfrac{4}{15}=\dfrac{-28}{15}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{4}{3};\dfrac{-28}{15}\right\}\)
\(c,7^{x+2}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.\left(7^3+2\right)=345\)
\(\Leftrightarrow7^{x-1}.\left(343+2\right)=345\)
\(\Leftrightarrow7^{x-1}.345=345\)
\(\Leftrightarrow7^{x-1}=345:345=1\)
\(\Leftrightarrow x-1=0\)
\(x=0+1=1\)
Vậy \(x=1\)
1: =>1/3:x=3/5-2/3=9/15-10/15=-1/15
=>x=-1/3:1/15=5
2: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)
=>x*2/3=-1
=>x=-3/2
3: \(\Leftrightarrow\dfrac{8}{3}:x=\dfrac{25}{12}:\dfrac{-3}{50}=\dfrac{25}{12}\cdot\dfrac{-50}{3}\)
hay x=-48/625
9: =>x=-2*3/1,5=-4
8: =>2/3:x=5/2:-3/10=5/2*(-10)/3=-50/6=-25/3
=>x=-2/3:25/3=-2/3*3/25=-2/25
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
I , tìm x :
a, \(\left|x\right|=1,21\)
Ta có : \(\left|x\right|=\left|1,21\right|\rightarrow\left|x\right|=\pm1,21\)
b, \(\dfrac{11}{12}-\left(\dfrac{2}{5}-x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}-x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}-x=\dfrac{1}{4}\) => \(x=\dfrac{2}{5}-\dfrac{1}{4}\)
=> \(x=\dfrac{3}{20}\)
c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}\div x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}\div x=\dfrac{-7}{20}\) => \(x=\dfrac{1}{4}\div\dfrac{-7}{20}\)
=> \(x=\dfrac{-5}{7}\)
d,\(3^x=81\)
Ta có 81= \(3^4\)
Vì : \(3^x=3^4\Rightarrow x=4\)
e,\(\dfrac{1}{2}.\left|x\right|-\dfrac{5}{2}=\dfrac{8}{3}\)
\(\left|x\right|-\dfrac{5}{6}=\dfrac{8}{3}:\dfrac{1}{2}\)
=> \(\left|x\right|-\dfrac{5}{2}=\dfrac{16}{3}\) => \(\left|x\right|=\dfrac{16}{3}+\dfrac{5}{2}\)
=> \(\left|x\right|=\dfrac{47}{6}\)
Vì \(\left|x\right|=\left|\dfrac{47}{6}\right|\Rightarrow x=\pm\dfrac{47}{6}\)
f, \(2^{x-3}=4\)
\(2^{x-3}=2^2\)
=> \(x-3=2\)
=> \(x=5\)
a, Ta có \(\left|x\right|=1,21\)
\(\Rightarrow\left[{}\begin{matrix}x=1,21\\x=-1,21\end{matrix}\right.\)
Vậy \(x\in\left\{1,21;-1,21\right\}\)
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
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
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
a: \(\Leftrightarrow7^x\cdot49+7^x\cdot\dfrac{2}{7}=345\)
\(\Leftrightarrow7^x=7\)
hay x=1
c: \(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{36}\)
\(\Leftrightarrow\left(-\dfrac{1}{6}\right)^{x-1}=\left(-\dfrac{1}{6}\right)^2\)
=>x-1=2
hay x=3
d: \(\dfrac{25}{5^x}=\dfrac{1}{125}\)
\(\Leftrightarrow5^x=5^2\cdot5^3=5^5\)
hay x=5