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giả thiết => \(\frac{M\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\frac{N\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\frac{32x-19}{\left(x+1\right)\left(x-2\right)}\)
=> M(x-2) + N(x+1) = 32x - 19
<=> M.x - 2.M + N.x + N = 32.x -19
=> (M+ N).x + (N - 2.M) = 32.x - 19
=> M+ N = 32 và -2M + N = -19
=> M = 17, N = 15
vậy M.N = 17. 15 =...
1) \(2x\left(x-5\right)+\left(x-2\right)\left(x+3\right)=2x^2-10x+x^2+3x-2x-6=3x^2-9x-6\)
2) \(\left(2x-5\right)\left(1-x\right)-\left(x-3\right)\left(-2x\right)=2x-2x^2-5+5x+2x^2-6x=x-5\)
3) \(\left(4x-3\right)\left(4x-3\right)-\left(3x+2\right)\left(3x-2\right)=\left(4x-3\right)^2-9x^2+4=16x^2-24x+9-9x^2+4\)
\(=7x^2-24x+13\)
4) \(\left(2x-1\right)\left(2x+1\right)\left(2x+1\right)-4\left(x^2+1\right)=\left(2x-1\right)[\left(2x+1\right)^2]-4x^2-4\)
\(=\left(2x-1\right)\left(4x^2+4x+4\right)-4x^2-4=8x^3+8x^2+8x-4x^2-4x-4-4x^2-4=8x^3+4x-8\)
5) \(3x\left(2x-8\right)-\left(2-6x\right)\left(5+x\right)=6x^2-24x-10-2x+30x+6x^2=12x^2+4x-10\)
6) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+6x+12x-24+8=-16\)
7) \(\left(x+2\right)\left(x^2-2x+4\right)-x^2\left(x-2\right)-2x^2=x^3+8-x^3+2x^2-2x^2=8\)
Bài 1:
a,\(3x\left(5x^2-2x-1\right)\)
\(=3x.5x^2-3x.2x-3x=15x^3-6x^2-3x\)
b,\(\left(x^2+2xy-3\right)\left(-xy\right)\)
\(=x^2.\left(-xy\right)+2xy.\left(-xy\right)-3.\left(-xy\right)\)
\(=-x^3y-2x^2y^2+3xy\)
c,\(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)
\(=\dfrac{1}{2}x^2y.\left(2x^3\right)-\dfrac{1}{2}x^2y.\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y\)
\(=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
Chúc bạn học tốt!!!
Bài 1:
a) \(3x\left(5x^2-2x-1\right)\\ =15x^3-6x^2-3x\)
b) \(\left(x^2+2xy-3\right)\left(-xy\right)\\ =-x^3y-2x^2y+3xy\)
c) \(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\\ =x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
\(a)\)
\(1-5x\le x^2-4\)
\(\Leftrightarrow x^2-4+5x-1\ge0\)
\(\Leftrightarrow x^2+5x-5\ge0\)
\(\Leftrightarrow x\le\frac{\left(-5-\sqrt{45}\right)}{2}\)hoặc \(x\ge\frac{\left(-5+\sqrt{45}\right)}{2}\)
\(c)\)
\(3x^2-6x+7\)
\(=3\left(x^2-2x+1\right)+4\)
\(=3\left(x-1\right)^2+4>0\)(Vô lý)
=> Bất phương trình vô nghiệm
\(d)\)
\(\frac{4-x}{x-9}>2\)
\(\Leftrightarrow\frac{\left(4-x\right)}{\left(x-9\right)}-2>0\)
\(\Leftrightarrow\frac{\left(-3x+22\right)}{x-9}>0\)
\(\Leftrightarrow\frac{22}{3}< x< 9\)
Bổ sung b)
2/7x-4 >1
<=> 2/( 7x - 4) - 1 > 0
<=> [ 2 - ( 7x -4)]/( 7x - 4) > 0
<=> ( 6-7x)/( 7x -4) > 0
<=> ( 7x - 6).( 7x - 4) < 0
<=> 4/7 < x < 6/7
a, \(\left(y-2\right)\left(y+2\right)\left(y^2+4\right)-\left(y+3\right)\left(y-3\right)\left(y^2+9\right)\)
\(=\left(y^2-4\right)\left(y^2+4\right)-\left(y^2-9\right)\left(y^2+9\right)\)
\(=y^4-16-y^4+81=65\)
b, \(2\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)-2\left(x^6-y^6\right)\)
\(=2\left(x^3-y^3\right)\left(x^3+y^3\right)-2\left(x^6-y^6\right)\)
\(=2\left(x^6-y^6\right)-2\left(x^6-y^6\right)=0\)
a)(x+2y-3z-t)(x+2y+3z+t)
=\(\left(x+2y\right)^2-\left(3z+t\right)^2\)
a)-2
b)-20
a, (2x+1)^2−4(x+2)^2=9
⇔4x^2+4x+1−4(x^2+4x+4)=9
⇔4x^2+4x+1−4x^2−16x−16=9
⇔ −12x−15=9
⇔−12x=24
⇔x=−2
b, (x+3)^2−(x−4)(x+8)=1
⇔x^2+6x+9−(x^2−4x+8x−32)=1
⇔x^2+6x+9−x^2−4x+32=1
⇔ 2x+41=1
⇔2x=−40
⇔x=−20
k cho mik nha