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\(\text{ 10-2.(4-3x)=-4 }\)
\(2\left(4-3x\right)=14\)
\(4-3x=14:2\)
\(4-3x=7\)
\(\Rightarrow3x=-3\)
\(\Rightarrow x=-1\)
\(\text{ 24:(3x-2)=-3}\)
\(3x-2=-8\)
\(\Rightarrow3x=-8+2\)
\(3x=-6\)
\(\Rightarrow x=-2\)
\(\text{-12+3(-x+7)=-18}\)
\(3\left(-x+7\right)=-6\)
\(-x+7=-6:3\)
\(-x+7=-2\)
\(\Rightarrow-x=-9\)
\(\Rightarrow x=9\)
\(\text{-45:5.(-3-2x)=3}\)
\(-9\left(-3-2x\right)=3\)
\(\Rightarrow-3-2x=\frac{-1}{3}\)
\(\Rightarrow-2x=\frac{-1}{3}+\left(-3\right)\)
\(-2x=\frac{-10}{3}\)
\(\Rightarrow x=\frac{5}{3}\)
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\(\left(2x-4\right)\left(3x+1\right)< 0\)
=> TH1: \(\begin{matrix}2x-4< 0\\3x+1>0\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}2x< 4\\3x>-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>-\dfrac{1}{3}\end{matrix}\right.\) (tm)
TH2: \(\begin{matrix}2x-4>0\\3x+1< 0\end{matrix}\)\(\Leftrightarrow\left\{{}\begin{matrix}2x>4\\3x< -1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< -\dfrac{1}{3}\end{matrix}\right.\) (vô lí)
=> \(2>x>-\dfrac{1}{3}\)
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\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{100}\right).\)
\(=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)
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\(27^3.9^4.243=\left(3^3\right)^3.\left(3^2\right)^4.3^5\)
\(=3^9.3^8.3^5\)
\(=3^{22}\)
Học tốt nha!!!
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a. ( x - 2 ) ( 6 - 2x ) = 0
=> x - 2 = 0 hoặc 6 - 2x = 0
x = 2 2x = 6
x = 3
Vậy x = 2; x =3
b. ( x + 1 )3 = -27
( x + 1 )3 = ( -3 )3
=> x + 1 = -3
x = -4
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Ta có : xy + x + y = 30
=> (xy + x) + y = 30
=> x(y + 1) + (y+ 1) = 31
=> (x + 1)(y + 1) = 31
Lại có 31 = 1.31 = 31.1
Lập bảng xét 2 trường hợp ta có :
x + 1 | 1 | 31 |
y + 1 | 31 | 1 |
x | 0 | 30 |
y | 30 | 0 |
Vậy các cặp số (x;y) thỏa mãn là : (30 ; 0) ; (0 ; 30)
Bài giải
\(xy+x+y=30\)
\(x\left(y+1\right)+y=30\)
\(x\left(y+1\right)+\left(y+1\right)=30+1\)
\(x\left(y+1\right)+\left(y+1\right)=31\)
\(\left(x+1\right)\left(y+1\right)=31\)
\(\Rightarrow\text{ }x+1\text{ , }y+1\inƯ\left(31\right)\)
Ta có bảng :
x + 1 | - 1 | 1 | - 31 | 31 |
y + 1 | - 31 | 31 | - 1 | 1 |
x | - 2 | 0 | - 32 | 30 |
y | - 32 | 30 | - 2 | 0 |
\(\text{Vậy các cặp }\left(x,y\right)=\left(-2\text{ ; }32\right),\left(0\text{ ; }30\right),\left(-32\text{ ; }-2\right),\left(30\text{ ; }0\right)\)
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a: =4*1/16+25*[(3/4:5/4)]^3:27/8
=1/4+25*(3/5:3/2)^3
=1/4+25*(2/5)^3
=1/4+8/5
=1,6+0,25=1,85
b: =8+3-1+8-8
=8+2=10
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a: \(=\dfrac{2}{7}\cdot\dfrac{-3}{4}\cdot\dfrac{4}{7}=\dfrac{-6}{49}\)
b: \(=\dfrac{3}{5}:\left(\dfrac{-5}{9}\cdot\dfrac{-3}{25}\right)=\dfrac{3}{5}:\dfrac{15}{225}=\dfrac{3}{5}\cdot15=9\)
c: \(=5+\dfrac{6}{7}-2-\dfrac{3}{8}-1-\dfrac{1}{8}=2+\dfrac{6}{7}-\dfrac{1}{2}=\dfrac{33}{14}\)
d: \(=\dfrac{-25}{12}-\dfrac{23}{12}-\dfrac{3}{2}=-4-\dfrac{3}{2}=-\dfrac{11}{2}\)
e: \(=\dfrac{-3}{5}\left(\dfrac{-1}{9}-\dfrac{5}{6}+\dfrac{5}{2}\right)=\dfrac{-3}{5}\cdot\dfrac{14}{9}=\dfrac{-42}{45}=\dfrac{-14}{15}\)
`45 : (3x - 4) = 3`
``\(\Rightarrow3x-4=45:3\)
\(\Rightarrow3x-4=15\)
\(\Rightarrow3x=15+4\)
\(\Rightarrow3x=19\)
\(\Rightarrow x=19:3\)
\(\Rightarrow x=\dfrac{19}{3}\)
Vậy \(x=\dfrac{19}{3}\)