chiều mai bn nộp thì làm luôn đi còn hỏi đáp nữa !!!!!!

mình làm bài 2 trước nha:

a) y.(a-b)+a.(y-b)=a.y-b.y+a.y-b.y

                        =(a.y+a.y)-(b.y+b.y)

                         =2.a.y-2.b.y

                        =2.y.(a-b)

b)x².(x+y)-y.(x²-y²)=x³+x².y-x²y+y³=x³+y³

\(x^2-y^2-ax+ay\)

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

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

\(2xy-x^2-y^2+16\)

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

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

\(=\left(4-x+y\right)\left(4+x-y\right)\)

\(x^2+5x+4\)

\(=\left(x^2+x\right)+\left(4x+4\right)\)

\(=x\left(x+1\right)+4\left(x+1\right)\)

\(=\left(x+1\right)\left(x+4\right)\)

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

Bài 6: 

a²+b²=(a+b)²-2ab

<=> 2010  =36-2ab   

<=>ab=-987

M=a³+b³

=(a+b)(a²-ab+b²)

=6(a²+987+b^2)

=6(2010+987)

=17982

1. Ta có: 

\(x^3-9x^2+27x-26=x^3-2x^2-7x^2+14x+13x-26\)

\(=x^2\left(x-2\right)-7x\left(x-2\right)+13\left(x-2\right)=\left(x-2\right)\left(x^2-7x+13\right)\)

Thay x = 23, ta có: \(C=\left(23-2\right)\left(23^2-7.23+13\right)=8001\)

2.

a) \(x^2+4y^2+6x-12y+18=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(4y^2-12y+9\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(2y-3\right)^2=0\)

Mà \(\left(x-3\right)^2\ge0\) với mọi x, \(\left(2y-3\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)và \(\left(2y-3\right)^2=0\Leftrightarrow2y-3=0\Leftrightarrow y=\frac{3}{2}\)

Vậy \(\left(x,y\right)=\left(3;\frac{3}{2}\right)\)

b) \(2x^2+2y^2+2xy-10x-8y+41=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)+\left(y^2-8y+16\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2+\left(y-4\right)^2=0\)

.....................................

Rồi giải tương tự như trên

Dề sai ko bạn

Chỉ cần ý b thôi 

mình xin lỗi sửa x-1 thành x+1

Ta có : \(\left(5x-3\right)^2-\frac{1^2}{64}=0\)

\(\Leftrightarrow\left(5x-3\right)^2=\frac{1}{64}\)

\(\Leftrightarrow\left(5x-3\right)^2=\left(\frac{1}{8}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=\frac{1}{8}\\5x-3=-\frac{1}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{1}{8}+3\\5x=-\frac{1}{8}+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{25}{8}\\5x=\frac{23}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{25}{8}.\frac{1}{5}\\x=\frac{23}{8}.\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\frac{23}{40}\end{cases}}\)

b) 3x - 7.(5x-1) = 6 - 2.(4-3x)

=> 3x - 35x + 7 = 6 - 8 + 6x

=> 3x - 35x - 6x = 6-8 -7

-38x = -9

x = 9/38

\(\frac{1}{5}\)ở đâu ra vậy bạn