Kết quả C. 3xyz²

thanks

\(=xyz^2\)

đúng không

\(\frac{3}{4}xyz^2+\frac{1}{2}xyz^2+\left(-\frac{1}{4}\right)xyz^2\)

=\(\left(\frac{3}{4}+\frac{1}{2}-\frac{1}{4}\right)xyz^2\)

=\(xyz^2\)

\(3xyz^2+\left(-\frac{4}{8}\right)xyz^5\cdot\frac{1}{3}xyz\)

\(=3xyz^2-\frac{1}{2}xyz\cdot\frac{1}{3}xyz\)

\(=3xyz-\frac{1}{6}x^2y^2z^2\)

\(xyz\left(3-\frac{1}{6}xyz\right)\)

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, \(3xyz^5\cdot\left(\frac{-1}{7}xyz^2\right)\cdot\frac{-1}{8}xyz^4=\left[3\cdot\left(\frac{-1}{7}\right)\cdot\left(\frac{-1}{8}\right)\right]xyz^5xyz^2xyz^4=\frac{3}{56}x^3y^3z^{11}\)

a) 3x²y³+x²y³=4x²y³

b)5x²y-1/2x²y=10/2x²y-1/2x²y=9/2x²y

c) \(\frac{3}{4}xyz^2+\frac{1}{2}xyz^2-\frac{1}{4}xyz^2\)

\(=\frac{3}{4}xyz^2+\frac{2}{4}xyz^2-\frac{1}{4}xyz^2\)

\(=\frac{5}{4}xyz^2-\frac{1}{4}xyz^2\)

\(=\frac{4}{4}xyz^2=xyz^2\)

\(a,3x^2y^3+x^2y^3=4x^2y^3\)

\(b,5x^2y-\frac{1}{2}x^2y=\frac{9}{2}x^2y\)

\(c,\frac{3}{4}xyz^2+\frac{1}{2}xyz^2-\frac{1}{4}xyz^2=\left(\frac{3}{4}xyz^2-\frac{1}{4}xyz^2\right)+\frac{1}{2}xyz^2=\frac{2}{4}xyz^2+\frac{1}{2}xyz^2=xyz^2\)

\(-\frac{3}{5}xyz^2\cdot\frac{1}{3}xy\cdot\left(-\frac{1}{4}\right)x^5yz\)

\(=\left(-\frac{3}{5}\cdot\frac{1}{3}\cdot\frac{-1}{4}\right)\left(x\cdot x\cdot x^5\right)\left(y\cdot y\cdot y\right)\left(z^2\cdot z\right)\)

\(=\frac{1}{20}x^7y^3z^3\)

1, \(\frac{x}{2}=\frac{2y}{3}=\frac{3z}{4}\)\(\Leftrightarrow\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=k\)\(\Leftrightarrow\hept{\begin{cases}x=2k\\y=\frac{3}{2}k\\z=\frac{4}{3}k\end{cases}}\)

Mà xyz = -108

\(\Leftrightarrow2k.\frac{3}{2}k.\frac{4}{3}k=-108\)

\(\Leftrightarrow4k^3=-108\)

<=> k³ = -27

<=> k = -3

\(\Leftrightarrow\hept{\begin{cases}x=2k=2.-3=-6\\y=\frac{3}{2}k=\frac{3}{2}.\left(-3\right)=\frac{-9}{2}\\z=\frac{4}{3}k=\frac{4}{3}.\left(-3\right)=-4\end{cases}}\)

2, \(\frac{x}{5}=\frac{y}{7}=\frac{z}{8}\)\(\Leftrightarrow\frac{2x}{10}=\frac{3y}{21}=\frac{4z}{32}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có: 

\(\frac{2x}{10}=\frac{3y}{21}=\frac{4z}{32}=\frac{2x+3y-4z}{10+21-32}=\frac{15}{-1}=-15\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=-15\\\frac{y}{7}=-15\\\frac{z}{8}=-15\end{cases}}\Rightarrow\hept{\begin{cases}x=-75\\y=-105\\z=-120\end{cases}}\)

3, 3x = 5y \(\Leftrightarrow\frac{x}{5}=\frac{y}{3}\)\(\Leftrightarrow\frac{x}{55}=\frac{y}{33}\)

    2y = 11z \(\Leftrightarrow\frac{y}{11}=\frac{z}{2}\) \(\Leftrightarrow\frac{y}{33}=\frac{z}{6}\)

\(\Rightarrow\frac{x}{55}=\frac{y}{33}=\frac{z}{6}\)\(\Rightarrow\frac{2x}{110}=\frac{5y}{165}=\frac{z}{6}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\frac{2x}{110}=\frac{5y}{165}=\frac{z}{6}=\frac{2x+5y-z}{110+165-6}=\frac{34}{269}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{55}=\frac{34}{269}\\\frac{y}{33}=\frac{34}{269}\\\frac{z}{6}=\frac{34}{269}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1870}{269}\\y=\frac{1122}{269}\\z=\frac{204}{269}\end{cases}}\)

4, \(\frac{x}{3}=\frac{2}{y}=\frac{z}{4}=k\)\(\Leftrightarrow\hept{\begin{cases}x=3k\\y=\frac{2}{k}\\z=4k\end{cases}}\)

Mà xyz = 240

<=> 3k . 2/k . 4k = 240

<=> 24k = 240

<=> k = 10

 \(\Leftrightarrow\hept{\begin{cases}x=3k=3.10=30\\y=\frac{2}{k}=\frac{2}{10}=\frac{1}{5}\\z=4k=4.10=40\end{cases}}\)

Tính tổng của các đơn thức: xyz²; xyz²; -xyz² là

xyz² + xyz² + (-xyz²) = ( \(\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{1}{4}\)) xyz² = xyz².

Hướng dẫn giải:

Tính tổng của các đơn thức: xyz²; xyz²; -xyz² là

xyz² + xyz² + (-xyz²) = ( + - ) xyz² = xyz².

Ta có:\(\frac{4}{x+1}=\frac{2}{y-2}=\frac{3}{z+2}\)\(\Rightarrow\frac{x+1}{4}=\frac{y-2}{2}=\frac{z+2}{3}\)

Đặt \(\frac{x+1}{4}=\frac{y-2}{2}=\frac{z+2}{3}=k\)

\(\Rightarrow x=4k-1,y=2k+2,z=3k-2\)

Theo đề ta có:xyz=12

\(\Rightarrow\left(4k-1\right)\left(2k+2\right)\left(3k-2\right)=12\)

\(\Rightarrow\left(8k^2+8k-2k-2\right)\left(3k-2\right)=12\)

\(\Rightarrow\left(8k^2+6k-2\right)\left(3k-2\right)=12\)

\(\Rightarrow\left(8k^2+6k\right)\left(3k-2\right)-2\left(3k-2\right)\)

\(\Rightarrow24k^3-16k^2+18k^2-12k-6k+4=12\)

\(\Rightarrow24k^3+2k^2-18k=8\)

\(\Rightarrow24k^3+2k^2-18k-8=0\)

\(\Rightarrow\left(k-1\right)\left(24k^2+26k+8\right)=0\)(làm hơi tắt)

TH1:k-1=0,k=1

TH2:\(\left(24k^2+26k+8\right)=0\)

\(24\left(k+\frac{13}{24}\right)^2+\frac{23}{24}>0\)(vô lí)

\(\Rightarrow k=1\)

\(\Rightarrow x=3,y=4,z=1\)

các bạn ko cần làm đâu mình bít giải rồi