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440 + 2 . ( 125 - x ) = 546
2 . ( 125 - x ) = 546 - 440
2 . ( 125 - x ) = 106
125 - x = 106 : 2
125 - x = 53
x = 125 - 53
x = 72
Vậy x = 72
440 + 2. ( 125 - x ) = 546
2 ( 125 - x ) = 546 - 440
2 ( 125 - x ) = 106
125 - x = 106 ÷ 2
125 - x = 53
x = 125 - 53
x = 72
2 . (125 - x) = 546 - 440
2. (125 - x) = 106
(125 - x) = 106 : 2
125 - x = 53
x = 125 - 53
x = 72
a, 440 + 2 . (125 - x) = 546
2 . (125 - x) = 546 - 440
2 . (125 - x) = 106
125 - x = 106 : 2
125 - x = 53
x = 125 - 53
x = 72
Ta có\(\frac{-543}{546}=\frac{3-546}{546}=\frac{3}{546}-1\)
\(\frac{-789}{792}=\frac{3-792}{792}=\frac{3}{792}-1\)
Vì \(546< 792\Rightarrow\frac{1}{546}>\frac{1}{792}\Rightarrow\frac{3}{546}>\frac{3}{792}\Rightarrow\frac{3}{546}-1>\frac{3}{792}-1\)
Do đó \(\frac{-543}{546}>\frac{-789}{792}\)
=440-20x(30-(2x2))
=440-20x(30-4)
=440-20x26
=440-520
=-80
440-20x{30-[2x(5-3)]}
=440-20x{30-[2x2])
=440-20x{30-4}
=440-20x26
=440-520=-80
a)\(\left(x-2,5\right)^2=\frac{4}{9}\\ \left(x-\frac{5}{2}\right)^2=\left(\pm\frac{2}{3}\right)^2\\\Leftrightarrow\left\{{}\begin{matrix}x-\frac{5}{2}=\frac{2}{3}\\x-\frac{5}{2}=\frac{-2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{19}{6}\\x=\frac{11}{6}\end{matrix}\right. \)
vậy....
b)\(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\\ \left(2x+\frac{1}{3}\right)^3=\left(\frac{-2}{3}\right)^3\\ 2x+\frac{1}{3}=\frac{-2}{3}\\ x=\frac{-1}{2}\)
vậy...
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Bài 1:
\(a.\left(-356+57\right)-\left(27-356\right)=-356+57-27+356=\left(-356+356\right)+\left(57-27\right)=30\) \(b.125.\left(-24+24.225\right)=125.\left(-24+5400\right)=125.\left(-24\right)+125.5400=-3000+675000=672000\)
\(c.26.\left(-125\right)-125.\left(-36\right)=-125.\left(26-36\right)=-125.\left(-10\right)=1250\)
Bài 2:
\(a.\left(2x-4\right)^2=0\)
\(\Rightarrow2x-4=0\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(b.\frac{x+5}{x+3}=\frac{x+3+2}{x+3}=\frac{x+3}{x+3}+\frac{2}{x+3}=1+\frac{2}{x+3}\)
Để (x+5) chia hết cho (x+3) thì 2 phải chia hết cho (x+3)
\(\Rightarrow x+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(x+3=1\Rightarrow x=-2\)
\(x+3=-1\Rightarrow x=-4\)
\(x+3=2\Rightarrow x=-1\)
\(x+3=-2\Rightarrow x=-5\)
Vậy \(x\in\left\{-2;-4;-1;-5\right\}\)
Bài 2:
a)\(\left(2x-4\right)^2=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\)
b)\(\frac{x+5}{x+3}=\frac{x+3+2}{x+3}=\frac{x+3}{x+3}+\frac{2}{x+3}=1+\frac{2}{x+3}\in Z\)
Suy ra \(2⋮x+3\Rightarrow x+3\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow x\in\left\{-2;-4;-1;-5\right\}\)