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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
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.
( x2 - 4x + 16 )( x + 4 ) - x( x + 1 )( x + 2 ) + 3x2 = 0
<=> x3 + 43 - x( x2 + 3x + 2 ) + 3x2 = 0
<=> x3 + 64 - x3 - 3x2 - 2x + 3x2 = 0
<=> 64 - 2x = 0
<=> 2x = 64
<=> x = 32
( 8x + 2 )( 1 - 3x ) + ( 6x - 1 )( 4x - 10 ) = -50
<=> 2x - 24x2 + 2 + 24x2 - 64x + 10 = -50
<=> -62x + 12 = -50
<=> -62x = -62
<=> x = 1
a: Ta có: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)
\(\Leftrightarrow-2x+1-x-2=8\)
\(\Leftrightarrow-3x=9\)
hay x=-3
b: Ta có: \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)
\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1
1)(x2-4x+16)(x+4)-x(x+1)(x+2)+3x2=0
\(\Rightarrow\)(x3+64)-x(x2+2x+x+2)+3x2=0
\(\Rightarrow\)x3+64-x3-2x2-x2-2x+3x2=0
\(\Rightarrow\)-2x+64=0
\(\Rightarrow\)-2x=-64
\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)
\(\Rightarrow x=32\)
2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50
\(\Rightarrow\)8x-24x2+2-6x+24x2-60x-4x+10=50
\(\Rightarrow\)-62x+12=50
\(\Rightarrow\)-62x=50-12
\(\Rightarrow\)-62x=38
\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)
a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0
TH2 3x-4=0 <=>x=4/3
KL:.....
b) (x+3)(x−1)+2x(x+3)=0.
<=> (x+3)(x-1+2x)=0
TH1: x+3=0 <=> x=-3
TH2 x-1=0 <=> x=1
KL:.....
c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 \)
KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)
KL:....
a) \(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(9x^2+6x+1=0\)
\(\Leftrightarrow\left(3x+1\right)^2=0\)
\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
d) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)
\(\Leftrightarrow x^2+x+3x+3-x^2+5x=11\)
\(\Leftrightarrow9x+3=11\)
\(\Leftrightarrow9x=11-3\)
\(\Leftrightarrow9x=8\)
\(\Leftrightarrow x=\dfrac{8}{9}\)
b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow\left(8x-24x^2+2-6x\right)+\left(24x^2-60x-4x+10\right)=-50\)
\(\Leftrightarrow2x-24x^2+2+24x^2-64x+10=-50\)
\(\Leftrightarrow-62x+12=-50\)
\(\Leftrightarrow-62x=-50-12\)
\(\Leftrightarrow-62x=-62\)
\(\Leftrightarrow x=\dfrac{-62}{-62}\)
\(\Leftrightarrow x=1\)
a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)
\(x^2+x+3x+3-x^2+5x=11\)
\(x+8x+3=11\)
\(x+8x=8\)
\(x\left(8+1\right)=8\)
\(x=\dfrac{8}{9}\)
b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)
\(-62x+12=-50\)
\(-62x=-62\)
\(x=1\)
a) Thực hiện rút gọn VT = -2x – 64
Giải phương trình -2x – 64 = 0 thu được x = -32.
b) Thực hiện rút gọn VT = -62 x +12
Giải phương trình -62x + 12 = -50 thu được x = 1.