a=(x-1)(x^2+x+1)+(x-2)^3-2(x+1)(x^2-x+1)+6(x-1)^3
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TN
29 tháng 6 2016
a)(x-1)(x2+x+1)-x(x+2)(x-2)=5
=>x3-1-4x-x3=5
=>x3-x3+4x-1=5
=>4x-1=5
=>4x=6
=>x=3/2
b)(x-2)^3-(x-3)(x^2+3x+9)+6(x+1)^2=15
=>x3-6x2+12x-8-x3+27+6x2+12x+6=15
=>(x3-x3)-(-6x2+6x2)+(12x+12x)-8+27+6=15
=>24x+25=15
=>24x=-10
=>x=-5/12
c)6(x+1)^2-2(x+1)^3+2(x-1)(x^2+x+1)=1
=>6x2+12x+6-2x3-6x2-6x-2+2x3-2=1
=>(6x2-6x2)+(12x-6x)-(-2x3+2x3)+6-2-2=1
=>6x+2=1
=>6x=-1
=>x=-1/6
H9
HT.Phong (9A5)
CTVHS
14 tháng 6 2023
a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)
b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)
\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)
c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)
\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)
\(\Rightarrow x=-2\)
H9
HT.Phong (9A5)
CTVHS
14 tháng 6 2023
d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)
\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{25}{9}\)
e) \(\dfrac{1}{2}x+650\%x-x=-6\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)
\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)
\(\Rightarrow6x=-6\)
\(\Rightarrow x=\dfrac{-6}{6}=-1\)
g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)
\(\Rightarrow2x-1-3+x=2-x\)
\(\Rightarrow3x-4=2-x\)
\(\Rightarrow3x+x=2+4\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)
20 tháng 10 2018
1
a) \(\left(3x+1\right)\left(3x-1\right)=9x^2-1\)
\(\left(x+5y\right)\left(x-5y\right)=x^2-25y\)
b) \(\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\)
\(\left(x-5\right)\left(x^2+5x+25\right)=x^3-125\)
26 tháng 10 2022
Bài 3:
a: \(\Leftrightarrow x^2+8x+16-x^2+1=16\)
=>8x+1=0
=>x=-1/8
b: \(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
=>2x+255=0
=>x=-255/2
c: \(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6x^2+12x+6=49\)
=>24x+62=49
=>24x=-13
=>x=-13/24
d: =>x^3+8-x^3-2x=15
=>-2x=15-8=7
=>x=-7/2
KV
30 tháng 11 2023
a (x + 2) - x(x + 3) = 2
x + 2 - x(x + 3) - 2 = 0
x + x(x + 3) = 0
x(1 + x + 3) = 0
x(x + 4) = 0
x = 0 hoặc x + 4 = 0
*) x + 4 = 0
x = -4
Vậy x = -4; x = 0
b) (x + 2)(x - 2) - (x + 1)² = 7
x² - 4 - x² - 2x - 1 = 7
-2x - 5 = 7
-2x = 7 + 5
-2x = 12
x = 12 : (-2)
x = -6
c) 6x² - (2x + 1)(3x - 2) = 1
6x² - 6x² + 4x - 3x + 2 = 1
x + 2 = 1
x = 1 - 2
x = -1
d) (x + 2)(x + 3) - (x - 2)(x + 1) = 2
x² + 3x + 2x + 6 - x² - x + 2x + 2 = 2
6x + 8 = 2
6x = 2 - 8
6x = -6
x = -6 : 6
x = -1
e) 6(x - 1)(x + 1) - (2x - 1)(3x + 2) + 3 = 0
6x² - 6 - 6x² - 4x + 3x + 2 + 3 = 0
-x - 1 = 0
x = -1
12 tháng 7 2023
a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)
=>(x+5)(x-3)+8=x^2-1
=>x^2+2x-15+8=x^2-1
=>2x-7=-1
=>x=3(loại)
b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)
=>(x-4)(x+1)+x^2+3+5(x-1)=0
=>x^2-3x-4+x^2+3+5x-5=0
=>2x^2+2x-6=0
=>x^2+x-3=0
=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)
e: =>x^2-2x+1+2x+2=5x+5
=>x^2+3=5x+5
=>x^2-5x-2=0
=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)
g: (x-3)(x+4)*x=0
=>x=0 hoặc x-3=0 hoặc x+4=0
=>x=0;x=3;x=-4
\(a=x^3-1+x^3-6x^2+12x-8-2\left(x^3+1\right)+6\left(x^3-3x^2+3x-1\right)\)
\(=2x^3-6x^2+12x-9-2x^3-2+6x^3-18x^2+18x-6\)
\(=6x^3-24x^2+30x-17\)
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