0.7×y+3/10×y=6.3
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Dấu \("."\) là dấu nhân
\(0,7.y+\frac{3}{10}.y=6,3\)
\(\Rightarrow0,7.y+0,3.y=6,3\)
\(\Rightarrow\left(0,7+0,3\right).y=6,3\)
\(\Rightarrow1.y=6,3\)
\(\Rightarrow y=6,3:1\)
\(\Rightarrow y=6,3\)
Vậy \(y=6,3\)
Chúc bạn học tốt !!!
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\(y=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(y=\frac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(y=\frac{2^9.3^9\left(2^3.3+120\right)}{2^{11}.3^{11}\left(2.3+1\right)}\)
\(y=\frac{6^9\left(2^3.3+120\right)}{6^{11}.7}\)
\(y=\frac{2^3.3+120}{6^2.7}\)
\(y=\frac{144}{252}\)
\(y=\frac{4}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)= \(\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
= \(\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)= \(\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)= \(\frac{2.6}{3.7}=\frac{4}{7}\)
c, theo đề bài ta có :
x2 = yz, y2 = xz , z2 = xy
\(\Rightarrow\frac{x}{y}=\frac{z}{x},\frac{y}{x}=\frac{z}{y},\frac{z}{x}=\frac{y}{z}\Rightarrow\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\)
AD t/c DTSBN, ta có
\(\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\Rightarrow\frac{X+z+y}{y+x+z}=1\)
x= 1y
z= 1x
y= 1z
=> x = y = x
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{3^5\cdot2^5\cdot3^6}{2^6\cdot3^{10}}-\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{1}{2}=1\)
0,7 * y + 3/10 * y = 6,3
0,7 * y + 0,3 * y = 6,3
y * [0,7+0,3] =6,3
y * 1 =6,3
y = 6,3 :1
y=6,3
0,7 x y + \(\frac{3}{10}\)x y = 6,3
( 0, 7 + \(\frac{3}{10}\)) x y = 6,3
1 x y = 6,3