Bài 4:

1: \(A=x^2\left(x+y\right)-y\left(x^2-y\right)+2020\)

\(=x^3+x^2y-x^2y+y^3+2020\)

\(=x^3+y^3+2020=1^3+\left(-1\right)^3+2020\)

=2020

2: \(B=5x\left(x-4y\right)-4y\left(y-5x\right)-\frac{11}{20}\)

\(=5x^2-20xy-4y^2+20xy-\frac{11}{20}\)

\(=5x^2-4y^2-\frac{11}{20}=5\cdot\left(-0,6\right)^2-4\cdot\left(-0,75\right)^2-\frac{11}{20}\)

\(=5\cdot0,36-4\cdot\frac{9}{16}-\frac{11}{20}\)

=1,8-2,25-0,55

=1,8-2,8

=-1

3: \(C=x\left(x-y+1\right)-y\left(y+1-x\right)\)

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

=(x-y)(x+y)+(x-y)

=(x-y)(x+y+1)

Khi x=-2/3; y=-1/3 thì \(C=\left(-\frac23+\frac13\right)\left(-\frac23-\frac13+1\right)=0\)

4: \(D=x\left(x^2-y\right)+y\left(x-y^2\right)\)

\(=x^3-xy+xy-y^3=x^3-y^3\)

\(=\left(-\frac12\right)^3-\left(-\frac12\right)^3=0\)

5: \(E=x^2\left(y^3-xy^2\right)+\left(-y+x+1\right)x^2y^2\)

\(=x^2y^3-x^3y^2-x^2y^3+x^3y^2+x^2y^2=x^2y^2\)

Khi x=-10; y=-10 thì \(E=\left(-10\right)^2\cdot\left(-10\right)^2=100\cdot100=10000\)

6: \(F=3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)\)

\(=3x^2-12xy-2,4y^2+12xy=3x^2-2,4y^2\)

\(=3\cdot4^2-2,4\cdot\left(-5\right)^2=3\cdot16-2,4\cdot25=48-60=-12\)

7: x=9

=>x+1=10

\(G=x^{10}-10x^9+10x^8-\cdots-10x+10\)

\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-\cdots-x\left(x+1\right)+\left(x+1\right)\)

\(=x^{10}-x^{10}-x^9+x^9+x^8+\cdots-x^2-x+x+1\)

=1

8: x=16

=>x+1=17

\(I=x^4-17x^3+17x^2-17x+20\)

\(=x^4-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+20\)

\(=x^4-x^4-x^3+x^3+x^2-x^2-x+20\)

=-x+20

=-16+20

=4

9: x=31

=>x-1=30

\(A=x^3-30x^2-31x+1\)

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

\(=x^3-x^3+x^2-x^2+1\)

=1

10: x=14

=>x+1=15; x+2=16; 2x+1=29; x-1=13

\(M=x^5-15x^4+16x^3-29x^2+13x\)

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

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

=-x

=-14

11: x=24

=>x+1=25

\(N=x^{10}-25x^9+25x^8-\cdots-25x^3+25x^2-25x+25\)

\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-\cdots-x^3\left(x+1\right)+x^2\left(x+1\right)\) -x(x+1)+(x+1)

\(=x^{10}-x^{10}-x^9+x^9+x^8-\cdots+x^3+x^2-x^2-x+x+1\)

=1

Bài 5:

1: \(A=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)+3x-10\)

\(=5x^2-3x-x^3+x^2+x^3-6x^2+3x-10\)

=-10

2: \(B=x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3\)

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

=3

3: \(C=5x\left(x^2-7x+2\right)-x^2\left(5x-8\right)+27x^2-10x+2\)

\(=5x^3-35x^2+10x-5x^3+8x^2+27x^2-10x+2\)

=2

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

\(=3x^2-15xy-3y^2+15xy-3x^2+3y^2-1\)

=-1

5: \(E=x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)

\(=x^3+x^2+x-x^3-x^2-x+5\)

=5

6: \(F=x\left(3x+2\right)-x\left(x^2+3x\right)+x^3-2x+3\)

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

=3

7: \(G=x\left(2x-3\right)+6x\left(\frac12-\frac13x\right)+1\)

\(=2x^2-3x+3x-2x^2+1\)

=1

8: \(H=3x\left(x^3-x+4\right)-\frac12x^2\left(6x^2-2\right)-2x\left(6-x\right)+1\)

\(=3x^4-3x^2+12x-3x^4+x^2-12x+2x^2+1\)

=1

Bài 1: 

Để B là số tự nhiên thì \(\left\{{}\begin{matrix}2x-10⋮x-1\\\left[{}\begin{matrix}x>5\\x< 1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\\x< 0\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{-1;-3;-7\right\}\)

không làm mà đòi có ăn thì ăn chubin nhé

Bài 2:

14.25x6,9+3,1x14,25

=14,25x(6,9+3,1)

=14,25x10

=142,5

Bài 1:

Diện tích mảnh vườn là;

\(80\times50=4000\left(m^2\right)\)

Diện tích đất đào ao cá là:

\(4000\times5\%=200\left(m^2\right)\)

Bài 3: 

Xét ΔIAB có 

\(\widehat{AIB}+\widehat{IAB}+\widehat{IBA}=180^0\)

\(\Leftrightarrow\widehat{IAB}+\widehat{IBA}=115^0\)

hay \(\widehat{DAB}+\widehat{ABC}=230^0\)

Xét tứ giác ABCD có 

\(\widehat{D}+\widehat{C}+\widehat{DAB}+\widehat{CBA}=360^0\)

\(\Leftrightarrow\widehat{D}+\widehat{C}=150^0\)

mà \(\widehat{C}-\widehat{D}=10^0\)

nên \(2\cdot\widehat{C}=160^0\)

\(\Leftrightarrow\widehat{C}=80^0\)

\(\Leftrightarrow\widehat{D}=70^0\)

5.
a) Dấu hiệu: điểm thi HKI môn Toán của học sinh lớp 7C
Có 10 giá trị của dấu hiệu
b) Giá trị(x) Tần số(n) 4 5,5 6 6,5 7 7,5 8 9 9,5 10 3 2 6 4 3 5 3 2 3 1 N=32

c)Số điểm thấp nhất: 4
Số điểm cao nhất: 10
Số điểm có nhiều nhất: 6
(bài 6 tương tự ạ)

Cảm ơn bạn

Bạn tham khảo code C++ nhé

10043: 2= 5021,5.

5021,5

k mình nha

Bài 1:

a)Vì \(m\perp CD\)

        \(n\perp CD\)

nên \(m//n\)

b)Vì \(m//n\) nên

\(\widehat{CFE}+\widehat{FED}=180^o\) (trong cùng phía)

\(110^o+\widehat{FED}=180^o\)

\(\widehat{FED}=70^o\)

Bài 2:

Vì \(AB\perp AD\)

        \(AB\perp CB\)

nên \(AD//BC\)

Vì \(AD//BC\) nên

\(\widehat{D_1}+\widehat{C_1}=180^o\) (trong cùng phía)

\(115^o+\widehat{C_1}=180^o\)

\(\widehat{C_1}=65^o\)

Tớ ko nhìn rõ nhé Xin lỗi!