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a, 3(6x−5)(4x+1)−(8x+3)(9x−2)=203
⇒3(24x2+6x−20x−5)−(72x2−16x+27x−6)=203
⇒72x2−42x−15−72x2−11x+6=203
⇒−53x=203−6+15=212
nhầm òi
3(6x−5)(4x+1)−(8x+3)(9x−2)=203
⇒3(24x2+6x−20x−5)−(72x2−16x+27x−6)=203
⇒72x2−42x−15−72x2−11x+6=203
⇒−53x=203−6+15=212
⇒x=−4
a ) \(4x\left(5x+2\right)-\left(10x-3\right)\left(2x+7\right)=133\)
\(\Leftrightarrow20x^2+8x-\left(20x^2-6x+70x-21\right)=133\)
\(\Leftrightarrow20x^2+8x-20x^2+6x-70x+21=133\)
\(\Leftrightarrow-56x+21=133\)
\(\Leftrightarrow-56x=112\)
\(\Leftrightarrow x=-2\)
Vậy \(x=-2\)
b ) \(3\left(6x-5\right)\left(4x+1\right)-\left(8x+3\right)\left(9x-2\right)=203\)
\(\Leftrightarrow\left(18x-15\right)\left(4x+1\right)-\left(72x^2+27x-16x-6\right)=203\)
\(\Leftrightarrow72x^2-60x+18x-15-72x^2-27x+16x+6=203\)
\(\Leftrightarrow\left(72x^2-72x^2\right)+\left(18x+16x-60x-27x\right)-\left(15-6\right)=203\)
\(\Leftrightarrow-53x-9=203\)
\(\Leftrightarrow-53x=212\)
\(\Leftrightarrow x=-4\)
Vậy \(x=-4\)
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
a: \(=\dfrac{4x\left(3x+1\right)}{\left(3x+1\right)\left(3x-1\right)}=\dfrac{4x}{3x-1}\)
b: \(=\dfrac{2\left(4x^2-4x+1\right)}{4x-30+2x}=\dfrac{4\left(2x-1\right)^2}{6x-30}=\dfrac{2\left(2x-1\right)^2}{3\left(x-5\right)}\)
d: \(=\dfrac{x\left(x-6\right)}{2\left(x-6\right)\left(x+6\right)}=\dfrac{x}{2x+12}\)
1. a) \(8x^3-32x=8x\left(x^2-4\right)=8x\left(x-4\right)\left(x+4\right)\)
b) \(y^3+64+\left(y+4\right)\left(y-16\right)=\left(y^3+4^3\right)+\left(y+4\right)\left(y-16\right)\)
\(=\left(y+4\right)\left(y^2-4y+16\right)+\left(y+4\right)\left(y-16\right)=\left(y+4\right)\left(y^2-4y+16+y-16\right)\)
\(=\left(y-4\right)\left(y^2-3y\right)=\left(y-4\right)y\left(y-3\right)\)
2) a)
\(4x^3-9x=0\)
\(\Leftrightarrow x\left(4x^2-9\right)=0\)
\(\Leftrightarrow x\left(2x+3\right)\left(2x-3\right)=0\)
<=> x=0 hoặc 2x+3=0 hoặc 2x-3=0
<=> x=0 hoặc x=-3/2 hoặc x=3/2
b) \(A=x^3-9x^2+27x-27=x^3-3.x^2.3+3.x.3^2-3^3=\left(x-3\right)^3\)
Tại x=203
A=(203-3)3=2003
Bài 1 :
a) \(8x^3-32x\)
\(=8x\left(x^2-4\right)\)
\(=8x\left(x-2\right)\left(x+2\right)\)
b) \(y^3+64+\left(y+4\right)\left(y-16\right)\)
\(=\left(y^3+4^3\right)+\left(y+4\right)\left(y-16\right)\)
\(=\left(y+4\right)\left(y^2-4y+16\right)+\left(y+4\right)\left(y-16\right)\)
\(=\left(y+4\right)\left(y^2-4x+16+y-16\right)\)
\(=\left(y+4\right)\left(y^2+y-4x\right)\)
Bài 2 :
a) \(4x^3-9x=0\)
\(x\left(4x^2-9\right)=0\)
\(x\left[\left(2x\right)^2-3^2\right]=0\)
\(x\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\2x-3=0\\2x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{3}{2}\\x=\frac{-3}{2}\end{cases}}}\)
P.s: ở trên dùng ngoặc vuông nhé
b) \(A=x^3-9x^2+27x-27\)
\(A=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\)
\(A=\left(x-3\right)^3\)
Thay x = 203 vào biểu thức ta có :
\(A=\left(203-3\right)^3\)
\(A=200^3\)
\(A=8000000\)
làm khuyến mại 1 câu;
a) = 12x2 -12x2 +20x -10x +17 =0
10x = -17
x = -17/10
a)\(A=x^5-36x^4+37x^3-69x^2+34x+15\)
=\(x^5-35x^4-x^4+35x^3+2x^2-70x^2+x^2-35x+x+15\)
=\(\left(x^4-x^3+x^2+x\right)\left(x-35\right)+x+15\)
=0+35+15=50(do x=35)
b, \(3\left(6x-5\right)\left(4x+1\right)-\left(8x+3\right)\left(9x-2\right)=203\)
\(\Rightarrow3\left(24x^2+6x-20x-5\right)-\left(72x^2-16x+27x-6\right)=203\)
\(\Rightarrow72x^2-42x-15-72x^2-11x+6=203\)
\(\Rightarrow-53x=203-6+15=212\)
\(\Rightarrow x=-4\)
Chúc bạn học tốt!!!