x+4x+7x+.......+58x=596
tìm x
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<=> x*(1+4+7+...+58)=595
<=> x*((58-1):3+1)*(58+1)=595
<=> x*590=595
<=> x=595:590
<=> x=119/118
Vậy x=119/118
\(\dfrac{4x-5}{7-4x}=\dfrac{7x-28}{26-7x}\)
\(\Leftrightarrow104x-28x^2-130+35x=49x-28x^2-196+112x\)
\(\Leftrightarrow-22x=-66\)
hay x=3
\(\Rightarrow\left(4x-5\right)\left(26-7x\right)=\left(7-4x\right)\left(7x-28\right)\\ \Rightarrow104x-28x^2-130+35x=49x-196-28x^2+112x\\ \Rightarrow-22x=-66\Rightarrow x=3\)
a) \(4x.\left(7x-5\right)-7x\left(4x-2\right)=-12\)
\(\Rightarrow28x^2-20x-28x^2+14x=-12\)
\(\Rightarrow x\left(14-20\right)=-12\)
\(\Rightarrow-6x=12\)
\(\Rightarrow x=-2\)
b) \(3x.\left(2x-4\right)-6x\left(x+5\right)=x-1\)
\(\Rightarrow3x.\left[\left(2x-4\right)-2.\left(x+5\right)\right]=x-1\)
\(\Rightarrow3x.\left(2x-4-2x-10\right)=x-1\)
\(\Rightarrow-42x=x-1\)
\(\Rightarrow-42x-x=-1\)
\(\Rightarrow-43x=-1\)
\(\Rightarrow x=\dfrac{1}{43}\)
a, \(4x\left(7x-5\right)-7x\left(4x-2\right)=-12\)
\(\Rightarrow28x^2-20x-28x^2+14=-12\)
\(\Rightarrow-20x=-12-14\)
\(\Rightarrow-20x=-26\Rightarrow x=1,3\)
Vậy \(x=1,3\)
b, \(3x\left(2x-4\right)-6x\left(x+5\right)=x-1\)
\(\Rightarrow6x^2-12x-6x^2-30-x=-1\)
\(\Rightarrow-13x=-1+30\)
\(\Rightarrow-13x=29\Rightarrow x=\dfrac{-29}{13}\)
Vậy \(x=\dfrac{-29}{13}\)
Chúc bạn học tốt!!!
A = - 3\(x\).(\(x-5\)) + 3(\(x^2\) - 4\(x\)) - 3\(x\) - 10
A = - 3\(x^2\) + 15\(x\) + 3\(x^2\) - 12\(x\) - 3\(x\) - 10
A = (- 3\(x^2\) + 3\(x^2\)) + (15\(x\) - 12\(x\) - 3\(x\)) - 10
A = 0 + (3\(x-3x\)) - 10
A = 0 - 10
A = - 10
Ta có : x + 4x + 7x + ...... + 58x = 596
<=> x(1 + 4 + 7 + ...... + 58) = 596
=> x 590 = 596
=> x = 596/590
x + 4x + 7x + ..... + 58x = 596
( x + 4x + 7x + ...+58x) = 596
( x + 58x) x [( 58x - x ) : 3 + 1 ] : 2 = 596
x = 590 = 596
x = 596 : 590
x = 596/590