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câu 1; \(-154+\left(x-9-18\right)=40\)
\(\Leftrightarrow-154+x-9-18=40\)
\(\Leftrightarrow x=40+154+9+18\)
\(\Leftrightarrow x=221\)
Câu 2: \(\left|9-x\right|=64+\left(-7\right)\)
\(\Leftrightarrow\left|9-x\right|=57\)
\(\Leftrightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=9-57\\x=-57-9\end{cases}\Rightarrow}\orbr{\begin{cases}x=-48\\x=-66\end{cases}}}\)
Vậy...
hok tốt!!
a) (102 − 15) − (15 − x) = 6
102 − 15 − 15 + x = 6
x = 6 − 102 + 15 + 15
x = −66.
b) −154 + (x − 9 − 18) = 40
−154 + x − 9 – 18 = 40
x = 40 + 154 + 9 + 18
x = 221.
a, -154+(x-9-18)=40
x-27=40-(-154)
x-27=194
x=194+27
x=221
b,\(\left|9-x\right|\)=64+(-7)
\(\left|9-x\right|\)=57
TH1:
9-x=57
x=9-57
x=-48
TH2:
9-x=-57
x=9-(-57)
x=66
a, -154+(x-9-18)=40
x-27=40-(-154)
x-27=194
x=194+27
x=221
b,|9−x||9−x|=64+(-7)
|9−x||9−x|=57
TH1:
9-x=57
x=9-57
x=-48
TH2:
9-x=-57
x=9-(-57)
x=66
\(a,159-\left(25-x\right)=43\)
\(\Leftrightarrow25-x=195-43\)
\(\Leftrightarrow25-x=152\)
\(\Leftrightarrow x=25-152\)
\(\Leftrightarrow x=-127\)
\(b,\left(79-x\right)-43=-\left(17-25\right)\)
\(\Leftrightarrow\left(79-x\right)-43=8\)
\(\Leftrightarrow79-x=8+43\)
\(\Leftrightarrow79-x=51\)
\(\Leftrightarrow x=79-51\Rightarrow x=28\)
\(-\left(-x+13-142\right)+18=55\)
\(\Leftrightarrow-\left(-x+13-142\right)=55-18\)
\(\Leftrightarrow-\left(-x+13-142\right)=37\)
\(\Leftrightarrow x+13-142=37\)
\(\Leftrightarrow x=37+142-13\)
\(\Leftrightarrow x=166\)
\(d,\left(102-15\right)-\left(15-x\right)=6\)
\(\Leftrightarrow87-\left(15-x\right)=6\)
\(\Leftrightarrow15-x=87-6\)
\(\Leftrightarrow15-x=81\Rightarrow x=15-81\)
\(\Leftrightarrow x=-66\)
200 - 18: (42 : 3 x X - 1) - 28 = 154
200 - 18 :(14 x X -1) = 182
18 : (14X - 1) = 182
18: 14X - 18 = 182
18 : 14X = 200
14X = 0,09
X = \(\frac{9}{1400}\)
Ta có 2/40 + 2/88 + 2/154 + ... + 2/x( x + 3) = 202
=> 2/5 x 8 + 2/8 x 11 + ... + 2/x( x + 3 ) = 202
=> 1/5 x 8 + 1/8 x 11 + ... + 1/x( x + 3 ) = 202 : 2
=> 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/x - 1/x + 3 = 101
=> 1/5 - 1/x + 3 = 101
=> 1/x + 3 = 1/5 - 101
=> 1/X + 3 = 504/5
=> 504(x + 3 ) = 5
\(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+...+\frac{2}{x\left(x+3\right)}=\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}\)
\(=\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Từ đó ta có:
\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(x+3=308\)
x = 305
\(\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(3\left(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{3\cdot101}{1540}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{308}{1540}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)
\(x+3=308\)
\(x=308-3=305\)
\(A=\frac{9}{10}+\frac{39}{40}+\frac{87}{88}+\frac{153}{154}+\frac{237}{238}+\frac{339}{340}=\frac{117}{20}\)
\(suyra:A<1\)