9/12=3/4

\(\dfrac{3}{4}=\dfrac{9}{12}=\dfrac{3}{4}\)

3/4=9/12

Đặt \(\left\{{}\begin{matrix}\dfrac{x}{3}=k\\\dfrac{y}{4}=k\\\dfrac{z}{11}=k\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=11k\end{matrix}\right.\)

Ta có: \(A=\dfrac{y+z-x}{x+z-y}\)

\(=\dfrac{4k+11k-3k}{3k+11k-4k}\)

\(=\dfrac{12k}{10k}=\dfrac{6}{5}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{11}=\dfrac{y+z-x}{12}=\dfrac{x+z-y}{10}\\ \Rightarrow\dfrac{y+z-x}{x+z-y}=\dfrac{12}{10}=\dfrac{6}{5}\)

\(3,=\left(\dfrac{13}{25}-\dfrac{38}{25}\right)+\left(\dfrac{14}{9}-\dfrac{5}{9}\right)=-1+1=0\\ 4,=\left(\dfrac{4}{9}\right)^5\cdot\left(\dfrac{9}{49}\right)^5=\left(\dfrac{4}{9}\cdot\dfrac{9}{49}\right)^5=\left(\dfrac{4}{49}\right)^5\\ 5,\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{x+y}{5+3}=\dfrac{2}{2}=\dfrac{x+y}{8}\Rightarrow x+y=8\\ 6,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\Rightarrow2\text{ giá trị}\\ 7,=\dfrac{3^{10}\cdot2^{30}}{2^9\cdot3^9\cdot2^{20}}=2\cdot3=6\)

Câu 7:

=6

\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{-2\left(-4k\right)-7k+5.3k}{2.\left(-4k\right)-3.\left(-7k\right)-6.3k}=\dfrac{16k}{-5k}=-\dfrac{16}{5}\)

Câu 1: 

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{15}{5}=3\)

Do đó: x=6; y=9

Câu 2: 

Đặt \(\dfrac{a}{2}=\dfrac{b}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2k\\b=5k\end{matrix}\right.\)

Ta có: \(a^3+b^3=133\)

\(\Leftrightarrow8k^3+125k^3=133\)

\(\Leftrightarrow k=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2k=2\\b=5k=5\end{matrix}\right.\)

Câu 1: 

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{15}{5}=3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=3\\\dfrac{y}{3}=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\end{matrix}\right.\)

Bài 2: 

\(\dfrac{a}{2}=\dfrac{b}{5}\Rightarrow\dfrac{a^3}{8}=\dfrac{b^3}{125}=\dfrac{a^3+b^3}{8+125}=\dfrac{133}{133}=1\)

\(\dfrac{a^3}{8}=1\)  và  \(\dfrac{b^3}{125}=1\)

\(\Rightarrow a=2\)  và  \(b=5\)

b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

a: \(A=31x^2y^3-2xy^3+\dfrac{1}{4}x^2y^2+2\)

\(B=2xy^3+\dfrac{3}{4}x^2y^2-31x^2y^3-x^2-5\)

P=\(A+B=x^2y^2-x^2-3\)

\(A-B=62x^2y^3-4xy^3-\dfrac{1}{2}x^2y^2+x^2+7\)

b: Khi x=6 và y=-1/3 thì \(P=\left(6\cdot\dfrac{-1}{3}\right)^2-6^2-3=4-36-3=1-36=-35\)