( 3x -y)2 -2(3x -y) (x +5y) +(x + 5y)2
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a)
\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{3x-2y}{3.5-2.2}=\dfrac{-55}{11}=-5\)
=> \(\left\{{}\begin{matrix}x=-5.5=-25\\y=-5.2=-10\end{matrix}\right.\)
b)
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{2x+5y}{2.3+5.2}=\dfrac{48}{16}=3\)
=> \(\left\{{}\begin{matrix}x=3.3=9\\y=3.2=6\end{matrix}\right.\)
c)
Có: \(\dfrac{x}{y}=-\dfrac{5}{2}\Leftrightarrow-\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x+y}{-5+2}=\dfrac{30}{-3}=-10\)
=> \(\left\{{}\begin{matrix}x=-10.-5=50\\y=-10.2=-20\end{matrix}\right.\)
d)
Có: \(\dfrac{x}{y}=\dfrac{4}{3}\Leftrightarrow\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{2x+3y}{2.4+3.3}=\dfrac{34}{17}=2\)
=> \(\left\{{}\begin{matrix}x=2.4=8\\y=2.3=6\end{matrix}\right.\)
a, \(3x=5y=7z=>\dfrac{3x}{105}=\dfrac{5y}{105}=\dfrac{7z}{105}=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)
áp dụng tính chất dãy tỉ số = nhau
\(=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}=\dfrac{x+y+z}{35+21+15}=\dfrac{10}{71}\)
\(=>\dfrac{x}{35}=\dfrac{10}{71}=>x=\dfrac{350}{71}\)
\(=>\dfrac{y}{21}=\dfrac{10}{71}=>y=\dfrac{210}{71}\)
\(=>\dfrac{z}{15}=\dfrac{10}{71}=>z=\dfrac{150}{71}\)
b, \(\)\(6x=5y=>\dfrac{x}{5}=\dfrac{y}{6}=>\dfrac{x}{20}=\dfrac{y}{24}\)
có \(7y=8z=>\dfrac{y}{8}=\dfrac{z}{7}=>\dfrac{y}{24}=\dfrac{z}{21}\)
\(=>\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}\)
áp dụng t/c dãy tỉ số = nhau
\(=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}=\dfrac{3x+2y+4z}{60+48+84}=\dfrac{12}{192}=\dfrac{1}{16}\)
\(=>\dfrac{3x}{60}=\dfrac{1}{16}=>x=1,25\)
\(=>\dfrac{2y}{48}=\dfrac{1}{16}=>y=1,5\)
\(=>\dfrac{4z}{84}=\dfrac{1}{16}=>z=1,3125\)
c, \(x:y:z=1:2:3=>\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)
\(=>x=\dfrac{y}{2},z=\dfrac{3y}{2}\)
thay x,z vào \(x^3+y^3+z^3=36=>\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)
\(=>y=2\)
\(=>x=\dfrac{y}{2}=\dfrac{2}{2}=1,z=\dfrac{3y}{2}=\dfrac{3.2}{2}=3\)
d, \(\dfrac{x}{2}=\dfrac{y}{3}=>x=\dfrac{2y}{3}\)
thay x vào \(3x^3+y^3=51=>3.\left(\dfrac{2y}{3}\right)^3+y^3=51=>y=3\)
\(=>x=\dfrac{2.3}{3}=2\)
c, từ đoạn này á
\(\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)
\(< =>\dfrac{y^3}{8}+\dfrac{8y^3}{8}+\dfrac{27y^3}{8}=36\)
\(=>\dfrac{36y^3}{8}=36=>36y^3=8.36=>y^3=8=>y=2\)
\(a,14x^2y-21xy^2+28x^2y^2=7xy\left(x-3y+4xy\right)\\ b,x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\\ c,10x\left(x-y\right)-8\left(y-x\right)=10x\left(x-y\right)+8\left(x-y\right)=\left(x-y\right)\left(10x+8\right)=2\left(x-y\right)\left(5x+4\right)\)
\(d,\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)\(e,x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
3: \(x^2\left(x-1\right)+2x\left(1-x\right)\)
\(=x^2\left(x-1\right)-2x\cdot\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x\right)=x\left(x-1\right)\left(x-2\right)\)
5: \(y^2\left(x^2+y\right)-z\cdot x^2-zy\)
\(=y^2\left(x^2+y\right)-z\left(x^2+y\right)\)
\(=\left(x^2+y\right)\left(y^2-z\right)\)
7: \(5\left(x+y\right)^2+15\left(x+y\right)\)
\(=5\left(x+y\right)\cdot\left(x+y\right)+5\left(x+y\right)\cdot3\)
=5(x+y)(x+y+3)
9: \(7x\left(y-4\right)^2-\left(4-y\right)^3\)
\(=7x\left(y-4\right)^2+\left(y-4\right)^3\)
\(=\left(y-4\right)^2\left(7x+y-4\right)\)
11: \(\left(x+1\right)\left(y-2\right)-\left(2-y\right)^2\)
\(=\left(x+1\right)\left(y-2\right)-\left(y-2\right)^2\)
=(y-2)(x+1-y+2)
=(y-2)(x+y+3)
2: \(5x\left(x-2\right)-3x^2\left(x-2\right)\)
\(=\left(x-2\right)\left(5x-3x^2\right)=x\left(5-3x\right)\left(x-2\right)\)
4: 3x(x-5y)-2y(5y-x)
=3x(x-5y)+2y(x-5y)
=(x-5y)(3x+2y)
6: b(a-c)+5c-5a
=b(a-c)-5(a-c)
=(a-c)(b-5)
8: 9x(x-y)-10(y-x)^2
=9x(x-y)-10(x-y)^2
=(x-y)(9x-10x+10y)
=(x-y)(-x+10y)
10: \(\left(a-b\right)^2-\left(a+b\right)\left(b-a\right)\)
\(=\left(a-b\right)^2+\left(a+b\right)\left(a-b\right)\)
=(a-b)(a-b+a+b)
=2a(a-b)
12: 2x(x-3)+y(x-3)+(3-x)
=(x-3)(2x+y)-(x-3)
=(x-3)(2x+y-1)