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a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
-x + 5 + 2x = 4 - x
-1x + 5 + 2x = 4 - x
x( - 1 + 2 ) + 5 = 4 - x
x + 5 = 4 - x
=> x + 5 - 4 + x = 0
2x + ( 5 - 4 ) = 0
2x + 1 = 0
2x = 1
x = 1/2
a) \(\frac{3}{4}x-\frac{1}{4}=2\left(x-3\right)+\frac{1}{4}x\)
\(\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\frac{3}{4}x-2x-\frac{1}{4}x=\frac{1}{4}-6\)
\(x\left(\frac{3}{4}-2-\frac{1}{4}\right)=-\frac{23}{4}\)
\(-\frac{3}{2}x=-\frac{23}{4}\)
\(x=-\frac{23}{4}\div\left(-\frac{3}{2}\right)\)
\(x=\frac{23}{6}\)
Subject: Homeworkquestion Name: Vicki Charron Question: I have two university degrees (somewhat dusty from age, I admit) and I'm embarrassed to admit I can't figure out how to help my grade 5 daughter determine the answer to this question... how can you calculate the total of the numbers one through fifty, without adding up the individual numbers? Hi Vicky,If you have seen this before you probably saw it as a "formula" and promply forgot it. I'm not surprised, I don't formulas. The idea here is very simple. Look at the list of numbers and imagine that you are going to add them, starting at both ends and working to the middle.1 + 50 = 51 2 + 49 = 51 3 + 48 = 51 4 + 47 = 51 Altogether you have 51, twenty-five times so the total of the integers from one to fifty is 51 x 25 = 1275 Cheers ~Hok tốt~ Có chị nè đang rảnh ! |
a: \(575-\left(2x+70\right)=445\)
=>\(2x+70=575-445=130\)
=>\(2x=130-70=60\)
=>x=60/2=30
b: \(575-2\left(x+70\right)=445\)
=>\(2\left(x+70\right)=575-445=130\)
=>x+70=130/2=65
=>x=65-70=-5
c: \(x^5=32\)
=>\(x^5=2^5\)
=>x=2
d: \(\left(3x-1\right)^3=8\)
=>\(\left(3x-1\right)^3=2^3\)
=>3x-1=2
=>3x=3
=>\(x=\dfrac{3}{3}=1\)
e: \(\left(x-2\right)^3=27\)
=>\(\left(x-2\right)^3=3^3\)
=>x-2=3
=>x=5
f: \(\left(2x-3\right)^2=9\)
=>\(\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
g: \(2x+5=3^4:3^2\)
=>\(2x+5=3^2\)
=>2x+5=9
=>2x=9-5=4
=>x=4/2=2
h: \(\left(4x-5^2\right)\cdot7^3=7^4\)
=>\(4x-25=\dfrac{7^4}{7^3}=7\)
=>4x=25+7=32
=>\(x=\dfrac{32}{4}=8\)
a: =>5x+25-3x+6=25+18
=>2x+41=43
=>2x=2
=>x=1
b: =>4x+8=3x+3+17
=>4x+8=3x+20
=>x=12
a: =>5x+25-3x+6=25+18
=>2x+41=43
=>2x=2
=>x=1
b: =>4x+8=3x+3+17
=>4x+8=3x+20
=>x=12
52x - 3 - 2.52 = 52.3
52x - 3 = 52.3 + 52.2
52x - 3 = 52.(3 + 2)
52x - 3 = 52.5
52x - 3 = 53
2x - 3 = 3
2x = 3 + 3 = 6
x = 6 : 2 = 3
Vậy x = 3