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a)
\(5x\left(4-8x\right)+40\left(x^2-1\right)=3\\ \Leftrightarrow20x-40x^2+40x^2-40=3\\ \Leftrightarrow20x-40=3\\ \Leftrightarrow20\left(x-2\right)=3\\ \Leftrightarrow x-2=\frac{3}{20}\\ \Leftrightarrow x=\frac{43}{20}\)
b)
\(\left(4x-5\right)\left(7-8x\right)+4x\left(3+8x\right)=4\\ \Leftrightarrow28x-32x^2-35+40x+12x+32x^2=4\\ \Leftrightarrow80x-35=4\\ \Leftrightarrow80x=39\\ \Leftrightarrow x=\frac{39}{80}\)
a) \(5x\left(4-8x\right)+40\left(x^2-1\right)=3\)
\(\Leftrightarrow\)\(20x-40x^2+40x^2-40=3\)
\(\Leftrightarrow20x-40=3\)
\(\Leftrightarrow20x=43\)
\(\Leftrightarrow x=\frac{43}{20}\)
b, \(\left(4x-5\right)\left(7-8x\right)+4x\left(3+8x\right)=4\)
\(\Leftrightarrow28x-32x^2-35+40x+12x+32x^2=4\)
\(\Leftrightarrow80x-35=4\)
\(\Leftrightarrow80x=39\)
\(\Leftrightarrow x=\frac{39}{80}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\) ( ĐKXĐ: \(x\ne0;x\ne2\) )
\(\Leftrightarrow\dfrac{x-1}{2x\left(x-2\right)}-\dfrac{7}{8x}=\dfrac{5-x}{4x\left(x-2\right)}-\dfrac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)4}{8x\left(x-2\right)}-\dfrac{7\left(x-2\right)}{8x\left(x-2\right)}=\dfrac{2\left(5-x\right)}{8x\left(x-2\right)}-\dfrac{1x}{8x\left(x-2\right)}\)
\(\Rightarrow4x-4-7x+14=10-2x-x\)
\(\Leftrightarrow-3x+2x+x=10+4-14\)
\(\Leftrightarrow0=0\)
Vậy pt đã cho có nghiệm đúng với mọi x
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(4x^5y^2-8x^4y^2+4x^3y^2\)
\(=4x^3y^2\left(x^2-2x+1\right)\)
\(=4x^3y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)
\(=4x^3y^2\left(x-1\right)^2\)
2) \(5x^4y^2-10x^3y^2+5x^2y^2\)
\(=5x^2y^2\left(x^2-2x+1\right)\)
\(=5x^2y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)
\(=5x^2y^2\left(x-1\right)^2\)
3) \(12x^2-12xy+3y^2\)
\(=3\left(4x^2-4xy+y^2\right)\)
\(=3\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)
\(=3\left(2x-y\right)^2\)
4) \(8x^3-8x^2y+2xy^2\)
\(=2x\left(4x^2-4xy+y^2\right)\)
\(=2x\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)
\(=2x\left(2x-y\right)^2\)
5) \(20x^4y^2-20x^3y^3+5x^2y^4\)
\(=5x^2y^2\left(4x^2-4xy+y^2\right)\)
\(=5x^2y^2\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)
\(=5x^2y^2\left(2x-y\right)^2\)
1: 4x^5y^2-8x^4y^2+4x^3y^2
=4x^3y^2(x^2-2x+1)
=4x^3y^2(x-1)^2
2: \(=5x^2y^2\left(x^2-2x+1\right)=5x^2y^2\left(x-1\right)^2\)
3: \(=3\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)^2\)
4: \(=2x\left(4x^2-4xy+y^2\right)=2x\left(2x-y\right)^2\)
5: \(=5x^2y^2\left(4x^2-4xy+y^2\right)=5x^2y^2\left(2x-y\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời:
\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)\(\left(đkxđ:x\ne0;x\ne2\right)\)
\(\Leftrightarrow\frac{x-1}{2x\left(x-2\right)}-\frac{7}{8x}=\frac{5-x}{4x\left(x-2\right)}-\frac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\frac{4\left(x-1\right)}{8x\left(x-2\right)}-\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{2\left(5-x\right)}{8x\left(x-2\right)}-\frac{x}{8x\left(x-2\right)}\)
\(\Rightarrow4\left(x-1\right)-7\left(x-2\right)=2\left(5-x\right)-x\)
\(\Leftrightarrow4x-4-7x+14=10-2x-x\)
\(\Leftrightarrow10-3x=10-3x\)
\(\Leftrightarrow-3x+3x=10-10\)
\(\Leftrightarrow0x=0\)( luôn thỏa mãn )
Vậy S = R với \(x\ne0;x\ne2\)
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b: Ta có: \(G=-4x^2+8x-2003\)
\(=-4\left(x^2-2x+\dfrac{2003}{4}\right)\)
\(=-4\left(x^2-2x+1+\dfrac{1999}{4}\right)\)
\(=-4\left(x-1\right)^2-1999\le-1999\forall x\)
Dấu '=' xảy ra khi x=1
F=5-4x2+4
=(5+4)-4x2
= 9- 4x2
Vì 4x2 ≥0 => -4x2 ≤0 => 9-4x2 ≤9 ∀x
Dấu = xảy ra khi x=0
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ĐKXĐ : \(\left\{{}\begin{matrix}4x^2-1\ne0\\8x^3+1\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm\dfrac{1}{2}\)
\(P=\dfrac{2x^5-x^4-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)
\(=\dfrac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)
\(=\dfrac{x^4-1}{2x+1}+\dfrac{2}{2x+1}=\dfrac{x^4+1}{2x+1}\)
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( 4x - 1 )3 + ( 3 - 4x )( 9 + 12x + 16x2 ) = ( 8x - 1 )( 8x + 1 ) - ( 3x - 5 )
<=> 64x3 - 48x2 + 12x - 1 + [ 33 - ( 4x )3 ] = ( 8x )2 - 12 - 3x + 5
<=> 64x3 - 48x2 + 12x - 1 + 27 - 64x3 = 64x2 - 1 - 3x + 5
<=> 64x3 - 48x2 + 12x - 64x3 - 64x2 + 3x = -1 + 5 + 1 - 27
<=> -112x2 + 15x = -22
<=> -112x2 + 15x + 22 = 0 (*) ( lại phải xài Delta :(( )
\(\Delta=b^2-4ac=15^2-4\cdot\left(-112\right)\cdot22=225+9856=10081\)
\(\Delta>0\)nên (*) có hai nghiệm phân biệt
\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-15+\sqrt{10081}}{-224}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-15-\sqrt{10081}}{-224}\end{cases}}\)
Nghiệm xấu quá -..-
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ta có 4 x 3 y 2 – 8 x 2 y 3 = 4 x 2 y 2 . x – 4 x 2 y 2 . 2 y = 4 x 2 y 2 ( x – 2 y )
Vậy 4x3y2 – 8x2y3 = 4x2y2(x – 2y)
Đáp án cần chọn là: C
bấm đúng cho mik đi
kết bạn ko^-^
= 8x ( 4x - 5 - 4x + 4 )
= 8x . ( -1 )
=-8x
học tốt