K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
6 tháng 1 2022
Câu 2:
1: \(\Leftrightarrow x\cdot\dfrac{7}{2}=\dfrac{9}{2}+3=\dfrac{15}{2}\)
hay x=15/7
2: \(\Leftrightarrow x=\dfrac{5}{2}\cdot\dfrac{8}{5}=4\)
3: \(\Leftrightarrow x=\dfrac{-11\cdot10}{5}=-11\cdot2=-22\)
4: =>2x=90
hay x=45
QP
2
9 tháng 12 2021
\(132-2x+8=46\Leftrightarrow140-2x=46\Leftrightarrow2x=94\Leftrightarrow x=47\)
9 tháng 12 2021
\(2,\)
\(a,132-2\left(x-4\right)=46\)
\(\Rightarrow-2\left(x-4\right)=-86\)
\(\Rightarrow x-4=43\)
\(\Rightarrow x=47\)
Vậy: \(x=47\)
\(b.9x-4x=6^{17}:6^{15}+48:12\)
\(\Rightarrow5x=6^2+4\)
\(\Rightarrow5x=40\)
\(\Rightarrow x=8\)
Vậy: \(x=8\)
\(c,2^{x+3}.4=64\)
\(\Rightarrow2^{x+3}=16\)
\(\Rightarrow2^{x+3}=2^4\)
\(\Rightarrow x+3=4\)
\(\Rightarrow x=1\)
Vậy: \(x=1\)
13 tháng 4 2022
a sai vì 4 ko phải số nguyên tố
b sai lun vì 2x3 =6 trong đó 2 và 3 đều là số nguyên tố mà tích là 6 là số chẵn
17 tháng 12 2023
m: \(\left(-7x+7\right)\left(2x+100\right)=0\)
=>\(\left[{}\begin{matrix}-7x+7=0\\2x+100=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}-7x=-7\\2x=-100\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-50\end{matrix}\right.\)
n: \(2x\left(x+2023\right)\left(-4x+8\right)=0\)
=>\(2\cdot x\left(x+2023\right)\cdot\left(-4\right)\left(x-2\right)=0\)
=>x(x-2)(x+2023)=0
=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x+2023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2023\end{matrix}\right.\)
o: \(-2024x\left(4x-4\right)\left(-2x+6\right)=0\)
=>\(x\left(4x-4\right)\left(-2x+6\right)=0\)
=>\(x\cdot4\left(x-1\right)\cdot\left(-2\right)\left(x-3\right)=0\)
=>x(x-1)(x-3)=0
=>\(\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)
p: -17-2x=199
=>2x=-17-199=-216
=>x=-216/2=-108
q: -24+2x=100
=>2x=100+24=124
=>\(x=\dfrac{124}{2}=62\)
r: \(119-\left(x+5\right)=-21\)
=>\(x+5=119-\left(-21\right)=119+21=140\)
=>x=140-5=135
s: \(-24+\left(-7+x\right)=45\)
=>\(\left(x-7\right)-24=45\)
=>x-31=45
=>x=45+31=76
t: \(-146-\left(-5+x\right)=-6\)
=>\(x-5=-146-\left(-6\right)=-140\)
=>x=-140+5=-135
u: \(-29+4\left(1-5x\right)=-45\)
=>4(1-5x)=-45+29=-16
=>1-5x=-4
=>5x=1+4=5
=>\(x=\dfrac{5}{5}=1\)
v: \(24-5\left(-3+2x\right)=59\)
=>\(24-5\left(2x-3\right)=59\)
=>5(2x-3)=24-59=-35
=>2x-3=-7
=>2x=-4
=>x=-4/2=-2
17 tháng 12 2023
-17-2x=199
<=>-2x=216
<=>x=-108
-24+2x=100
<=>2x=124
<=>x=62
-24+(-7+x)=45
<=>-7+x=69
<=>x=76
-146-(-5+x)=-6
<=>-146+5-x=-6
<=>x=-135
-29+4(1-5x)=-45
<=>-29+4+20x=-45
<=>20x=-20
<=>x=-1
24-5(-3+2x)=59
<=>24+15-10x=59
<=>-10x=20
<=>x=-2
cần giúp gấp vs ạ, em cảm ơn nhìu
25 tháng 12 2023
a: \(\left(-256\right)\cdot45-256\cdot56+256\)
\(=256\left(-45-56+1\right)\)
\(=256\left(-100\right)=-25600\)
b: \(\left(-2\right)^3\cdot1975\cdot\left(-4\right)\cdot\left(-5\right)^3\cdot25\)
\(=\left(-8\right)\cdot\left(-125\right)\cdot\left(-4\right)\cdot25\cdot1975\)
\(=1000\cdot\left(-100\right)\cdot1975=-197500000\)
c: \(2076-1976\cdot65-1976\cdot35\)
\(=2076-1976\left(65+35\right)\)
\(=2076-1976\cdot100=2076-197600=-195524\)
