3 + 3 x 24 -13= ????
giúp mk với
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\(A=2\dfrac{3}{13}\times\dfrac{13}{58}\times8\times2\dfrac{15}{24}\times\dfrac{8}{21}\)
\(A=\dfrac{29}{13}\times\dfrac{13}{58}\times8\times\dfrac{21}{8}\times\dfrac{8}{21}\)
\(A=\dfrac{29\times13\times8\times21\times8}{13\times58\times8\times21}\)
\(A=\dfrac{1\times1\times1\times1\times8}{1\times2\times1\times1}\)
\(A=\dfrac{8}{2}\)
\(A=4\)
\(A=\dfrac{29}{13}\cdot\dfrac{13}{58}\cdot8\cdot\dfrac{21}{8}\cdot\dfrac{8}{21}=\dfrac{1}{2}\cdot8=4\)
c) \(x^2-9=2\cdot\left(x+3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left[x-3-2\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3-2x-6\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
d) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x^2-8x\right)+\left(3x-24\right)=0\)
\(\Leftrightarrow x\left(x-8\right)+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)
a) \(x^2-9=2\left(x+3\right)^2\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)=2\left(x+3\right)^2\)
\(\Leftrightarrow2\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[2\left(x+3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[2x+6-x+3\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+9\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+9=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x-8\right)x+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
c) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
a/ ta thấy tích trên có 4 thừa số âm , nên (-16).(-8).(-4).(-3) = 1 số dương=> (-16).1253.(-8).(-4).(-3)= 1 số dương=> (-16).1253.(-8).(-4).(-3)>0
b/ ta thấy tích trên có 3 thừa số âm, nên 13.(-24).(-15).(-8) là 1 số âm,=>13.(-24).(-15).(-8).4 là 1 số âm=>13.(-24).(-15).(-8).4 <0
\(\sqrt{x-2}=3\left(x\ge2\right)\\ \Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\\ \sqrt{4x^2}+4x+1=3\Leftrightarrow\left|2x\right|=2-4x\\ \Leftrightarrow\left[{}\begin{matrix}2x=2-4x\left(x\ge0\right)\\2x=4x-2\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{3}\)
a/ (3x -12 ):8+10=13
<=>(3x-12):8=3
<=>3x-12=24
<=>3x=36
<=>x=12
b/(x+5)x8-20=60
<=>(x+5)x8=80
<=>x+5=80:8
<=>x+5=10
x=5
c/3x[35+(x-10)]-375=0
<=>3x[35+(x-10)]=375
<=>35+(x+10)=125
<=>x+10=90
x=80
d/(158x+173-24):35=81
<=>158x+173-24=2835
<=>158x+173=2859
<=>158x=2686
<=>x=17
e/x-140:35=0
<=>x-140=0
<=>x=140+0
x=140
\(1\frac{13}{15}.0,75-\left(\frac{8}{15}+25\%\right).\frac{24}{47}-3\frac{12}{13}:3\)
\(=\frac{28}{15}.\frac{3}{4}-\left(\frac{8}{15}+\frac{1}{4}\right).\frac{24}{47}-\frac{51}{13}:3\)
\(=\frac{7}{5}-\frac{47}{60}.\frac{24}{47}-\frac{17}{13}\)
\(=\frac{7}{5}-\frac{2}{5}-\frac{17}{13}\)
\(=\frac{-4}{13}\)
\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Leftrightarrow\frac{13}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{-11}{12}\)
\(\Leftrightarrow\frac{-13}{9}\le x\le\frac{-11}{18}\)
\(\Leftrightarrow x=-1\)
3 + 3 x 24 - 13
= 3 x 1 + 3 x 24 - 13
= 3 x (1 + 24) - 13
= 3 x 25 - 13
= 75 - 13
= 62
~Study well~
#SJ
3+3.24-13=?
3+3.24-13
=3+72-13
=75-13
=62,
Học tốt !