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a) 5.(x^2-3x+1)+x.(1-5x)=x-2
\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
\(\Leftrightarrow-14x-x=-2-5\)
\(\Leftrightarrow-15x=-7\)
\(\Leftrightarrow x=\frac{7}{15}\)
b\(,3x.\left(\frac{4}{3}+1\right)-4x\left(x-2\right)=10\)
\(\Leftrightarrow4x+3x-4x^2+8x-10=0\)
\(\Leftrightarrow-4x^2+15x-10=0\)
Đề sai???
\(c,12x^2-4x\left(3x-5\right)=10x-17\)
\(\Leftrightarrow12x^2-12x^2+20x-10x=-17\)
\(\Leftrightarrow10x=-17\)
\(\Leftrightarrow x=-\frac{17}{10}\)
\(d,4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)
\(\Leftrightarrow4x^2-20x-7x^2+28x+3x^2=12\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\frac{3}{2}\)
làm khuyến mại 1 câu;
a) = 12x2 -12x2 +20x -10x +17 =0
10x = -17
x = -17/10
\(a,5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)
\(\Leftrightarrow5x^2-15x+5+x-5x^2-x+2=0\)
\(\Leftrightarrow-15x+7=0\)
\(\Leftrightarrow-15x=-7\)
\(\Leftrightarrow x=-\frac{7}{-15}\)
\(\Leftrightarrow x=\frac{7}{15}\)
Trả lời:
a, ( x2 - 4x + 16 )( x + 4 ) - x ( x + 1 )( x + 2 ) + 3x2 = 0
<=> x3 + 4x2 - 4x2 - 16x + 16x + 64 - x ( x2 + 3x + 2 ) + 3x2 = 0
<=> x3 + 64 - x3 - 3x2 - 2x + 3x2 = 0
<=> 64 - 2x = 0
<=> 2x = 64
<=> x = 32
Vậy x = 32 là nghiệm của pt.
b, ( 8x + 2 )( 1 - 3x ) + ( 6x - 1 )( 4x - 10 ) = - 50
<=> 8x - 24x2 + 2 - 6x + 24x2 - 60x - 4x + 10 = - 50
<=> - 62x + 12 = - 50
<=> - 62x = - 62
<=> x = 1
Vậy x = 1 là nghiệm của pt.
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)
\(\Leftrightarrow-62x=-92\)
hay \(x=\dfrac{46}{31}\)
b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)
\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)
\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)
\(\Leftrightarrow8x^2+49x-15=0\)
\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)
$ a/ 12x(x – 5) – 3x(4x - 10) = 120$
`<=>12x^2-60x-12x^2+30x=120`
`<=>-30x=120`
`<=>x=-4`
Vậy `x=-4`
$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$
`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`
`<=>-6x^2+26x=112-6x^2-2x`
`<=>28x=112`
`<=>x=4`
Vậy `x=4`
$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$
`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`
`<=>-32x-18x^2=154+45x-18x^2`
`<=>77x=-154`
`<=>x=-2`
Vậy `x=-2`
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
4x2 + 3x - 4x2 + 8x - 10 = 0
11x - 10 = 0
x = 10/11
\(3x\left(\dfrac{4}{3}x+1\right)-4x\left(x-2\right)=10\)
\(\Leftrightarrow4x^2+3x-4x^2+8x=10\)
\(\Leftrightarrow x=\dfrac{10}{11}\)