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10 tháng 9 2018
b)(x-2)3-(x-3)(x2+3x+9)+6(x+1)2=49
(=) x3- 6x2 +12 x -8 - ( x3 - 27 ) + 6( x2 + 2x +1)
(=) x3 - 6x2 +12x -8 - x3 +27 + 6x2 +12x +6
(=) 24x + 25 = 49
(=) 24x = 49 - 25 = 24
(=) x = 24/24 =1
BA
18 tháng 10 2019
\(a,\left(x+1\right)^2=x+1\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\cdot\left(x+1-1\right)=0\)
\(x\cdot\left(x+1\right)=0\)
\(\hept{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}}\)
KN
18 tháng 10 2019
\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x+25=49\)
\(\Leftrightarrow24x=24\Leftrightarrow x=1\)
MT
22 tháng 6 2015
(x - 2)3 - (x - 3)(x2 + 3x + 9) + 6(x + 1)2 = 49
<=>x3-6x2+12x-8-(x3-27)+6(x2+2x+1)=49
<=>x3-6x2+12x-8-x3+27+6x2+12x+6=49
<=>24x+25=49
<=>24x=24
<=>x=1
x(x + 5)(x - 5) - (x + 2)(x2 - 2x + 4) = 42
<=>x(x2-25)-(x3+8)=42
<=>x3-25x-x3-8=42
<=>-25x-8=42
<=>-25x=50
<=>x=-2
LD
17 tháng 8 2017
(x - 2)3 - (x - 3)(x2 + 3x + 9) + 6(x + 1)2 = 49
<=>x3-6x2+12x-8-(x3-27)+6(x2+2x+1)=49
<=>x3-6x2+12x-8-x3+27+6x2+12x+6=49
<=>24x+25=49
<=>24x=24
<=>x=1 x(x + 5)(x - 5) - (x + 2)(x2 - 2x + 4) = 42
<=>x(x2-25)-(x3+8)=42
<=>x3-25x-x3-8=42
<=>-25x-8=42
<=>-25x=50
<=>x=-2
22 tháng 7 2017
1. \(125x^3+y^6=\left(5x\right)^3+\left(y^2\right)^3\)
\(=\left(5x+y^2\right)\left[\left(5x\right)^2-5x.y^2+\left(y^2\right)^2\right]\)
\(=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
2. \(4x\left(x-2y\right)+8y\left(2y-x\right)\)
\(=4x\left(x-2y\right)-8y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(4x-8y\right)\)
3. \(25\left(x-y\right)^2-16\left(x+y\right)^2\)
\(=\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2\)
\(=\left[5\left(x-y\right)-4\left(x+y\right)\right]\left[5\left(x-y\right)+4\left(x+y\right)\right]\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)\)
\(=\left(x-9y\right)\left(9x-y\right)\)
4. \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
5. \(a^3x-ab+b-x\)
\(=a^3x-x-ab+b\)
\(=x\left(a^3-1\right)-b\left(a-1\right)\)
\(=x\left(a-1\right)\left(a^2+a+1\right)-b\left(a-1\right)\)
\(=\left(a-1\right)\left[x\left(a^2+a+1\right)-b\right]\)
6. \(x^3-64=x^3-4^3\)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
7. \(0,125\left(a+1\right)^3-1\)
\(=\left[0,5\left(a+1\right)\right]^3-1^3\)
\(=\left[0,5\left(a+1\right)-1\right]\left\{\left[0,5\left(a+1\right)\right]^2+\left[0,5\left(a+1\right).1\right]+1^2\right\}\)
\(=\left[0,5\left(a+1-2\right)\right]\left[0,25a^2+0,5a+0,25+0,5a+0,5+1\right]\)
\(=\left[0,5\left(a-1\right)\right]\left(0,25a^2+a+1,75\right)\)
8. \(9\left(x+5\right)^2-\left(x-7\right)^2\)
\(=\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2\)
\(=\left(3x+15-x+7\right)\left(3x+15+x-7\right)\)
\(=\left(2x+22\right)\left(4x+8\right)\)
9. \(49\left(y-4\right)^2-9\left(y+2\right)^2\)
\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)
\(=\left(4y-34\right)\left(10y-22\right)\)
10. \(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(xy-1\right)\)
11. \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
12. \(x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-1\right)\)
H
4 tháng 8 2019
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
12 tháng 8 2022
a: =>\(x^3+8-x^3-2x=15\)
=>2x=-7
hay x=-7/2
c: =>(5x-3)(5x+3)=0
=>x=3/5 hoặc x=-3/5
d: =>\(x^2+8x+16-x^2+1=16\)
=>8x+1=0
hay x=-1/8
NH
tìm x:
(x+3)3 - (x(3x+1)2 + (2x+1)4x2 - (2x+1) - 3x2=54
(x-3)3 - (x-3)(x2+3x+9)+ 6(x+1)2 + 3x2 = -33
1
HC
26 tháng 7 2017
1.
(x + 3)3 - x(3x + 1)2 + (2x + 1)(4x2 - 2x + 1) - 3x2 = 54
x3 + 9x2 + 27x + 27 - x(9x2 + 6x + 1) + 8x3 + 1 - 3x2 = 54
9x3 + 6x2 + 27x - 9x3 - 6x2 - x
= 54 - 27 - 1
26x = 26
x = 1
=> x3-6x2+12x-8-(x3-27) + 6(x2+2x+1) = 49
=> >x3 -6x2+12x-8-x3+27+6x2+12x+6=49
=>24x+25=49
<=>24x=24
<=>x=1
Xong rồi đó dòng thứ 2 do chị đang bận nên em tự hiểu là dấu mũ nha
\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Rightarrow x^3-3x^22+3x2^2-2^3-x^3+3^3+6\left(x^2+2x+1\right)=49\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Rightarrow24x+25=49\)
\(\Rightarrow24x=49-25=24\)
\(\Rightarrow x=1\)