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* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
\(\left(\dfrac{3x}{4}+5\right)-\left(\dfrac{2x}{3}-4\right)-\left(\dfrac{x}{6}+1\right)=\left(\dfrac{1}{3}+4\right)-\left(\dfrac{1}{3}x-3\right)\)
\(\Leftrightarrow\dfrac{3x}{4}-\dfrac{2x}{3}-\dfrac{x}{6}+5+4-1=\dfrac{13}{3}-\dfrac{1}{3}x+9\)
\(\Leftrightarrow\dfrac{9x-8x-2x}{12}+8=\dfrac{13-x}{3}+\dfrac{27}{3}\)
\(\Leftrightarrow\dfrac{-x}{12}+\dfrac{96}{12}=\dfrac{40-x}{3}\Leftrightarrow\dfrac{96-x}{12}=\dfrac{160-4x}{12}\)
\(\Rightarrow96-160=-4x+x\Leftrightarrow-64=-3x\Leftrightarrow x=\dfrac{64}{3}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
3/4x + 5-(2/3x-4)-(1/6+1)=(1/3x+4)-(1/3x-3)
=3/4x+5-2/3x-4-1/6x+1=1/3x+4-1/3x-3
=-1/12=7
x=84
Đ/S...
a: =>2-12x=8+2x
=>-14x=6
=>x=-3/7
b: \(\dfrac{1}{3}-\dfrac{2}{5}x=\dfrac{1}{10}+\dfrac{4}{7}x\)
=>-2/5x-4/7x=1/10-1/3
=>-14/35x-20/35x=3/30-10/30
=>-34/35x=-7/30
=>x=7/30:34/35=49/204
a) \(x^2+2x+3=0\)
\(\Rightarrow x^2+2x+3-3=0-3\)
\(\Rightarrow x^2+2x=-3\)
\(\Rightarrow x^2+2x+1=-3+1\)
\(\Rightarrow\left(x+1\right)^2=-2\)
Điều này là vô lý vì bình phương của 1 số luôn lớn hơn hoặc bằng 0 mà -2 < 0.
Vậy đa thức vô nghiệm.
[ 6 x (\(\frac{-1}{3}\)) 3 - 3 x ( \(\frac{-1}{3}\)) + 1 ] : ( \(\frac{-1}{3}\) - 1 )
= [ 6 x \(\frac{-1}{3}\)3 - 3 x \(\frac{-1}{3}\)+ 1 ] : ( \(\frac{-1}{3}\)- 1 )
= [ 6 x \(\frac{-1}{27}\)- ( - 1 ) + 1 ] : ( \(\frac{-1}{3}\)- \(\frac{3}{3}\))
= [ \(\frac{-2}{9}\)- \(\frac{-9}{9}\)+ \(\frac{9}{9}\)] : \(\frac{-4}{3}\)
= \(\frac{16}{9}\)x \(\frac{-3}{4}\)
= \(\frac{4}{3}\)x \(\frac{-1}{1}\)= \(\frac{-4}{3}\)= \(1\frac{-1}{3}\).