a, \(\left|2x+1\right|=3\)

=> 2x + 1 = 3 hoặc 2x + 1 = -3

=> 2x = 3 - 1 hoặc 2x = -3 - 1

=> 2x = 2 hoặc 2x = -4

=> x = 1 hoặc x = -2

b, \(2\frac{1}{2}x+1\frac{1}{2}=2\frac{2}{3}\)

=> \(\frac{5}{2}x+\frac{3}{2}=\frac{8}{3}\)

=> \(\frac{5}{2}x=\frac{8}{3}-\frac{3}{2}\)

=> \(\frac{5}{2}x=\frac{16-9}{6}\)

=> \(\frac{5}{2}x=\frac{7}{6}\)

=> \(x=\frac{7}{6}:\frac{5}{2}=\frac{7}{6}\cdot\frac{2}{5}=\frac{7}{3}\cdot\frac{1}{5}=\frac{7}{15}\)

c, \(3\cdot5^{x-3}+1=16\)

=> 3 . 5^x-3 = 16 - 1

=> 3 . 5^x-3 = 15

=> 5^x-3 = 15 : 3

=> 5^x-3 = 5

=> x - 3 = 5 : 5

=> x - 3 = 1

=> x = 1 + 3 = 4

d, \((x-1)^2=25\)

=> \((x-1)^2=5^2\)

=> x - 1 = 5 hoặc x - 1 = -5

=> x = 6 hoặc x = -4

e, \((-2)^2+\left|3x+1\right|=(-28)\cdot7\)

=> 4 + |3x + 1| = -196

=> |3x + 1| = -196 - 4 = -200

=> |3x + 1| = -200

Không thỏa mãn điều kiện

\(\text{a) }3x^2+x-2\)

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

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

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

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

b) \(5x^2-18x-8\)

\(=5x^2-20x+2x-8\)

\(=5x\left(x-4\right)+2\left(x-4\right)\)

\(=\left(5x+2\right)\left(x-4\right)\)

a) \(x^2-x-xy-2y^2+2y\)

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

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

\(=2y\left(-y-x+1\right)+x\left(y+x-1\right)\)

\(=-2y\left(y+x-1\right)+x\left(y+x-1\right)\)

\(=\left(x-2y\right)\left(y+x-1\right)\)

b) \(x^2+4xy+2x+3y^2+6y\)

\(=x^2+3xy+xy+2x+3y^2+6y\)

\(=\left(3y^2+3xy+6y\right)+\left(xy+x^2+2x\right)\)

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

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

đề là gì vậy

Tìm Xy ok 

\(c=5+5^2+5^3+...+5^{20}\)

\(c=5.1+5.5+5.5^2+...+5.5^{19}\)

\(c=5.\left(1+5+5^2+...+5^{19}\right)\)chia hết cho 5

#Học tốt

Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

=> \(\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\)

=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\).

6,5 - \(\frac{9}{4}\) :     x + 

sorry lỡ bấm gửi trả lời