sửa đề câu a  và câu b  nhá  , mik nghĩ đề như này :

  \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

 \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

= \(\frac{1}{1}-\frac{1}{215}\)

\(=\frac{214}{215}\)

b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)

    \(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)

\(A\cdot2=\frac{214}{215}\)

\(A=\frac{214}{215}:2\)

\(A=\frac{107}{215}\)

@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều

trả lời hiền thương đề bài của bạn ấy là đúm gòi nha

=> 213 + (x + 84).(-2 + 1) = 0

=> 213 - (x + 84) = 0

=> 213 - x - 84 = 0

=> 129 - x = 0

=> x = 129

a) 129 048 : X = 456                b) X x 97 = 77 794

   X = 129 048 : 456                      X = 77 794 : 97

   X =  283                                       X = 802    

a, x = 129048 : 456 

x = 283

b, x = 77794 : 97 

x = 802

\(a)\) \(-\left(x+84\right)+213=-16\)

\(\Leftrightarrow\)\(-x-84+213=-16\)

\(\Leftrightarrow\)\(x=213-84+16\)

\(\Leftrightarrow\)\(x=145\)

Vậy \(x=145\)

\(b)\) \(\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)

\(\Leftrightarrow\)\(\left(x-1\right)^2=\left|-1\right|\)

\(\Leftrightarrow\)\(\left(x-1\right)^2=1\)

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

Vậy \(x=0\) hoặc \(x=2\)

Chúc bạn học tốt ~ 

a) \(-\left(x+84\right)+213=-16\)

                   \(-\left(x+84\right)=-16-213\)

                   \(-\left(x+84\right)=-229\)

\(\Rightarrow x+84=229\)

\(\Rightarrow x=229-84=145\)

Vậy \(x=145\)

b) \(\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)

    \(\left(x-1\right)^2=\left|\frac{-1}{4}-\frac{3}{4}\right|\)

    \(\left(x-1\right)^2=\left|-1\right|\)

    \(\left(x-1\right)^2=1\)

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

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

Vậy \(x\in\left\{0;2\right\}\)

a) \(-\left(x+84\right)+213=-16\)

\(-x-(84-213)=-16\)

\(-x+129=-16\)

\(-x=-16-129\)

\(-x=-145\)

\(x=145\)

b) \(\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)

\(\left(x-1\right)^2=\left|-1\right|\)

\(\left(x-1\right)^2=1\)

\(\Rightarrow\left(x-1\right)^2=1^2\)

\(\Rightarrow x-1=1\Rightarrow x=2\)

CHÚC BN HỌC TỐT!!

a) /-9-x/=12+5=17

=> -9-x=17 hoac -9-x=-17

=> x=-26 hoac x=8

b) x-45=387-416

=> x-45=-29

=> x=-29+45=16

c) -x-84+213=-16

=> -x+129=-16

=> -x=-16-129=-145

=>x=145

d)11-x-84=97

=> -x-73=97

=> -x=97+73=170

=> x=-170

2) 6a+1⋮3a-1

=>2(3a-1)+3⋮3a-1

Vi 2(3a-1)⋮3a-1 nen 3⋮3a-1

=> 3a-1 ∈U(3)=(1,-1,3,-3)

=> 3a∈(2,0,4,-2)

=> a∈(2/3,0,4/3,-2/4)

a, - ( x + 84 ) + 213 = -16

=> - ( x + 84 ) = -16 - 213

=>  - ( x + 84 ) =-229

=> x + 84 = 229

=> x = 145

b.  x + ( -35 ) = 18

=> x = 18 - ( -35 )

=> x = 18 + 35

=> x = 53

c. -2x - ( - 17 ) = 15

=> -2x = 15 + ( -17)

=> -2x = -2 

=> x = 1

\(a,-\left(x+84\right)+213=-16\)

                             \(x+84=213+16\)

                             \(x+84=229\)

                                        \(x=229-84\) 

                                        \(x=145\)

\(b,x+\left(-35\right)=18\)

                        \(x=18+35\)

                        \(x=53\)

\(c,-2x-\left(-17\right)=15\)

               \(-2x+17=15\)

                              \(2x=17-15\)         

                              \(2x=2\)

                                 \(x=1\)

Bài 1:

Ta có: \(4-2\left(x+1\right)=2\)

\(\Leftrightarrow2\left(x+1\right)=2\)

\(\Leftrightarrow x+1=1\)

hay x=0

Bài 2: 

Ta có: \(\left|2x-3\right|-1=2\)

\(\Leftrightarrow\left|2x-3\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

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