nhấn vào đúng 0 sẽ ra

x=11;y=3;z=0

a) Để (x + 2) . (3 - x) = 0 

=> x + 2 = 0  hoặc 

     3 - x = 0

Ta có :

Với \(\hept{\begin{cases}x+2=0\\3-x=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\x=3\end{cases}}}\)

truong hop1;x+2=0suy ra x=-2                 truong hop2;3-x=0suy ra x=3

(x + 2).(3 - x) = 0

x + 2 = 0

hoặc

3 - x = 0

x =0 - 2

hoặc

x =3 - 0

x = -2

hoặc 

x = 3

Ta thấy : VT >= 0

Dấu "=" xảy ra <=> 3x-5=0 ; y^2-1=0 ; x-z=0

<=> x=z=5/3 ; y=-1 hoặc x=z=5/3 ; y=1

Vậy .........

Tk mk nha

\(\left(3x-5\right)^{2016}\ge0\)

\(\left(y^2-1\right)^{2018}\ge0\)

\(\left(x-z\right)^{2100}\ge0\)

suy ra \(\left(3x-5\right)^{2016}+\left(y^2-1\right)^{2018}+\left(x-z\right)^{2100}\ge0\)

Dấu bằng xảy ra khi và chỉ khi

\(\hept{\begin{cases}\left(3x-5\right)^{2016}=0\\\left(y^2-1\right)^{2018}=0\\\left(x-z\right)^{2100}=0\end{cases}}\)

\(\hept{\begin{cases}3x-5=0\\y^2-1=0\\x-z=0\end{cases}}\)

\(\hept{\begin{cases}3x=5\\y^2=1\\x=z\end{cases}}\)

\(\hept{\begin{cases}x=\frac{5}{3}\\y=\pm1\\z=\frac{5}{3}\end{cases}}\)

T I C K nha

/2/x - ( - 11 ) = 21

 2.x - ( - 11 ) = 21

       2.x       = 21 + ( - 11 )

          2.x    = 10

              x  = 10 : 2

              x = 5

Vậy x = 5

\(\Leftrightarrow\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)-x+\dfrac{221}{231}=\dfrac{4}{3}\)

\(\Leftrightarrow\dfrac{10}{231}+\dfrac{221}{231}-x=\dfrac{4}{3}\)

=>1-x=4/3

hay x=-1/3

bn l-ike cho mk trước đã

Ta có : A=20/11×13 + 20/13×15 +20/15×17+...+20/53×55

A = 10 ×( 2/11×13+2/13×15+...12/53×55)

A = 10 ×(1/11-1/13+1/13-1/15+1/15-1/17+...+1/53-1/55)

A = 10 × (1/11-1/55)

A =10 × 4/55

A = 8/11

Ta có:

\(\frac{2}{7}x-\frac{1-x}{3}=\frac{-11}{21}\)

\(\frac{2}{7}x-\left(\frac{1}{3}-\frac{1}{3}x\right)=\frac{-11}{21}\)

\(\frac{2}{7}x+\frac{1}{3}x-\frac{1}{3}=-\frac{11}{21}\)

\(\left(\frac{6}{21}+\frac{7}{21}\right)x=-\frac{11}{21}+\frac{7}{21}\)

\(\frac{13}{21}x=-\frac{4}{21}\)

\(x=-\frac{4}{21}:\frac{13}{21}=-\frac{4}{21}.\frac{21}{13}=-\frac{4}{13}\)

Vậy \(x=-\frac{4}{13}\)

\(\frac{2}{7}x-\frac{1-x}{3}=\frac{-11}{21}\Leftrightarrow6x-7+7x=-11\Leftrightarrow13x=-4\Leftrightarrow x=\frac{-4}{13}\)

nhưng mình mới học lớp 6

a) \(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

x = 1

b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\) ( nhân cho cả tử và mẫu của các số hạng trên ( ngoại trừ 2/x.(x+1) ) là 2)

\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

=> x + 1 = 18

x = 17

\(a,x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Rightarrow x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)

\(\Rightarrow x=1\)

\(b,\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{18}=\frac{1}{x+1}\)

\(\Rightarrow x+1=18\Leftrightarrow x=17\)