a) 3(x + 5) - 8 = x + 17

=> 3x + 15 - 8 = x + 17

=> 3x + 7 = x + 17

=> 3x - x = 17 - 7

=> 2x = 10

=> x = 10 : 2

=> x = 5

b) 5(x + 3) - 10 = 2x - 19

=> 5x + 15 - 10 = 2x - 19

=> 5x + 5 = 2x - 19

=> 5x - 2x = -19 - 5

=> 3x = -24

=> x = -24 : 3

=> x = -8

a. 3x+15-8=x+17

3x-x=17-15+8

2x=10

x=5

b.5x+15-10=2x-19

5x+5=2x-19

5x-2x=-19-5

3x=-24

x=-8

thank

dễ lắm kết quả =kimochii 

\(1\frac{1}{2}=\frac{3}{2}\)

Trong \(1\frac{1}{2}\) ngày Mai dệt được : \(\frac{\frac{3}{2}}{\frac{1}{2}}\cdot3=9\) ( cái chiếu )

Đ/s : 9 cái chiếu

\(3x\left(x-2\right)-x+2=0\)

\(\Leftrightarrow3x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)

\(B1:\)

\(3x\left(x-2\right)-\left(x-2\right)=0\)

\(\left(3x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)

\(x^2-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x-2\right)^3=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

\(A=x^2-6x+15\)

\(A=x^2-2\cdot x\cdot3+3^2+6\)( biến đổi về dạng HĐT )

\(A=\left(x-3\right)^2+6\)

vì ( x - 3 )² luôn >= 0 với mọi x

\(\Rightarrow A\ge6\)với mọi x

Dấu "=" xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy Amin = 6 <=> x = 3

\(B=2x^2-10x+8\)

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

\(B=2\left(x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{9}{4}\right)\)

\(B=2\left[\left(x-\frac{5}{2}\right)^2-\frac{9}{4}\right]\)

\(B=2\left(x-\frac{5}{2}\right)^2-\frac{9}{2}\)

Vì 2( x - 5/2 )² luôn >= 0 với mọi x

\(\Rightarrow B\ge\frac{-9}{2}\)với mọi x

Dấu "=" xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Vậy Bmin = -9/2 <=> x = 5/2

\(\left(x-8\right)\left(3x-15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\3x-15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=5\end{cases}}\)

Vậy x = 5 hoặc x = 8

\(\left(x-8\right)\left(3x-15\right)=0\)

Hoặc \(x-8=0\Leftrightarrow x=8\)

Hoặc\(3x-15=0\Leftrightarrow x=5\)

Vậy \(x=8\)hoặc \(x=5\)

\(\left(x-8\right)\left(3x-15\right)=0\)

Hoặc \(x-8=0\Leftrightarrow x=8\)

Hoặc\(3x-15=0\Leftrightarrow x=5\)

Vậy \(x=8\)hoặc \(x=5\)

a) 4/7 + 7/20 + 1/4                                       b) 21,15 + 3/5 + 3/4

= 4/7 + 7/20 + 5/20                                     =   21,15 + 0,6 + 0,75

= 4/7 + 3/5                                                  = 22,5

= 20/35 + 21/35

= 41/35

a,\(\frac{4}{7}\) \(+\frac{7}{20}\) \(+\frac{1}{4}\)

\(=\frac{4}{7}\) \(+\frac{7}{20}\) \(+\frac{5}{20}\)

\(=\frac{4}{7}\) \(+\frac{3}{5}\)

\(=\frac{20}{35}\)\(+\frac{21}{35}\)

\(=\frac{41}{35}\) 

#Hemingson

b, \(21,15\) \(+\frac{3}{5}\) \(+\frac{3}{4}\)

\(=21,15+0,6+0,75\)

\(=22,5\)

#Hemingson

a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)

b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)