chụp rõ hơn một xíu đc ko bạn

Em lớp 5 thôi nhưng em nghĩ bài này nhân vế bình thường thôi ạ 

Ví dụ câu a : \(\left(2x^2+x-1\right)\left(-3x^2-7x-5\right)\)

\(=-6x^4-17x^3-14x^2+2x+5\)

Tương tự chị nhé 

Câu b : 

\(\left(2x^2-3xy+y^2\right)\left(x+y\right)\)

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

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

Bài 1: 

a) Ta có: \(M=\left(\dfrac{x+2}{x^2+2x+1}+\dfrac{x-2}{1-x^2}\right)\cdot\dfrac{x+1}{x}\)

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

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

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

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

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

Bài 2:

1: Ta có: \(\left(x-5\right)^2+\left(x+3\right)^2=2\left(x-4\right)\left(x+4\right)-5x+7\)

\(\Leftrightarrow x^2-10x+25+x^2+6x+9=2\left(x^2-16\right)-5x+7\)

\(\Leftrightarrow2x^2-4x+34=2x^2-32-5x+7\)

\(\Leftrightarrow2x^2-4x+34-2x^2+5x+25=0\)

\(\Leftrightarrow x+59=0\)

hay x=-59

Vậy: S={-59}

Bài 2: 

Ta có: \(3n^3+10n^2-5⋮3n+1\)

\(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)

\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow3n\in\left\{0;-3;3\right\}\)

hay \(n\in\left\{0;-1;1\right\}\)

BAI 3 :quy đồng lên ta được a^3/abc+b^3/abc+c^3/abc=(a^3+b^3+c^3)/abc

ta có (a+b)^3=a^3+3a^2b+3ab^2+b^3=>a^3+b^3=(a+b)^3-3ab(a+b)

=>a^3+b^3+c^3=(a+b+c)^3+3(a+b)c(a+b+c)=0+0=0

=>A=0/ABC=0

BAI 4:

theo dinh ly py ta go ta co ah^2=ac^2-hc^2

va ah^2 cung bang ab^2-bh^2

=>2ah^2=ac^2-hc^2+ab^2-bh^2=ab^2+ac^2-hb^2-hc^2=ac^2-bh^2-hc^2

=(bh+ch)^2-bh^2-ch^2=bh^2+2.bh.ch+ch^2-bh^2-ch^2

=2.bh.ch=2ah^2

==>ah^2=bhxch

d. DE cat AM tai O

vi tam giac ahm vuong tai h co ho la trung tuyen nen ho=am/2

ma am=de nen oh=de/2

==>tam giac dhe vuong tai h

2.1 

\(\Leftrightarrow x^3+3x^2+2x-3x^2-9x-6=0\)

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

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

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)

2.2

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

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

\(\Leftrightarrow\left(x-1\right)\left(x^2-2x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\pm\sqrt{3}\end{matrix}\right.\)

2.3

\(\Leftrightarrow3x^3-3x^2+2x+3x^2-3x+2=0\)

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

\(\Leftrightarrow\left(x+1\right)\left(3x^2-3x+2\right)=0\)

\(\Leftrightarrow x=-1\)

2.5

\(\Leftrightarrow2x^3+x^2+3x-4x^2-2x-6=0\)

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

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

\(\Leftrightarrow x=2\)

Câu 1: 

Ta có: \(\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)

\(=6x^2+9x+14x+21-\left(6x^2+33x-10x-55\right)\)

\(=6x^2+23x+21-6x^2-23x+55\)

=76

cám ơn ạ

a/lay cac so nhan cong lai voi nhau :( a+b+c)x(ta cong cac so tru va bi tru lai voi nhau)(b^2+c^2+a^2)-(c^2+a^2+b^2)=a+b+c

tuong tu ban lam cac bai kia cung vay