Tìm x biết
5.2x-1 =40
3x+37 =118
Tính tổng
S = 1+ 9 +92 +....+ 92017
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1.
145+55=200
13+50+37=100
92+56-48=100
2.\(6x+x^2=-9\Leftrightarrow x\left(6+x\right)=-9\Leftrightarrow\hept{\begin{cases}x=-9\\6+x=-9\end{cases}\Leftrightarrow\hept{\begin{cases}x=-9\\x=-15\end{cases}}}\)
k mk nhé
1) \(\dfrac{1}{4}x-\dfrac{1}{3}=\dfrac{-5}{9}\)
\(\Rightarrow\dfrac{1}{4}x=-\dfrac{2}{9}\Rightarrow x=-\dfrac{8}{9}\)
2) \(2^{x-3}-3.2^x=-92\)
\(\Rightarrow2^x\left(2^{-3}-3\right)=-92\)
\(\Rightarrow2^x.\dfrac{-23}{9}=-92\)
\(\Rightarrow2^x=32\Rightarrow x=5\)
f) \(-\dfrac{1}{2}+\dfrac{11}{4}-\left(\dfrac{11}{4}-\dfrac{1}{2}\right)=-\dfrac{1}{2}+\dfrac{11}{4}-\dfrac{11}{4}+\dfrac{1}{2}=0\)
g) \(\left(\dfrac{92}{9}+\dfrac{13}{5}\right)-\dfrac{47}{9}=\left(\dfrac{92}{9}-\dfrac{47}{9}\right)+\dfrac{13}{5}=5+\dfrac{13}{5}=\dfrac{25}{5}+\dfrac{13}{5}=\dfrac{38}{5}\)
h) \(\left(\dfrac{44}{7}-\dfrac{32}{9}\right)-\left(\dfrac{37}{7}+\dfrac{4}{9}\right)=\dfrac{44}{7}-\dfrac{32}{9}-\dfrac{37}{7}-\dfrac{4}{9}=\left(\dfrac{44}{7}-\dfrac{37}{7}\right)-\left(\dfrac{32}{9}+\dfrac{4}{9}\right)=1-4=-3\)
Bài 5 :
S = 1 + 3 - 5 - 7 + 9 + 11 - ... - 397 - 399
S = 1 + (3 - 5 - 7 + 9) + (11 - 13 - 15 + 17) + ... + (387 - 389 - 391 + 393) + (395 - 397 - 399)
S = 1 + 0 + 0 + ... + 0 + (- 401)
S = 1 - 401
S = - 400
Bài 5
A= 1+3-5-7+9+11-13-15+...-397-399
A= ( 1+3-5-7)+( 9+11-13-15)+...+( 393+395-397-399)
A= -8 -8 -...-8
A = -8.50 ( từ 1 đến 399 có 200 số, chia làm 4 cặp)
A= -400
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
5.2x-1 = 40
=>2x-1 = 8
=>2x-1 = 23
=>x - 1 = 3
=>x = 4
3x + 37 = 118
=>3x = 81
=>3x = 34
=>x = 4
5 . 2x-1 = 40
- > 2x-1 = 40 : 5 = 8
2x-1 = 23
- > x - 1 = 3 - > x = 4
3x + 37 = 118
3x = 118 - 37 = 81
3x = 34 - > x = 4
S = 1 + 9 + 92 + ... + 92017
9S = 9 + 92 + 93 + ... + 92018
- > 9S - S = 8S = ( 9 + 92 + ... + 92018 ) - ( 1 + 9 + ... + 92017 )
8S = 92018 - 1
= > S = ( 92018 - 1 ) : 8