Ta có: \(x+y=2,5=\frac{5}{2}\)

\(\Rightarrow\left(x+y\right)^2=\frac{25}{4}\)

Ta  có: 

\(A=\frac{x}{y}+\frac{y}{x}\)

\(=\frac{x.x}{x.y}+\frac{y.y}{y.x}\)

\(=\frac{x^2}{xy}+\frac{y^2}{xy}\)

\(=\frac{x^2+y^2}{xy}\)

\(=\frac{\frac{25}{4}}{1}=\frac{25}{4}\)

x=25/4

x - 2 x + 5x - 9x +11 3 2 x 2 x - 2x 3 2 7x 2 - 9x + 11 + 7x 7x - 14x 2 5x +11 + 5 5x - 10 21

1. 2x²y(2x²+3x+3)

    = 4x⁴y +6x³y+6x²y

2) ( 4x²y + 6xy² - 2x )1/2xy²

   = 2x³y³ +3x²y⁴-x²y²

Với x = 2,y = -1 ta có :

\(2^2\left[2-(-1)\right]-2\cdot2^3\left[1+(-1)\right]\)

\(=4\cdot3-2\cdot2^3\cdot0\)

\(=12-16\cdot0=12-0=12\)

2^2(2+1)-2.2^3(1-1)=4.3-0=12

g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)

\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)

\(\Leftrightarrow-5\left(4x+3\right)=0\)

\(\Leftrightarrow4x+3=0\)

\(\Leftrightarrow4x=-3\)

\(\Leftrightarrow x=\frac{-3}{4}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)

h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)

\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)

\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)

\(\Leftrightarrow-9x+2x-3-10x=30\)

\(\Leftrightarrow-17x-3=30\)

\(\Leftrightarrow-17x=33\)

\(\Leftrightarrow x=\frac{-33}{17}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)

Mình chịu 

\(1+a^2=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)

Tương tự, ta có: \(1+b^2=\left(a+b\right)\left(b+c\right)\)\(;\)\(1+c^2=\left(b+c\right)\left(c+a\right)\)

\(\Rightarrow\)\(\frac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) ( do a, b, c dương ) 

\(\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}=\frac{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

... 

a,\(xy+3x-7y-21\)

\(=x\left(y+3\right)-7\left(y+3\right)\)

\(=\left(y+3\right)\left(x-7\right)\)

\(b,2xy-15-6x+5y\)

\(=\left(2xy-6x\right)+\left(-15+5y\right)\)

\(=2x\left(y-3\right)-5\left(3-y\right)\)

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

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

#)Giải :

Bài 1 :

a) \(9\left(4x+3\right)^2=16\left(3x-5\right)^2\)

\(\Leftrightarrow144x^2+216x+81=144x^2-480x+400\)

\(\Leftrightarrow144x^2+216=144x^2-480x+319\)

\(\Leftrightarrow696x=319\)

\(\Leftrightarrow x=\frac{11}{24}\)

b) \(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^2+2\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)

\(\Leftrightarrow x=1\)

c) \(x^5+x^4+x^3+x^2+x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0\)

\(\Leftrightarrow x=-1\)

a) 9(4x + 3)² = 16(3x - 5)²

=> [3(4x + 3)]² - [4(3x - 5)]² = 0

=> (12x + 9)² - (12x - 20)² = 0

=> (12x + 9 - 12x + 20)(12x + 9 + 12x - 20) = 0

=> 29.(24x - 11) = 0

=> 2x - 11 = 0

=> 2x = 11

=>  x = 11 : 2 = 11/2

b) (x³ - x²)² - 4x² + 8x - 4 = 0

=> (x³ - x²)² - (2x - 2)² = 0

=> (x³ - x² - 2x + 2)(x³ - x² + 2x - 2) = 0

=> [x²(x - 1) - 2(x - 1)][x²(x - 1) + 2(x - 1)] = 0

=> (x² - 2)(x - 1)(x² + 2)(x - 1) = 0

=> (x² - 2)(x² + 2)(x - 1)² = 0

=> x² - 2 = 0

hoặc : x² + 2 = 0

hoặc : (x - 1)² = 0

=> x² = 2

 hoặc : x² = -2 (vl)

hoặc : x - 1 = 0

=> \(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)

hoặc : x = 1

Vậy ...

c) x⁵  + x⁴ + x³ + x² + x + 1 = 0

=> x⁴(x +1) + x²(x + 1) + (x + 1) = 0

=> (x⁴ + x² + 1)(x + 1) = 0

=> \(\orbr{\begin{cases}x^4+x^2+1=0\\x+1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x^4+x^2=-1\left(vl\right)\\x=-1\end{cases}}\) (vì x⁴ \(\ge\)0 \(\forall\)x; x² \(\ge\)0 \(\forall\)x => x⁴ + x² \(\ge\)0 \(\forall\)x)

=> x = -1

\(a,\frac{x+1}{x-2}-\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x^2+4}{\left(x-2\right)\left(x+2\right)}\)

\(\Rightarrow x^2+2x+x+2-\left(x^2-2x-x+2\right)=2x^2+4\)

\(\Leftrightarrow x^2+3x+2-x^2+2x+x-2=2x^2+4\)

\(\Leftrightarrow6x=2x^2+4\)

\(\Leftrightarrow2x^2+4-6x=0\)

\(\Leftrightarrow2x^2+4-6x=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)

\(b,\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)

\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x-1\right)\left(x-1\right)\)

\(\Leftrightarrow2x^2+2x+x+1=5\left(x^2-2x+1\right)\)

\(\Leftrightarrow2x^2+3x+1=5x^2-10x+5\)

\(\Leftrightarrow5x^2-2x^2-10x-3x+5-1=0\)

\(\Leftrightarrow3x^2-13x+4=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{1}{3}\end{cases}}}\)

\(a,\left(6x+1\right)\left(x+2\right)-2x\left(3x-5\right)\)

\(=6x^2+12x+x+2-6x^2+10x\)

\(=23x+2\)

a) (6x + 1)(x + 2) - 2x(3x - 5)

= 6x² + 12x + x + 2 - 6x² + 10x

= (6x² - 6x²) + (12x + x + 10x) + 2

= 23x + 2

b) (2x - 1)² - (2x - 3)(2x + 3)

= 4x² - 4x + 1 - 4x² + 9

= (4x² - 4x²) - 4x + (1 + 9)

= -4x + 10

c) (2x - 3)³  - (3x  + 1)(5 - 4x) - 16x²

= 8x³ - 36x² + 54x - 15x + 12x² - 5 + 4x - 16x²

= 8x³ - (36x² - 12x² + 16x²) + (54x - 15x + 4x) - 5

= 8x³ - 40x² + 43x - 5

d) (3x + 2) - (x - 5) - x(3x - 13)

= 3x  + 2 - x + 5 - 3x² + 13x

= (3x - x + 13x) + (2 + 5) - 3x²

= 15x + 7 - 3x²