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\(-\frac{3}{5}xyz^2\cdot\frac{1}{3}xy\cdot\left(-\frac{1}{4}\right)x^5yz\)
\(=\left(-\frac{3}{5}\cdot\frac{1}{3}\cdot\frac{-1}{4}\right)\left(x\cdot x\cdot x^5\right)\left(y\cdot y\cdot y\right)\left(z^2\cdot z\right)\)
\(=\frac{1}{20}x^7y^3z^3\)
\(\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+\frac{3}{5}\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+0,6\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=1\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=1-\frac{3}{4}\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{4}\\x-\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{4}\end{cases}}}\)
Vậy ...
\(1)\) Ta có :
\(3x=4y\)\(\Leftrightarrow\)\(\frac{x}{4}=\frac{y}{3}\)\(\Leftrightarrow\)\(\frac{x}{8}=\frac{y}{6}\)
\(5y=6z\)\(\Leftrightarrow\)\(\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow\)\(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}\)
Đặt \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=k\)\(\Rightarrow\)\(\hept{\begin{cases}x=8k\\y=6k\\z=5k\end{cases}}\) \(\left(1\right)\)
Thay \(\left(1\right)\) vào \(xyz=30\) ta được :
\(8k.6k.5k=30\)
\(\Leftrightarrow\)\(240k^3=30\)
\(\Leftrightarrow\)\(k^3=\frac{30}{240}\)
\(\Leftrightarrow\)\(k^3=\frac{1}{8}\)
\(\Leftrightarrow\)\(k^3=\left(\frac{1}{2}\right)^3\)
\(\Leftrightarrow\)\(k=\frac{1}{2}\)
Suy ra :
\(x=8k=8.\frac{1}{2}=\frac{8}{2}=4\)
\(y=6k=6.\frac{1}{2}=\frac{6}{2}=3\)
\(z=5k=5.\frac{1}{2}=\frac{5}{2}\)
Vậy \(x=4\)\(;\)\(y=3\) và \(z=\frac{5}{2}\)
Chúc bạn học tốt ~
Câu 1:
\(3\left(x-1\right)=2\left(y-2\right)\Leftrightarrow3x-3=2y-4\Leftrightarrow3x=2y-1\)
\(4\left(y-2\right)=3\left(z-3\right)\Leftrightarrow4y-8=3z-9\Leftrightarrow4y=3z-1\)
Lại có:
\(3x=2y-1\Leftrightarrow6x=4y-2=3z-1-2=3z-3\)
\(\Rightarrow6x=4y-2=3z-3\)
\(\Rightarrow6x=3z-3\Leftrightarrow2x=z-1\)
\(\Rightarrow2x+3y-z=z-1+3y-z=3y-1=50\Leftrightarrow3y=51\Leftrightarrow y=17\)\(\Rightarrow\left\{{}\begin{matrix}x=11\\z=23\end{matrix}\right.\)
Câu 3:
\(\frac{a}{b}=\frac{8}{5}\Leftrightarrow\frac{a}{8}=\frac{b}{5}\Leftrightarrow\frac{1}{2}.\frac{a}{8}=\frac{1}{2}.\frac{b}{5}\Leftrightarrow\frac{a}{16}=\frac{b}{10}\) (1)
\(\frac{b}{c}=\frac{2}{7}\Leftrightarrow\frac{b}{2}=\frac{c}{7}\Leftrightarrow\frac{1}{5}.\frac{b}{2}=\frac{1}{5}.\frac{c}{7}\Leftrightarrow\frac{b}{10}=\frac{c}{35}\) (2)
Từ (1) và (2)
\(\Rightarrow\frac{a}{16}=\frac{b}{10}=\frac{c}{35}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=16k\\b=10k\\c=35k\end{matrix}\right.\)
\(\Rightarrow a+b+c=16k+10k+35k=61k=61\Rightarrow k=1\)
\(\Rightarrow\left\{{}\begin{matrix}a=16k=16\\b=10k=10\\c=35k=35\end{matrix}\right.\)
a)p+(x2 -2y2)=x2 -y2 +3y2 -1
\(\Rightarrow\)p=(x2 -2y2 +3y2 -1)-(x-2y).(x+2y)
\(\Rightarrow\)p=(x2-2y2 +3y2 -1)-(x2 +2xy-2xy)
\(\Rightarrow\)p=x2 -2y2+3y2 -1-x2 -2xy+2xy
\(\Rightarrow\) p=2y2-1
Vậy P=2y2-1
b)Q-(5x2-xyz)=xy+2x2-3xyz+5
\(\Rightarrow\)Q=(xy+2x2-3xyz+5)+(5x2-xyz)
\(\Rightarrow\)Q=7x2-4xyz+xy+5
Vậy Q=7x2-4xyz+xy+5
\(3xyz^2+\left(-\frac{4}{8}\right)xyz^5\cdot\frac{1}{3}xyz\)
\(=3xyz^2-\frac{1}{2}xyz\cdot\frac{1}{3}xyz\)
\(=3xyz-\frac{1}{6}x^2y^2z^2\)
\(xyz\left(3-\frac{1}{6}xyz\right)\)
b) \(3xyz^5\cdot\left(-\frac{1}{7}\right)xyz\cdot\frac{-1}{8}xyz^4\)
\(=\left[3\cdot\left(-\frac{1}{7}\right)\cdot\left(-\frac{1}{8}\right)\right]\left(x\cdot x\cdot x\right)\left(y\cdot y\cdot y\right)\left(z^5\cdot z\cdot z^4\right)\)
\(=\frac{3}{56}x^3y^3z^{10}\)
a, \(3xyz^2+\left(\frac{-4}{8}xyz^5\right)\cdot\frac{1}{3}xyz=3xyz^2+\left[\left(\frac{-4}{8}\right)\cdot\frac{1}{3}\right]xyz^5xyz\)\(=3xyz^2-\frac{1}{2}x^2y^2z^6\)
b, \(3xyz^5\cdot\left(\frac{-1}{7}xyz^2\right)\cdot\frac{-1}{8}xyz^4=\left[3\cdot\left(\frac{-1}{7}\right)\cdot\left(\frac{-1}{8}\right)\right]xyz^5xyz^2xyz^4=\frac{3}{56}x^3y^3z^{11}\)