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1. Tìm x:
a) \(\left(x+36\right)^2=1936\Leftrightarrow x+36=\pm44.\) Vậy x = 8 hoặc x = -80
b) \(\left(\dfrac{3}{5}\right)^{x+2}=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}\right)^{x+2}=\left(\dfrac{3}{5}\right)^4\Leftrightarrow x+2=4\Leftrightarrow x=2\)
c) Xem lại đề
d) \(\left(\dfrac{9}{16}\right)^{x-5}=\left(\dfrac{4}{3}\right)^4\Leftrightarrow\left(\dfrac{3}{4}\right)^{2\left(x-5\right)}=\left(\dfrac{3}{4}\right)^{-4}\Leftrightarrow2\left(x-5\right)=-4\Leftrightarrow x=3\)
e) \(\left(\dfrac{3}{5}\right)^x.\left(\dfrac{125}{27}\right)^x=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}.\dfrac{125}{27}\right)^x=\left(\dfrac{3}{5}\right)^4\Leftrightarrow\left(\dfrac{5}{3}\right)^{2x}=\left(\dfrac{5}{3}\right)^{-4}\Leftrightarrow2x=-4\) Vậy x = -2
3. Tính giá trị của biểu thức:
\(A=\left\{-\left[\left(\dfrac{1}{x}\right)^2\right]^3\right\}^5.\left\{-\left[\left(-x\right)^5\right]^2\right\}^3\) \(\left(x\notin0\right)\)
\(=\left\{-\left[-\dfrac{1}{x^2}\right]^3\right\}^5.\left\{-\left[-\left(-x\right)^5\right]^2\right\}^3=\left\{-\left[-\dfrac{1}{x^6}\right]\right\}^5.\left\{-\left[x^5\right]^2\right\}^3\)
\(=\left\{\dfrac{1}{x^6}\right\}^5.\left\{-x^{10}\right\}^3=\dfrac{1}{x^{30}}.\left(-x^{30}\right)=-1\)
a/dễ --> tự lm
b/ \(\left(x-\dfrac{4}{7}\right)\left(1\dfrac{3}{5}+2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=0\\1\dfrac{3}{5}+2x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\end{matrix}\right.\)
Vậy...............
c/ \(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)
TH1: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{4}{7}\\x>-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{4}{7}\)
TH2: \(\left\{{}\begin{matrix}x-\dfrac{4}{7}< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{4}{7}\\x< -\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow x< -\dfrac{1}{2}\)
Vậy \(x>\dfrac{4}{7}\) hoặc \(x< -\dfrac{1}{2}\) thì thỏa mãn đề
d/ \(\left(2x-3\right):\left(x+1\dfrac{3}{4}\right)< 0\)
TH1: \(\left\{{}\begin{matrix}2x-3>0\\x+1\dfrac{3}{4}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1,5\\x< -\dfrac{7}{4}\end{matrix}\right.\)(vô lý)
TH2: \(\left\{{}\begin{matrix}2x-3< 0\\x+1\dfrac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< 1,5\\x>-\dfrac{7}{4}\end{matrix}\right.\)\(\Rightarrow-\dfrac{7}{4}< x< 1,5\)
Vậy...................
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
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
Bài 1:
a, \(2y.\left(y-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)
Vậy \(y\in\left\{0;\dfrac{1}{7}\right\}\)
b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{-4}{15}+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)
\(\Rightarrow y=\dfrac{4}{25}\)
Vậy \(y=\dfrac{4}{25}\)
Chúc bạn học tốt!!!
Bài 1:
a, \(2y\left(y-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)
Vậy...
b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)
\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)
\(\Rightarrow y=\dfrac{4}{25}\)
Vậy...
Bài 2:
a, \(x\left(x-\dfrac{4}{7}\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\)
\(\Rightarrow x>\dfrac{4}{7}\left(x\ne0\right)\) hoặc \(x< \dfrac{4}{7}\left(x\ne0\right)\)
Vậy...
Các phần còn lại tương tự nhé
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
mình làm lại câu b) nha
b) |x-3|=-4
th1: x-3=-4
x=3+(-4)
x=-1
th2: x-3=4
x=3+4
x=7
b) \(\left|x-3\right|=-4\)
t/h1:\(x-3=-4\)
\(x=3-\left(-4\right)\)
\(x=7\)
t/h2:\(x-3=4\)
\(x=3-4\)
\(x=-1\)
a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
T giải thử thôi nhé :w
a) \(1\frac{1}{4}x^2y\left(\frac{-5}{6}xy\right)^0.\left(-2\frac{1}{3}xy\right)\)
\(=\frac{5}{4}x^2y\left(\frac{-5}{6}xy\right)^0.\left(-\frac{5}{2}xy\right)\)
\(=1.\frac{5}{4}x^2y\left(-\frac{5}{2}xy\right)\)
\(=-\frac{5}{4}x^2y.1.\frac{5}{2}xy\)
\(=-1.\frac{5}{4}.\frac{5}{2}x^3y^2\)
\(=-1.\frac{25x^3y^2}{8}\)
\(=-\frac{25x^3y^2}{8}\)