a)(x-1/x)^2=x^2-......+1/x^2 b) (1/2x+.....)(1/4x^2-.....+1/9y^2)=1/8x^3+1/27y^3
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a) x2 + xy + y - 1 = (x2 - 1) + (xy + y) = (x - 1)(x + 1) + y(x + 1) = (x + 1)(x + y - 1)
b) 4 - x2 + 2xy - y2 = 4 - (x2 - 2xy + y2) = 4 - (x - y)2 = (x - y + 2)(4 - x + y)
c) 8x2 - 18y2 = 2(4x2 - 9y2) = 2[(2x)2 - (3y)2] = 2(2x - 3y)(2x + 3y)
d) 8x3 - 4x2 - 6xy - 9y2 - 27y3
= (8x3 - 27y3) - (4x2 + 6xy + 9y2)
= (2x - 3y)(4x2 + 6xy + 9y2) - (4x2 + 6xy + 9y2)
= (2x - 3y - 1)(4x2 + 6xy + 9y2)
e) 4x2 - x - 3 = 4x2 - 4x + 3x - 3 = 4x(x - 1) + 3(x - 1) = (x - 1)(4x + 3)
f) 4x2 - 8x + 3 = 4x2 - 2x - 6x + 3 = 2x(2x - 1) - 3(2x - 1) = (2x - 3)(2x - 1)
a) (x2-6xy+9y2):(3y-x)
= (x-3y)2:(3y-x)
=(3y-x)2:(3y-x)
= 3y-x
b) (8x3-1):(4x2+2x+1)
=[(2x)3-1]:(4x2+2x+1)
= (2x-1)(4x2+2x+1):(4x2+2x+1)
= 2x-1
c) (4x4-9):(2x2-3)
=(2x2-3)(2x2+3):(2x2-3)
=2x2+3
d) (8x3-27):(4x2+6x+9)
=(2x-3)(4x2+6x+9):(4x2+6x+9)
=2x-3
Bài 1) A=(8x3+27x3):2x+3y
=[(2x)3+(3y)3]:2x+3y
=(2x)2+(3y)2
=4x2+9y2
B=(x3-27):(x-3)
=(x3-33):(x-3)
=x2-32
=x2-9
`@` `\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`
a) (1 - 2x) (2x + 1) = 1 - 4x2 __ hằng đẳng thức số 3 (A + B) (A - B) = A2 - B2 (ở đây A = 1 , B = 2x)
câu b) có sai đề ko bn
Bài 2:
a: \(\Leftrightarrow4x^2-14x+10x-35-\left(4x+3\right)^2=16\)
\(\Leftrightarrow4x^2-4x-35-16x^2-24x-9-16=0\)
\(\Leftrightarrow-12x^2-28x-60=0\)
\(\Leftrightarrow3x^2+7x+15=0\)
\(\text{Δ}=7^2-4\cdot3\cdot15=-131< 0\)
Do đó: Phương trình vô nghiệm
b: Ta có: \(\left(8x^2+3\right)\left(8x^2-3\right)-\left(8x^2-1\right)^2=22\)
\(\Leftrightarrow64x^4-9-64x^4+16x^2-1=22\)
\(\Leftrightarrow16x^2=32\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
c: Ta có: \(49x^2+14x+1=0\)
=>\(\left(7x+1\right)^2=0\)
hay x=-1/7
ĐK: \(\hept{\begin{cases}x\ge2\\y\ge-\frac{1}{3}\end{cases}}\)
\(\sqrt{x-2}+x^3-6x^2+12x=\sqrt{3y+1}+27y^3+27y^2+9y+9\)
<=> \(\sqrt{x-2}+x^3-6x^2+12x-8=\sqrt{3y+1}+27y^3+27y^2+9y+1\)
<=> \(\sqrt{x-2}+\left(x-2\right)^3=\sqrt{3y+1}+\left(3y+1\right)^3\)
<=> \(\left(\sqrt{x-2}-\sqrt{3y+1}\right)+\left[\left(x-2\right)^3-\left(3y+1\right)^3\right]=0\)
<=> \(\frac{x-3y-3}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-3y-3\right)\left[\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2\right]=0\)
<=> \(\left(x-3y-3\right)\left(\frac{1}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2\right)=0\)
<=> \(x-3y-3=0\)
vì \(\frac{1}{\sqrt{x-2}+\sqrt{3y+1}}+\left(x-2\right)^2+\left(x-2\right)\left(3y+1\right)+\left(3y+1\right)^2>0\)
<=> x = 3y + 3
Thế vào phương trình trên ta có:
\(2+2\left(3y+3\right)^2-2y^2+3\left(3y+3\right)y-4\left(3y+3\right)-3y=0\)
<=> \(25y^2+30y+8=0\Leftrightarrow\orbr{\begin{cases}y=-\frac{2}{5}\\y=-\frac{4}{5}\end{cases}}\)không thỏa mãn đk
Vậy hệ vô nghiệm.
bài đây nhé mn nhanh nhanh giúp mik với
a: (x-1/x)^2=x^2-2+1/x^2
b: (1/2x+1/3y)(1/4x^2-1/6xy+1/9y^2)=1/8x^3+1/27y^3