\(\dfrac{21}{11}\times\dfrac{22}{17}\times\dfrac{68}{63}\) = \(\dfrac{42}{17}\times\dfrac{68}{63}\) = \(\dfrac{8}{3}\)

 Vì \(\dfrac{21}{11}\times\dfrac{22}{17}=\dfrac{21\times22}{11\times17}=\dfrac{42}{17}\)

 Vậy:

 \(\dfrac{21}{11}\times\dfrac{22}{17}\times\dfrac{63}{68}\) =\(\dfrac{42}{17}\times\dfrac{68}{63}\) =\(\dfrac{21\times2\times17\times4}{17\times21\times3}\) =\(\dfrac{8}{3}\) = 

\(\frac{21}{11}\)x \(\frac{22}{17}\)x\(\frac{68}{63}\)

= \(\frac{42}{17}\)x\(\frac{68}{63}\)

= \(\frac{8}{3}\)

18 - x : 2 = 16

x : 2        = 18 - 16

x             = 2 x 2

x             = 4

\(\frac{21}{11}\times\frac{22}{17}\times\frac{68}{63}\)

\(=\frac{21}{11}\times\frac{22\times68}{17\times63}\)

\(=\frac{21}{11}\times\frac{22\times4}{1\times63}\)

\(=\frac{21}{11}\times\frac{88}{63}\)

\(=\frac{21\times88}{11\times63}\)

\(=\frac{1\times8}{1\times3}\)

\(=\frac{8}{3}\)

\(18-x:2=16\)

\(x:2=18-16\)

\(x:2=2\)

\(x=2\times2\)

\(x=4\)

650 x 34 + 34 x 2 x 135 + 34 x 80 

= 650 x 34 + 34 x 270 + 34 x 80

= 34 x ( 650 + 270 + 80 )

= 34 x 1000

= 34000

Tick nha 

650 x 34 + 34 x 2 x 135 + 34 x 80 

= 650 x 34 + 34 x 270 + 34 x 80

= 34 x ( 650 + 270 + 80 )

= 34 x 1000

= 34000

1/5 + 3/12 - 3/4 = 5/60 + 15/60 - 15/60 = 5/60 = 1/12

28/17 x 68/14 = 14x2x17x4/17x14 = 2x4 = 8

a)-3/10
b)8
____________

\(\dfrac{21}{11}\times\dfrac{22}{17}\times\dfrac{68}{63}:\dfrac{4}{3}=\dfrac{21}{11}\times\dfrac{22}{17}\times\dfrac{68}{63}\times\dfrac{3}{4}=\dfrac{462}{187}\times\dfrac{204}{252}=\dfrac{42}{17}\times\dfrac{17}{21}=\dfrac{2}{1}=2\)

21/11 x 22/17 x 68/63 : 4/3

=( 21/11 x 22/17) x (68/63 : 4/3)

=       42/17          x        17/21

=          2/1 = 2

ĐKXĐ: x khác -2;-1;0;1.

\(\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{3x-3}=\frac{1}{5x}\)

\((\frac{1}{x+1}-\frac{1}{5x})+(\frac{1}{x+2}+\frac{1}{3x-3})=0\)

\(\frac{4x-1}{5x(x+1)}+\frac{4x-1}{(x+2)(3x-3)}=0\)

hoặc \(4x-1=0\) hoặc \(5x(x+1)=(x+2)(3x-3)\)

Phương trình thứ nhất có nghiệm x=0,25 (t/m đkxđ)

Phương trình thứ 2 vô nghiệm.

Vậy pt có tập nghiệm S={0,25}.

Chúc bạn học tốt!

a) \(x^2-6x-17=\left(x^2-6x+9\right)-26=\left(x-3\right)^3-26\ge-26\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\)

                        \(\Leftrightarrow x-3=0\)

                        \(\Leftrightarrow x=3\)

b)\(x^2-10x=\left(x^2-10x+25\right)-25=\left(x-5\right)^2-25\ge-25\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-5\right)^2=0\)

                        \(\Leftrightarrow x-5=0\)

                        \(\Leftrightarrow x=5\)

c)\(3x^2-12x+5=3\left(x^2-4x+4\right)-7=3\left(x-2\right)^2-7\ge-7\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\)

                        \(\Leftrightarrow x-2=0\)

                        \(\Leftrightarrow x=2\)

d)\(2x^2-x+1=2\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{7}{8}=2\left(x-\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-\frac{1}{4}\right)^2=0\)

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

                        \(\Leftrightarrow x=\frac{1}{4}\)

e)\(x^2+y^2-8x+4y+27=\left(x^2-8x+16\right)+\left(y^2+4y+4\right)+7=\left(x-4\right)^2+\left(y+2\right)^2+7\ge7\forall x\)Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-4\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-4=0\\y+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=4\\y=-2\end{cases}}\)

f)\(x\left(x-6\right)=x^2-6x=\left(x^2-6x+9\right)-9=\left(x-3\right)^2-9\ge-9\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\)

                        \(\Leftrightarrow x-3=0\)

                        \(\Leftrightarrow x=3\)

h)\(\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)\left(x-5\right)\left(x-2\right)\left(x-5\right)=\left[\left(x-2\right)\left(x-5\right)\right]^2\ge0\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-2\right)\left(x-5\right)=0\)

                        \(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}}\)

                        \(\Leftrightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}\)