d: \(-437-25\cdot78+25\cdot178\)
\(=-437+25\left(178-78\right)\)
\(=-437+2500=2063\)
AH
Akai Haruma
Giáo viên
3 tháng 2
c/
$C=\frac{11}{2}(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{91.93})$
$=\frac{11}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{93-91}{91.93}\right)$
$=\frac{11}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{91}-\frac{1}{93}\right)$
$=\frac{11}{2}(1-\frac{1}{93})$
$=\frac{11}{2}.\frac{92}{93}=\frac{506}{93}$
AH
Akai Haruma
Giáo viên
3 tháng 2
d/
$D=5\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{675}\right)$
$=\frac{5}{2}\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{675}\right)$
$=\frac{5}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{27-25}{25.27}\right)$
$=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{27}\right)$
$=\frac{5}{2}\left(1-\frac{1}{27}\right)$
$=\frac{5}{2}.\frac{26}{27}=\frac{65}{27}$
KT
1
NT
(đề bài: tìm x biết)
1
15 tháng 12 2023
a: \(\left(x+10\right)\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x+10=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=5\end{matrix}\right.\)
b: \(\left(2x+10\right)\left(4+x\right)=0\)
=>\(\left[{}\begin{matrix}2x+10=0\\4+x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-4\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)
c: \(\left(4x+20\right)\left(12x-24\right)=0\)
=>\(\left[{}\begin{matrix}4x+20=0\\12x-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-20\\12x=24\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
d: \(\left(x-2024\right)\left(4x+4\right)=0\)
=>\(\left[{}\begin{matrix}x-2024=0\\4x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2024\\4x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-1\\x=2024\end{matrix}\right.\)
e: \(\left(2x-6\right)\left(7+x\right)=0\)
=>\(\left[{}\begin{matrix}2x-6=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\x=-7\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
g: (4x+8)(6-x)=0
=>\(\left[{}\begin{matrix}4x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x=6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\)
h: (2x+2)(4x-8)=0
=>2(x+1)*4*(x-2)=0
=>(x+1)(x-2)=0
=>\(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
i: (2x-2024)(8x-16)=0
=>\(2\left(x-1012\right)\cdot8\cdot\left(x-2\right)=0\)
=>\(\left(x-1012\right)\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-1012=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1012\\x=2\end{matrix}\right.\)
2
24 tháng 3 2022
\(\dfrac{3}{5}x-\dfrac{1}{2}x=\dfrac{3}{2}-0\)
\(\left(\dfrac{3}{5}-\dfrac{1}{2}\right)x=\dfrac{3}{2}\)
\(\dfrac{1}{10}\cdot x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}:\dfrac{1}{10}\)
\(x=15\)
O
24 tháng 3 2022
\(\dfrac{3}{5x}-\dfrac{1}{2x}=\dfrac{3}{2-0}\)
\(\dfrac{3}{5x}+\dfrac{-1}{2x}=\dfrac{3}{2}\)
\(\dfrac{3}{5}+\dfrac{-1}{2}=\dfrac{3}{2}\cdot x\)
\(\dfrac{1}{10}=\dfrac{3}{2}\cdot x\)
\(\dfrac{3}{2}\cdot x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}\div\dfrac{3}{2}\)
\(x=\dfrac{2}{30}\)
\(x=\dfrac{1}{15}\)
A