16x -5x mũ 2 -3
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20 tháng 9 2020
\(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
=> \(2x^2+3\left(x^2-1\right)=5x^2+5x\)
=> \(2x^2+3x^2-3-5x^2-5x=0\)
=> \(-3-5x=0\)
=> \(5x=-3\Rightarrow x=-\frac{3}{5}\)
\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+\frac{9}{2}\right)=\frac{7}{2}\)
=> \(x\left[2x\left(x+5\right)-1\left(x+5\right)\right]-2x^2\left(x+\frac{9}{2}\right)-1\left(x+\frac{9}{2}\right)=\frac{7}{2}\)
=> \(x\left(2x^2+10x-x-5\right)-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)
=> \(2x^3+10x^2-x^2-5x-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)
=> \(\left(2x^3-2x^3\right)+\left(10x^2-x^2-9x^2\right)+\left(-5x-x\right)-\frac{9}{2}=\frac{7}{2}\)
=> \(-6x-\frac{9}{2}=\frac{7}{2}\)
=> \(-6x=8\Rightarrow x=-\frac{8}{6}=-\frac{4}{3}\)
\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
=> 12x(4x - 1) - 5(4x - 1) + 3x(1 - 16x) - 7(1 - 16x) = 81
=> 48x2 - 12x - 20x + 5 + 3x - 48x2 - 7 + 112x = 81
=> -12x - 20x + 3x + 112x + 5 - 7 = 81
=> 83x + 5 - 7 = 81
=> 83x = 81 + 7 - 5
=> 83x = 83
=> x = 1
20 tháng 9 2020
1) \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow5x=-3\)
\(\Rightarrow x=-\frac{3}{5}\)
2) \(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+\frac{9}{2}\right)=\frac{7}{2}\)
\(\Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-\frac{9}{2}=\frac{7}{2}\)
\(\Leftrightarrow-6x=8\)
\(\Rightarrow x=-\frac{4}{3}\)
3) \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-32x+5-48x^2+115x-7=81\)
\(\Leftrightarrow83x=83\)
\(\Rightarrow x=1\)
MD
1
H9
HT.Phong (9A5)
CTVHS
22 tháng 8 2023
\(A=x^3-15x^4+16x^3-29x^2\)
\(A=\left(x^3+16x^3\right)-15x^4-29x^2\)
\(A=17x^3-15x^4-29x^2\)
\(A=-15x^4+17x^3-29x^2\)
\(A=-x^2\left(15x^2-17x+29\right)\)
TN
1
24 tháng 4 2023
a: =>9(2x+1)=6(3-x)
=>3(2x+1)=2(3-x)
=>6x+3=6-2x
=>8x=3
=>x=3/8
b: =>-3x^2-2+3x^2-18x=-26
=>-18x=-24
=>x=4/3
H9
HT.Phong (9A5)
CTVHS
3 tháng 10 2023
\(16x+40=10\cdot3^2+5\cdot\left(1+2+3\right)\)
\(\Rightarrow16x+40=10\cdot9+5\cdot6\)
\(\Rightarrow16x+40=90+30\)
\(\Rightarrow16x+40=120\)
\(\Rightarrow16x=120-40\)
\(\Rightarrow16x=80\)
\(\Rightarrow x=\dfrac{80}{16}\)
\(\Rightarrow x=5\)
30 tháng 9 2021
Ta có: \(16x+40=10\cdot3^2+5\cdot\left(1+2+3\right)\)
\(\Leftrightarrow16x+40=90+5\cdot6=120\)
hay x=5
ND
26 tháng 5 2018
1. \(x^4-2x^2+1=\left(x^2-1\right)^2\)
2. \(x^2+5x+\dfrac{25}{4}=x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
3. \(16x^2-8x+1=\left(4x-1\right)^2\)
4. \(x^2+x-y^2+y=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x-y+1\right)\left(x+y\right)\)
5. \(\dfrac{1}{4}x^2-\dfrac{4}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)\left(\dfrac{1}{2}x+\dfrac{2}{3}y\right)\)
6. \(a^2-2ab+b^2-x^2=\left(a-b\right)^2-x^2=\left(a-b-x\right)\left(a-b+x\right)\)
7. \(4x^2-20x+25-y^2=\left(2x-5\right)^2-y^2=\left(2x-5-y\right)\left(2x-5+y\right)\)
3 tháng 8 2018
Câu a : \(4x^3-5x^2+6x+9\)
\(=4x^3+3x^2-8x^2-6x+12x+9\)
\(=\left(4x^3+3x^2\right)-\left(8x^2+6x\right)+\left(12x+9\right)\)
\(=x^2\left(4x+3\right)-2x\left(4x+3\right)+3\left(4x+3\right)\)
\(=\left(4x+3\right)\left(x^2-2x+3\right)\)
Câu b : \(5x^3-12x^2+14x-4\)
\(=5x^3-10x^2-2x^2+10x+4x-4\)
\(=\left(5x^3-2x^2\right)-\left(10x^2-4x\right)+\left(10x-4\right)\)
\(=x^2\left(5x-2\right)-2x\left(5x-2\right)+2\left(5x-2\right)\)
\(=\left(5x-2\right)\left(x^2-2x+2\right)\)
Câu c : \(x^3-5x^2+2x+8\)
\(=x^3+x^2-6x^2-6x+8x+8\)
\(=\left(x^3+x^2\right)-\left(6x^2+6x\right)+\left(8x+8\right)\)
\(=x^2\left(x+1\right)-6x\left(x+1\right)+8\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+8\right)\)
\(=\left(x+1\right)\left[x^2-2x-4x+8\right]\)
\(=\left(x+1\right)\left[x\left(x-2\right)-4\left(x-2\right)\right]\)
\(=\left(x+1\right)\left(x-2\right)\left(x-4\right)\)
Câu d : \(4x^3+5x^2+10x-12\)
\(=4x^3+8x^2-3x^2+16x-6x-12\)
\(=\left(4x^3-3x^2\right)+\left(8x^2-6x\right)+\left(16x-12\right)\)
\(=x^2\left(4x-3\right)+2x\left(4x-3\right)+4\left(4x-3\right)\)
\(=\left(4x-3\right)\left(x^2+2x+4\right)\)
\(=-5x^2+15x+x-3\)
=(x-3)(-5x+1)