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5x2 - 7 = 38 => x2 = 9 => x = \(\pm\)3
Từ đây thay x vào \(\dfrac{3x-2}{4}\) để tìm y,z
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a, Có \(\dfrac{3x-2y}{7}=\dfrac{4x+3y}{5}\)
=> 5(3x-2y)=7(4x+3y)
=> 15x-10y=28x+21y
=> 15x-28x=21y+10y
=> -13x=31y
=> \(\dfrac{x}{y}=\dfrac{31}{-13}=\dfrac{-31}{13}\)
b,\(\dfrac{5x-2y}{3x+4y}=\dfrac{-3}{4}\)
=> 4(5x-2y)=-3(3x+4y)
=> 20x-8y= -9x-12y
=> 20x+9x=-12y+8y
=> 29x=-4y
=> \(\dfrac{x}{y}=\dfrac{-4}{29}\)
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a) x - 7 = -5
=> x = -5 + 7
=> x = 2
b) 128 - 3 . (x + 4) = 23
=> 3.( x + 4) = 128 - 23
=> 3.( x + 4) = 105
=> x + 4 = 105 : 3
=> x + 4 = 35
=> x = 35 - 4
=> x = 31
c) [ ( 6x - 39 ) : 7 ].4 = 12
( 6x -39) : 7 = 12 :4
=> ( 6x - 39 ): 7 = 3
=> 6x - 39 = 3 x 7
=> 6x - 39 = 21
=> 6x = 21 + 39
=> 6x = 60
=> x = 60: 6
=> x = 10
d) ( x: 3 - 4) , 5 = 15
=> x: 3 - 4 = 15 : 5
=> x : 3 - 4 = 3
=> x: 3 = 3+4
=> x: 3 = 7
=> x = 7x 3
=> x =21
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áp dụng công thức \(\frac{n\left(n-1\right)}{2}\)
<=>\(\frac{114\cdot\left(114-1\right)}{2}\)
<=> A =6441
A=1+2-3-4+5+6-7-8+...-111-112+113+114
A=1+(2-3-4+5)+(6-7-8+9)+...+(110-111-112+113)+114
A=1+ 0 +0 +.........+0+114
A=115
Ta có : \(42+\frac{3}{7}\left|3x-1\right|=12\)
=> \(\left|3x-1\right|=-70\)
TH1 : \(3x-1\ge0\left(x\ge\frac{1}{3}\right)\)
=> \(3x-1=-70\)
=> \(3x=-69\)
=> \(x=-23\) ( KTM )
TH2 : \(3x-1< 0\left(x< \frac{1}{3}\right)\)
=> \(1-3x=-70\)
=> \(-3x=-71\)
=> \(x=\frac{71}{3}\) ( KTM )
Vậy phương trình trên vô nghiệm .
oko