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Bài 1:
\(a,x^2-y^2-2x+2y=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)
\(b,2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)
\(c,3a^2-6ab+3b^2-12c^2=3\left(a-b\right)^2-12c^2=3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)
\(d,x^2-25+y^2+2xy=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\)
Bài 1:
\(e,a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b-c\right)\left(a+b\right)\)
\(f,x^2-2x-4y^2-4y=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-2y-2\right)\left(x+2y\right)\)
\(g,x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)
\(h,x^2\left(x-1\right)+16\left(1-x\right)=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
1) \(\left(x+\dfrac{1}{3}\right)^3=x^3+3.x^2.\dfrac{1}{3}+3.x.\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3\)
\(=x^3+x^2+\dfrac{x}{3}+\dfrac{1}{27}\)
2) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.y^2+3.2x.\left(y^2\right)^2+\left(y^2\right)^3\)
\(=8x^3+12x^2y^2+6xy^4+y^6\)
3) \(\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y\right)^3=\left(\dfrac{1}{2}x^2\right)^3+3.\left(\dfrac{1}{2}x^2\right)^2.\dfrac{1}{3}y+3.\dfrac{1}{2}x^2.\left(\dfrac{1}{3}y\right)^2+\left(\dfrac{1}{3}y\right)^3\)
\(=\dfrac{1}{8}x^6+\dfrac{1}{4}x^4y+\dfrac{1}{6}x^2y^2+\dfrac{1}{27}y^3\)
4) \(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.3x^2.\left(2y\right)^2-\left(2y\right)^3\)
\(=27x^6-54x^4y+36x^2y^2-8y^3\)
5) \(\left(\dfrac{2}{3}x^2-\dfrac{1}{2}y\right)^3=\left(\dfrac{2}{3}x^2\right)^3-3.\left(\dfrac{2}{3}x^2\right)^2.\dfrac{1}{2}y+3.\dfrac{2}{3}x^2.\left(\dfrac{1}{2}y\right)^2-\left(\dfrac{1}{2}y\right)^3\)
\(=\dfrac{8}{27}x^6-\dfrac{1}{3}x^4y+\dfrac{1}{2}x^2y^2-\dfrac{1}{8}y^3\)
6) \(\left(2x+\dfrac{1}{2}\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.\dfrac{1}{2}+3.2x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)
\(=8x^3+6x^2+\dfrac{3}{2}x+\dfrac{1}{8}\)
7) \(\left(x-3\right)^3=x^3-3.x^2.3+3.x.3^2-3^3\)
\(=x^3-9x^2+27x-27\)
8) \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x+1\right)\left(x^2-x.1+1^2\right)\)
\(=x^3+1^3\)
\(=x+1\)
9) \(\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-3\right)\left(x^2+x.3+3^2\right)\)
\(=x^3-3^3\)
\(=x^3-27\)
10) \(\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x-2\right)\left(x^2+x.2+2^2\right)\)
\(=x^3-2^3\)
\(=x^3-8\)
11) \(\left(x+4\right)\left(x^2-4x+16\right)\)
\(=\left(x+4\right)\left(x^2-x.4+4^2\right)\)
\(=x^3+4^3\)
\(=x^3+64\)
12) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
13) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)
\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+x^2.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\right]\)
\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)
\(=x^6-\dfrac{1}{27}\)
14) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)
\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)
\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\dfrac{1}{27}x^3+8y^3\)
a: \(\Leftrightarrow2x\left(x+2\right)+4>x^2+4x+4\)
\(\Leftrightarrow2x^2+4x-x^2-4x>0\)
=>x<>0
b: \(\Leftrightarrow3\left(1-2x\right)-24x< 4\left(1-5x\right)\)
=>3-6x-24x<4-20x
=>-30x+3<4-20x
=>-10x<1
hay x>-1/10
c: \(\Leftrightarrow x^2+6x+8>x^2+10x+16+26\)
=>6x+8>10x+42
=>-4x>34
hay x<-17/2
3:
a: \(M=x+2x-4y-y-3=3x-5y-3\)
bậc là 1
b: \(N=-x^2t+13t^3+xt^2+5t^3-4\)
bậc là 3
5:
S=(3x+4y)*2*2z=4z(3x+4y)
V=3x*4y*2z=24xyz
Khi x=4;y=2;z=1 thì S=4*1*(3*4+4*2)=4*20=80cm2
V=24*4*2*1=192cm3
\(A=\left(\dfrac{4}{\left(x-2\right)\left(x+2\right)}+2\right)\cdot\dfrac{x+2}{2x}+\dfrac{2}{x-2}\)
\(=\dfrac{4+2x^2-8}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2x}+\dfrac{2}{x-2}\)
\(=\dfrac{2\left(x^2-2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2x}+\dfrac{2}{x-2}\)
\(=\dfrac{\left(x^2-2\right)}{x\left(x-2\right)}+\dfrac{2}{x-2}\)
\(=\dfrac{x^2+2x-2}{x\left(x-2\right)}\)
VD4:
a: A+x^2-y^2=x^2-2y^2+3xy-2
=>A-y^2=-2y^2+3xy-2
=>A=-2y^2+3xy-2+y^2
=>A=-y^2+3xy-2
b: B-(5x^2-2xyz)=2x^2+2xyz+1
=>B=2x^2+2xyz+1+5x^2-2xyz
=>B=7x^2+1
Ví dụ 5:
a) Ta có: \(C=A+B\)
\(\Leftrightarrow C=\left(4x^2+3y^2-5xy\right)+\left(3x^2+2y^2+2x^2y^2\right)\)
\(\Leftrightarrow C=\left(4x^2+3x^2\right)+\left(3y^2+2y^2\right)-5xy+2x^2y^2\)
\(\Leftrightarrow C=7x^2+5y^2-5xy+2x^2y^2\)
b) Ta có: \(C+A=B\)
\(\Leftrightarrow C=B-A\)
\(\Leftrightarrow C=\left(3x^2+2y^2+2x^2y^2\right)-\left(4x^2+3y^2-5xy\right)\)
\(\Leftrightarrow C=3x^2+2y^2+2x^2y^2-4x^2-3y^2+5xy\)
\(\Leftrightarrow C=\left(3x^2-4x^2\right)+\left(2y^2-3y^2\right)+2x^2y^2+5xy\)
\(\Leftrightarrow C=-x^2-y^2+2x^2y^2+5xy\)
a: Xét ΔABC có
M là trung điểm của BC
N là trung điểm của AC
Do đó: MN là đường trung bình của ΔACB
Suy ra: MN//AB
hay ABMN là hình thang
a, Vì M,N là trung điểm BC,AC nên MN là đtb tg ABC
Do đó MN//AB hay ABMN là hthang
b, Vì N là trung điểm AC và MD (t/c đối xứng) nên ADCM là hbh
c, Vì ADCM là hbh nên AD=CM=BM và AD//CM hay AD//BM
Do đó ADMB là hbh
Mà I là trung điểm AM nên cũng là trung điểm BD
Vậy B,I,D thẳng hàng
a: \(\dfrac{5x^2+10xy+5y^2}{3x^3+3y^3}\)
\(=\dfrac{5\left(x+y\right)^2}{3\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\dfrac{5\left(x+y\right)}{3\left(x^2-xy+y^2\right)}\)
b: \(3x\left(2x-1\right)-\left(2x+1\right)\left(3x+2\right)\)
\(=6x^2-3x-6x^2-4x-3x-2\)
\(=-10x-2\)
bạn có thể giúp mik bài 2 đc ko