7^3 :7 -7^2
các anh ,chị giải giúp em với ạ!!!!!!
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\(\left|x+\dfrac{1}{7}\right|-\dfrac{2}{3}=0\)
\(\Rightarrow\left|x+\dfrac{1}{7}\right|=0+\dfrac{2}{3}\\ \Rightarrow\left|x+\dfrac{1}{7}\right|=\dfrac{2}{3}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{7}=\dfrac{2}{3}\\x+\dfrac{1}{7}=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}-\dfrac{1}{7}\\x=-\dfrac{2}{3}-\dfrac{1}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{21}\\x=-\dfrac{17}{21}\end{matrix}\right.\)
câu 5:
x=3,6
y=6,4
câu 6: chụp lại đề
câu 7:
a)ĐKXĐ: \(x\ge0\)
\(3\sqrt{x}=\sqrt{12}\\ \Rightarrow9x=12\\ \Rightarrow x=\dfrac{4}{3}\)
b) ĐKXĐ: \(x\ge6\)
\(\sqrt{x-6}=3\\ \Rightarrow x-6=9\\ \Rightarrow x=15\)
Question 2: David has volunteered for 2 years
Question 3: I think collecting stamps is interesting
\(x\cdot\dfrac{3}{7}-x\cdot\dfrac{1}{2}=\dfrac{3}{5}\)
\(x\left(\dfrac{3}{7}-\dfrac{1}{2}\right)=\dfrac{3}{5}\)
\(x\cdot\dfrac{-1}{14}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{-1}{14}\)
\(x=\dfrac{-42}{5}\)
5:
=10x^3*1/2xy-2/5y*1/2xy+1/2z*1/2xy
=5x^3y-1/5xy^2+1/4xyz
6: =x^2y*4xy+x^2y*3y-5x*x^2y
=4x^3y^2+3x^2y^2-5x^3y
7: =-4/3xy*3x^2y+4/3xy*6xy-4/3xy*9x
=-4x^3y^2+8x^2y^2-12x^2y
\(7^3:7-7^2\)
\(=7^{3-1}-7^2\)
\(=7^2-7^2\)
= 0
7^3:7-7^2
=7^2-7^2
=7^0=1