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đặt \(\frac{x}{5}=\frac{y}{3}\text{ }=k\)
\(\Rightarrow\text{ }x=5k\text{ };\text{ }y=3k\)
\(\Rightarrow\left(5k\right)^2-\left(3k\right)^2=4\)
\(\Rightarrow\text{ }25k^2-9k^2=4\)
\(\Rightarrow\text{ }k^2.\left(25-9\right)=4\)
\(\Rightarrow\text{ }k^2.16=4\)
\(\Rightarrow\text{ }k^2=\frac{1}{4}=\left(\frac{1}{2}\right)^2\)
\(\Rightarrow\text{ }\orbr{\begin{cases}k=\frac{1}{2}\\k=-\frac{1}{2}\end{cases}}\)
Nếu k = \(\frac{1}{2}\)thì \(x=\frac{5}{2}\text{ };\text{ }y=\frac{3}{2}\)
Nếu k = \(-\frac{1}{2}\)thì \(x=\frac{-5}{2}\text{ };\text{ }y=\frac{-3}{2}\)
10x = 6y
\(\Rightarrow\text{ }\frac{x}{6}=\frac{y}{10}\)
đặt \(\frac{x}{6}=\frac{y}{10}=k\)
\(\Rightarrow\text{ }x=6k\text{ };\text{ }y=10k\)
\(\Rightarrow\text{ }2.\left(6k\right)^2-\left(10k\right)^2=-28\)
\(\Rightarrow\text{ }72k^2-100k^2=-28\)
\(\Rightarrow\text{ }\left(72-100\right).k^2=-28\)
\(\Rightarrow\text{ }\left(-28\right).k^2=\left(-28\right)\text{ }\)
\(\Rightarrow\text{ }k^2=\left(-28\right)\text{ }:\text{ }\left(-28\right)\)
\(\Rightarrow\text{ }k^2=1\)
\(\Rightarrow\text{ }\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)
Nếu k = 1 thì x = 10 ; y = 6
Nếu k = -1 thì x = -10 ; y = -6
x/2=y/3;y/2=z/5 => x/2=2y/6;3y/6=z/5 => x/4=y/6=z/15
adtcdtsbn:
x/4=y/6=z/15=x+y+z/4+6+15=50/25=2
suy ra : x/4=2=>x=4.2=8
y/6=2=>y=2.6=12
z/15=2 => z=15.2=30
a)\(P+Q=\left(x^2y+xy^2-5x^2y^2+x^3\right)+\left(3xy^2-x^2y+x^2y^2\right)\)
=\(x^2y+xy^2-5x^2y^2+x^3+3xy^2-x^2y+x^2y^2\)
=\(x^2y-x^2y+xy^2+3xy^2-5x^2y^2+x^2y^2+x^3\)
=\(4xy^2-4x^2y^2+x^3\)
b)\(M+N=\left(x^3+xy+y^2-x^2y^2-2\right)+\left(x^2y^2+5-y^2\right)\)
=\(x^3+xy+y^2-x^2y^2-2+x^2y^2+5-y^2\)
=\(x^3+xy+y^2-y^2-x^2y^2+x^2y^2-2+5\)
=\(x^3+xy+3\)
Bài dài nên chắc sẽ có sai sót, nếu đúng bạn nha
a) Ta có: P = x2y + xy2 – 5x2y2 + x3 và Q = 3xy2 – x2y + x2y2
=> P + Q = x2y + xy2 – 5x2y2 + x3 + 3xy2 – x2y + x2y2
= x3 – 5x2y2 + x2y2 + x2y – x2y + xy2 + 3xy2
= x3 – 4x2y2 + 4xy2
b) Ta có: M = x3 + xy + y2 – x2y2 – 2 và N = x2y2 + 5 – y2.
=> M + N = x3 + xy + y2 – x2y2 – 2 + x2y2 + 5 – y2
= x3 – x2y2 + x2y2 + y2 – y2 + xy - 2 + 5
= x3 + xy + 3.
a)
P + Q = x2y + xy2 – 5x2y2 + x3 + 3xy2 – x2y + x2y2
= x3 – 5x2y2 + x2y2 + x2y – x2y + xy2 + 3xy2
= x3 – 4x2y2 + 4xy2
b)
M + N = x3 + xy + y2 – x2y2 – 2 + x2y2 + 5 – y2
= x3 – x2y2 + x2y2 + y2 – y2 + xy - 2 + 5
= x3 + xy + 3.
Đặt x/5=y/2=k
=>x=5k; y=2k
3x^2-2y^2=268
=>3*25k^2-2*4k^2=268
=>k^2=4
TH1: k=2
=>x=10; y=4
TH2: k=-2
=>x=-10; y=-4