1/

a)5x – 20y=5(x-4y)

b) 5x.(x –  1) –  3x(x – 1)=2x(x-1)

c) x.(x+y) – 5x – 5y=c) x.(x+y) – 5(x+y)=(x-5)(x+y)

2/

a)x² + xy + x = x(x+y+1)=77.(77+22+1)=77.100=7700

b)  x . ( x – y ) + y . ( y – x )=(x-y)(x-y)=(x-y)²=(53-3)²=2500

3/

a) X + 5x² = 0

⇒x(x+5)=0

⇒hoặc x=0

x+5=0⇒x=-5

b)x + 1 = ( x + 1 )² 

⇒(x + 1)-( x + 1 )² =0

⇒x(x+1)=0

⇒ hoặc x=0

hoặc x+1=0⇒x=-1

4/

a) 97 . 13 + 130 . 0,3 = 97.13+13.10.0,3=97.13+13.3=100.13=1300

b)86 . 153 – 530 . 8,6=86.153–53.10.8,6=86.153-53.86=86.100=8600

C) 85 .12,7 + 5,3 . 12,7= 12,7(85+5,3)=12,7.90,3=1146,81

D)52.143 – 52 . 39 – 8.26=52(143-39)-8,26=52.104-8,26=5399,74

Bài 1:  

a) 5x-20y=5(x-4y)

b) \(5x\left(x-1\right)-3x\left(x-1\right)=2x\left(x-1\right)\)

c) \(x\left(x+y\right)-5x-5y=\left(x+y\right)\left(x-5\right)\)

Bài 2: 

a) \(x^2+xy+x\)

\(=x\left(x+y+1\right)\)

\(=77\cdot\left(77+22+1\right)\)

=7700

b) \(x\left(x-y\right)+y\left(y-x\right)\)

\(=x\left(x-y\right)-y\left(x-y\right)\)

\(=\left(x-y\right)^2\)

\(=50^2=2500\)

Bài 10:

a) (x+2)² -x(x+3) + 5x = -20

=> x² + 4x + 4 - x² - 3x + 5x = -20

=> 6x = -20 + (-4)

=> 6x = -24

=> x = -4

b) 5x³-10x²+5x=0   

=>5x(x²-2x+1)=0

=>5x(x-1)² =0

=> 5x=0 hoặc (x-1)²=0

=>x=0 hoặc x=1

c) (x² - 1)³ - (x⁴ + x² + 1)(x² - 1) = 0

=> (x² - 1)[(x² - 1)² -  (x⁴ + x² + 1)] = 0

<=> (x² - 1)(x⁴ - 2x² + 1 - x⁴ - x² - 1) = 0

<=>  (x² - 1)(-3x²) = 0

<=> (x² - 1)=0 hoặc (-3x²) =0

<=> x²=1 hoặc x²=0

<=> x=−1;1 hoặc x=0

d)

⇔x=-23/12

Học tốt nhé:333

a) \(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

b) \(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\Rightarrow\left(x-1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) x= + - 5
b) x\(\in\)\(\left\{-1;-7\right\}\)

a/ \(x^2-25=0\)

\(\Rightarrow\left(x+5\right)\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+5=0\Rightarrow x=-5\\x-5=0\Rightarrow x=5\end{matrix}\right.\)

b/ \(x\left(x+7\right)+x+7=0\)

\(x\left(x+7\right)+\left(x+7\right)=0\)

\(\left(x+7\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+7=0\Rightarrow x=-7\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)

a) x^2 - 25 = 0 <=> (x + 5)(x - 5) = 0

<=> x = -5 hoặc x = 5

b) x(x + 7) + x + 7 = 0 <=> (x + 7)(x + 1) = 0

<=> x = -7 hoặc x = -1

Bài 1: 

a: \(11x^2-6xy-5y^2\)

\(=11x^2-11xy+5xy-5y^2\)

\(=11x\left(x-y\right)+5y\left(x-y\right)\)

\(=\left(x-y\right)\left(11x+5y\right)\)

b: \(4x^3-16x^2+19x-6\)

\(=4x^3-8x^2-8x^2+16x+3x-6\)

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

\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)

Bài 1: 

a: \(11x^2-6xy-5y^2\)

\(=11x^2-11xy+5xy-5y^2\)

\(=11x\left(x-y\right)+5y\left(x-y\right)\)

\(=\left(x-y\right)\left(11x+5y\right)\)

b: \(4x^3-16x^2+19x-6\)

\(=4x^3-8x^2-8x^2+16x+3x-6\)

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

\(=\left(x-2\right)\left(2x-3\right)\left(2x-1\right)\)

\(a,=11x^2-11xy+5xy-5y^2=\left(11x+5y\right)\left(x-y\right)\\ b,=4x^3-8x^2-8x^2+16x+3x-6\\ =\left(x-2\right)\left(4x^2-8x+3\right)\\ =\left(x-2\right)\left(4x^2-2x-6x+3\right)\\ =\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)

Bài 1: 

a: \(11x^2-6xy-5y^2\)

\(=11x^2-11xy+5xy-5y^2\)

\(=11x\left(x-y\right)+5y\left(x-y\right)\)

\(=\left(x-y\right)\left(11x+5y\right)\)

b: \(4x^3-16x^2+19x-6\)

\(=4x^3-8x^2-8x^2+16x+3x-6\)

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

\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)

a) Ta có: \(\dfrac{P}{x+2}=\dfrac{x^2+5x+6}{x^2+4x+4}\)

\(\Leftrightarrow\dfrac{P}{x+2}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)

hay P=x+3

b) Ta có: \(\dfrac{\left(a+1\right)^2}{P}=\dfrac{a+1}{a-1}\)

\(\Leftrightarrow P=\left(a+1\right)\left(a-1\right)\)

\(\Leftrightarrow P=a^2-1\)

a: Ta có: \(x^2\ge0\forall x\)

\(\left(y-\dfrac{1}{10}\right)^4\ge0\forall y\)

Do đó: \(x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(0;\dfrac{1}{10}\right)\)