a,(x+2y)^2=x^2+4xy+4y^2

b,(x-3y)(x+3y)=x^2-9y^2

c,(5-x)^2=25-10x+x^2

b: \(=3\left[\left(x+y\right)^2-2xy\right]-2\left[\left(x-y\right)^3+3xy\left(x-y\right)\right]\)

\(=3\left(1-2xy\right)-2\left(1+3xy\right)\)

\(=3-6xy-2-6xy=-12xy+1\)

c: \(=\left(x+y\right)^3-3\left(x^2+y^2+2xy\right)+3\left(x+y\right)+2012\)

\(=101^2-3\cdot101^2+3\cdot101+2012\)

=1002013

Lời giải:
Đặt $\frac{x}{3}=\frac{y}{5}=t(t\neq 0)$

$\Rightarrow x=3t; y=5t$

Khi đó:

$\frac{5x^2+3y^2}{10x^2-3y^2}=\frac{5(3t)^2+3(5t)^2}{10(3t)^2-3(5t)^2}=\frac{120t^2}{15t^2}=8$

Ta có : \(A=x^3y+xy^3=xy\left(x^2+y^2\right)=xy\left[\left(x+y\right)^2-2xy\right]\)

Thay x+y=3 và xy=1 vào ta có : \(A=3^2-2=7\)

Vậy A=7

Ta có: \(A=x^3y+xy^3=xy\left(x^2+y^2\right)\)

              \(=xy\left[\left(x+y\right)^2-2xy\right]\)

Thay \(x+y=3\)và \(xy=1\)vào, ta đc:

\(A=3^2-2=7\)

Vậy ta tìm đc \(A=7\)

Rất vui vì giúp đc bạn !!!

thay x =10 và y = 5 vào A, ta có

A=2.10 + 3.5 =20+15= 35

A=2x+3y=2.10+3.5=20+15=35

=2.10 + 3.5

=20 +15

=35

Vậy A=35

\(\dfrac{5x+5y}{3x-3y}:\dfrac{5x}{x^2-y^2}.\)

\(=\dfrac{5\left(x+y\right)}{3\left(x-y\right)}.\dfrac{\left(x-y\right)\left(x+y\right)}{5x}.\)

\(=\dfrac{x+y}{3}.\dfrac{x+y}{x}.\)

\(=\dfrac{\left(x+y\right)^2}{3x}.\)

\(A=\left(x+3y-5\right)^2-6xy+26\)

\(=x^2+9y^2+25+6xy-30y-10x+26-6xy\)

\(=x^2+9y^2-10x+51-30y\)

\(=\left(x^2-10x+25\right)+\left(9y^2-30y+25\right)+1\)

\(=\left(x-5\right)^2+\left(3y-5\right)^2+1\ge1\)

Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\3y-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\frac{5}{3}\end{matrix}\right.\)

Vậy \(Min_A=1\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\frac{5}{3}\end{matrix}\right.\)

Ta có : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{7}\)

=> \(\frac{5x}{10}=\frac{y}{3}=\frac{2z}{14}=\frac{5x+y-2z}{10+3-14}=\frac{4}{-1}=-4\)

=> 5x = -40 => x = -8

;     y = -12

; 2x = -56 => z = -28

Do đó x - 3y + 4z = -8 - 3.(-12) + 4.(-28) = -8 - (-36) + (-112) = -84

Ta có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{7}\)

\(\Rightarrow\frac{5x}{10}=\frac{y}{3}=\frac{2z}{14}=\frac{5x+y-2z}{10+3-14}=\frac{4}{-1}=-4\)

\(\Rightarrow x=-4.2=-8\)

\(y=-4.3=-12\)

\(z=-4.7=-28\)

\(\Rightarrow x-3y+4z=-8-3.\left(-12\right)+4.\left(-28\right)=-84\)