Tính thuận tiện :

5,6 x 8 + 2,8 x 8 - 2 x 5,6

Ta có 2,8 x 2 = 5,6

Rút 2,8 x 8 = 5,6 x 4

5,6 x 8 + 5,6 x 4 - 2 x 5,6

= 5,6 x ( 8 + 4 - 2 ) 

= 5,6 x 10

= 56

5,6*8+2,8*8-2*5,6

=5,6*8+5,6*4-2*5,6

=5,6*(8+4-2)

=5,6*10

=56

bang 214,8

Giải:

(71,5 - 17,8) . 4

= 70,72 . 4

= 282,88

Giải:

a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)

\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\) 

\(=\dfrac{107}{220}+\dfrac{43}{20}\)

\(=\dfrac{29}{11}\)

b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\) 

\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\) 

\(=\dfrac{31}{7}:\dfrac{31}{5}\) 

\(=\dfrac{5}{7}\) 

c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\) 

\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\) 

\(=\dfrac{60}{23}:\dfrac{5}{23}\) 

\(=12\)

\(\left|3x-4\right|=\left|x-5\right|\)

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

+)\(x< \frac{4}{3}\)

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

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

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

\(\Leftrightarrow-2x-1=0\)

\(\Leftrightarrow-2x=1\)

\(\Leftrightarrow x=-\frac{1}{2}\left(tm\right)\)

+)\(5>x\ge\frac{4}{3}\)

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

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

\(\Leftrightarrow3x-4+x-5=0\)

\(\Leftrightarrow4x-9=0\)

\(\Leftrightarrow x=\frac{9}{4}\left(tm\right)\)

+)\(x\ge5\)

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

\(\Leftrightarrow\)\(\Leftrightarrow3x-4-x+5=0\)

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow x=-\frac{1}{2}\left(ktm\right)\)

| 3x - 4 | = | x - 5 |

<=> | 3x - 4 | - | x - 5 | = 0

Với x < 4/3

pt <=> -( 3x - 4 ) - [ -( x - 5 ) ] = 0

<=> -3x + 4 + x - 5 = 0

<=> -2x = 1 <=> x = -1/2 (tm)

Với 4/3 ≤ x < 5

pt <=> 3x - 4 - [ -( x - 5 ) ] = 0

<=> 3x - 4 + x - 5 = 0

<=> 4x = 9 <=> x = 9/4 (tm)

Với x ≥ 5

pt <=> 3x - 4 - ( x - 5 ) = 0

<=> 3x - 4 - x + 5 = 0 

<=> 2x = -1 <=> x = -1/2 (ktm)

Vậy tập nghiệm của phương trình là S = { -1/2 ; 9/4 }

Ta có: \(x^2-y+\frac{1}{4}=y^2-x+\frac{1}{4}=0\)

\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Rightarrow}x=y=\frac{1}{2}\)

Vậy \(x=y=\frac{1}{2}\)