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Bài 3: 

a: \(x^2+2x+1=\left(x+1\right)^2\)

b: \(4x^2-4x+1=\left(2x-1\right)^2\)

c: \(a^2-6a+9=\left(a-3\right)^2\)

d: \(x^2+10x+25=\left(x+5\right)^2\)

18 tháng 7 2023

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) \(\left(x-3\right)^3=x^3-3.x^2.3+3.x.3^2-3^3\)

\(=x^3-9x^2+27x-27\)

18 tháng 7 2023

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) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)

\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)

\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)

\(=\dfrac{1}{27}x^3+8y^3\)

HQ
Hà Quang Minh
Giáo viên
1 tháng 8 2023

Bài 1:

\(a,x^2-y^2-2x+2y=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)=\left(x-y\right)\left(x+y-2\right)\)

\(b,2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)

\(c,3a^2-6ab+3b^2-12c^2=3\left(a-b\right)^2-12c^2=3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)

\(d,x^2-25+y^2+2xy=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\)

HQ
Hà Quang Minh
Giáo viên
1 tháng 8 2023

Bài 1:

\(e,a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b-c\right)\left(a+b\right)\)

\(f,x^2-2x-4y^2-4y=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-2y-2\right)\left(x+2y\right)\)

\(g,x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)

\(h,x^2\left(x-1\right)+16\left(1-x\right)=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

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

Bài 1: 

a: \(x^3-10x^2+25x\)

\(=x\left(x^2-10x+25\right)\)

\(=x\left(x-5\right)^2\)

b: \(3x-3y-x^2+2xy-y^2\)

\(=3\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left(3-x+y\right)\)

c: \(x^3+x-y^3-y\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2+1\right)\)

13 tháng 11 2016

Câu 1:

\(2x^3-3x^2+x+a\)

\(=2\left(x^3-6x^2+12x-8\right)+9\left(x^2-4x+4\right)+13\left(x-2\right)+\left(6+a\right)\)

\(=2\left(x-2\right)^3+9\left(x-2\right)^2+13\left(x-2\right)+\left(6+a\right)\)chia hết cho \(x-2\)khi và chỉ khi :

\(6+a=0\Leftrightarrow a=-6\). Vậy \(a=-6\).

Câu 2:

\(\left(x+1\right)\left(2x-x\right)-\left(3x+5\right)\left(x+2\right)=4x^2+1\)

\(\Leftrightarrow x^2+x-\left(3x^2+11x+10\right)=-4x^2+1\)

\(\Leftrightarrow x^2+x-3x^2-11x-10+4x^2-1=0\)

\(\Leftrightarrow2x^2-10x-11=0\)

\(\Delta'=\left(-5\right)^2-2\left(-11\right)=47>0\)

\(\Rightarrow\)Phương trình có 2 nghiệm phân biệt:

\(x=\frac{5+\sqrt{47}}{2}\)hoặc \(x=\frac{5-\sqrt{47}}{2}\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{5+\sqrt{47}}{2};\frac{5-\sqrt{47}}{2}\right\}\)

c: \(\dfrac{1}{x^3-1}=\dfrac{1}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\dfrac{3}{x^2+x+1}=\dfrac{3x-3}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\dfrac{-2}{x-1}=\dfrac{-2x^2-2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)

NV
5 tháng 3 2023

Gọi chiều rộng của mảnh đất là x (m) với x>0

Chiều dài của mảnh đất là: \(x+6\) (m)

Do chu vi mảnh đất là 128m nên ta có pt:

\(2\left(x+x+6\right)=128\)

\(\Leftrightarrow4x+12=128\)

\(\Leftrightarrow x=29\)

Diện tích mảnh đất là: \(29.\left(29+6\right)=1015\left(m^2\right)\)

Số rau thu hoạch được: \(1015.1,2=1218\left(kg\right)\)

Số tiền thu được:

\(1218.20000=24360000\left(đ\right)\)