a , x + 109 - 213=407 [ x-129] : 2 =84
b 597 - x : 2=538 X:5x7=77
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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 \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
=> 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
\(a)\) \(-\left(x+84\right)+213=-16\)
\(\Leftrightarrow\)\(-x-84+213=-16\)
\(\Leftrightarrow\)\(x=213-84+16\)
\(\Leftrightarrow\)\(x=145\)
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) /-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: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
a. x + 109 - 213 = 407 [ x -109 ] : 2 = 84
x + 109 = 407 + 213 [ x -109 ] = 84 x2
x + 109 = 620 [ x - 109 ] = 168
x = 620 - 109 x = 168 + 109
x = 511 x = 277
-Học tốt-
a , x + 109 - 213 = 407
x + 109 = 407 + 213
x + 109 = 620
x = 620 - 109
x = 511
b , ( x - 129 ) : 2 = 84
( x - 129 ) = 84 x 2
x - 129 = 168
x = 168 + 129
x = 297