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1) \(\left(\dfrac{-13}{17}-\dfrac{31}{52}\right)-\left(\dfrac{73}{52}-\dfrac{13}{17}+\dfrac{5}{6}\right)-\dfrac{3}{4}\)
\(=\dfrac{-13}{17}-\dfrac{31}{52}-\dfrac{73}{52}+\dfrac{13}{17}-\dfrac{5}{6}-\dfrac{3}{4}\)
\(=\left(\dfrac{-13}{17}+\dfrac{13}{17}\right)-\left(\dfrac{31}{52}+\dfrac{73}{52}\right)-\left(\dfrac{5}{6}+\dfrac{3}{4}\right)\)
\(=0-2-\dfrac{19}{12}\)
\(=-2-\dfrac{19}{12}\)
\(=\dfrac{-43}{12}\)
Bài 5:
a) \(x:\left(-\dfrac{1}{2}\right)^3=-\dfrac{1}{2}\)
\(\Rightarrow x=\left(-\dfrac{1}{2}\right)^3\cdot\left(-\dfrac{1}{2}\right)\)
\(\Rightarrow x=\left(\dfrac{1}{2}\right)^4=\dfrac{1}{16}\)
b) \(\left(\dfrac{3}{4}\right)^5\cdot x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^7:\left(\dfrac{3}{4}\right)^5\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2=\dfrac{9}{16}\)
c) \(\left(\dfrac{2}{3}\right)^5:x=\left(\dfrac{2}{3}\right)^3\)
\(\Rightarrow x=\left(\dfrac{2}{5}\right)^5:\left(\dfrac{2}{3}\right)^3\)
\(\Rightarrow x=\left(\dfrac{2}{5}\right)^2=\dfrac{4}{25}\)
d) \(\dfrac{25}{5^x}=5\)
\(\Rightarrow5^x\cdot5=25\)
\(\Rightarrow5^{x+1}=25\)
\(\Rightarrow5^{x+1}=5^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=2-1=1\)
e) \(\left(-\dfrac{1}{3}\right)^{x-5}=\dfrac{1}{81}\)
\(\Rightarrow\left(-\dfrac{1}{3}\right)^{x-5}=\left(-\dfrac{1}{3}\right)^4\)
\(\Rightarrow x-5=4\)
\(\Rightarrow x=4+5=9\)
g) \(\left(2x-3\right)^2=9\)
\(\Rightarrow\left(2x-3\right)=3^2\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=\dfrac{6}{2}=3\)
h) \(\left(x+5\right)^3=64\)
\(\Rightarrow\left(x+5\right)^3=4^3\)
\(\Rightarrow x+5=4\)
\(\Rightarrow x=4-5=-1\)
i) \(2^{3x+2}=4^{x+5}\)
\(\Rightarrow2^{3x+2}=\left(2^2\right)^{x+5}\)
\(\Rightarrow2^{3x+2}=2^{2x+10}\)
\(\Rightarrow3x+2=2x+10\)
\(\Rightarrow3x-2x=10-2\)
\(\Rightarrow x=8\)
k) \(5^{x+1}-5^x=500\)
\(\Rightarrow\left(5-1\right)\cdot5^x=500\)
\(\Rightarrow4\cdot5^x=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
g: =>-4/5x-1/4+x=-13/3
=>1/5x=-13/3+1/4=-52/12+3/12=-49/12
=>x=-49/12*5=-245/12
h: =>12/7:x-1/2=0 hoặc 2/5x-3/2=0
=>12/7:x=1/2 hoặc 2/5x=3/2
=>x=12/7:1/2=24/7 hoặc x=3/2:2/5=3/2*5/2=15/4
\(-\dfrac{4}{5}x-\left(0,25-x\right)=-\dfrac{13}{3}\\ -\dfrac{4}{5}x-0,25+x=-\dfrac{13}{3}\\ x-\dfrac{4}{5}x=-\dfrac{13}{3}+\dfrac{1}{4}=-\dfrac{49}{12}\\ \dfrac{1}{5}x=-\dfrac{49}{12}\\ x=-\dfrac{49}{12}:\dfrac{1}{5}=-\dfrac{245}{12}\\ ----\\ \left(\dfrac{12}{7}:x-0,5\right)\left(\dfrac{2}{5}.x-1\dfrac{1}{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{12}{7}:x-0,5=0\\\dfrac{2}{5}.x-\dfrac{3}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{12}{7}:x=0,5\\\dfrac{2}{5}.x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{12}{7}:\dfrac{1}{2}=\dfrac{24}{7}\\x=\dfrac{3}{2}:\dfrac{2}{5}=\dfrac{15}{4}\end{matrix}\right.\)
a: Xét ΔBAD có BA=BD
nên ΔBAD cân tại B
hay \(\widehat{BAD}=\widehat{BDA}\)
b: \(\widehat{HAD}+\widehat{BDA}=90^0\)
\(\widehat{CAD}+\widehat{BAD}=90^0\)
mà \(\widehat{BAD}=\widehat{BDA}\)
nên \(\widehat{HAD}=\widehat{CAD}\)
hay AD là tia phân giác của góc HAC
c: Xét ΔADH vuông tại H và ΔADK vuông tại K có
AD chung
\(\widehat{HAD}=\widehat{KAD}\)
Do đó:ΔADH=ΔADK
Suy ra: AH=AK
a) \(4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
b) \(\left(x-1\right)^2=9\Leftrightarrow x-1=3\Leftrightarrow x=4\)
c: Áp dung tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{9}=\dfrac{x+y}{4+9}=3\)
Do đó: x=12; y=27