là sao ạ

Áp dụng hằng đẳng thức \(\left(a+b\right)^2\) với \(a=x+y\) và \(b=z\)

Đã cẩn thận khoanh ngoặc cho bạn nhìn đỡ phải hỏi rồi mà vẫn đi hỏi :D

\(\Rightarrow x^2+2x+1-y^2-4y-4-7=0\\ \Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=16\\\left(y+2\right)^2=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x+1=4\\y+2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-4\\y+2=-3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Bạn làm như thế này là sai rồi nhé bạn dùng HDT số 3 rồi xét các ước của pt=> nghiệm nha

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

\(=x^2+xy-6x\)

\(=x.\left(x+y-6\right)\)

\(3\left(x-1\right)^2-3x\left(2-5\right)=21\)

\(\Leftrightarrow3x^2-6x+3+9x-21=0\)

\(\Leftrightarrow3x^2+3x-18=0\)

\(\Leftrightarrow3\left(x^2+x-6\right)=0\)

\(\Leftrightarrow3\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Vậy \(S=\left\{2;-3\right\}\)

11)

\(\dfrac{2x}{x+2}\) \(\times\) \(\dfrac{x^{2^{ }}-4}{4}\) - \(\dfrac{x}{2}\) 

= \(\dfrac{2x}{x+2}\) \(\times\) \(\dfrac{x^2-2^2}{4}\) - \(\dfrac{x}{2}\) 

= \(\dfrac{2x}{x+2}\) \(\times\) \(\dfrac{\left(x-2\right)\left(x+2\right)}{4}\) - \(\dfrac{x}{2}\) 

= \(\dfrac{x\left(x-3\right)}{2}\)

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

\(=\dfrac{2x^2+4x+2}{3\left(x^2-x+1\right)}=\dfrac{2\left(x+1\right)^2}{3\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}}\ge0\)

Do đó: \(\dfrac{1}{3}\le\dfrac{x^2+x+1}{x^2-x+1}\)(1)

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

\(=\dfrac{-2x^2+4x-2}{x^2-x+1}=\dfrac{-2\left(x-1\right)^2}{x^2-x+1}\le0\)

Do đó: \(\dfrac{x^2+x+1}{x^2-x+1}\le3\)(2)

Từ (1)và (2) suy ra ĐPCM

IV: Để M nguyên thì \(3x^3-2x^2-6x+5\) ⋮3x-2

=>\(x^2\left(3x-2\right)-6x+4+1\) ⋮3x-2

=>1⋮3x-2

=>3x-2∈{1;-1}

=>3x∈{3;1}

=>x∈{1;1/3}

mà x nguyên

nên x=1

III:

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

=>(x-4)(x-4-x-3)=7

=>-7(x-4)=7

=>x-4=-1

=>x=3

2: \(\left(x-4\right)^2-x+4=0\)

=>\(\left(x-4\right)^2-\left(x-4\right)=0\)

=>(x-4)(x-4-1)=0

=>(x-4)(x-5)=0

=>\(\left[\begin{array}{l}x-4=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=5\end{array}\right.\)

1: \(3ab-6a^2b+3a^3b=3ab\cdot1-3ab\cdot2a+3ab\cdot a^2\)

\(=3ab\left(1-2a+a^2\right)=3ab\left(1-a\right)^2\)

2: \(x^2-7x-y^2-7y\)

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

=(x+y)(x-y)-7(x+y)

=(x+y)(x-y-7)

3: 5a(a-5)-2a+10

=5a(a-5)-2(a-5)

=(a-5)(5a-2)

4: \(5x\left(x-3\right)-x^2+9\)

=5x(x-3)-(x-3)(x+3)

=(x-3)(5x-x-3)

=(4x-3)(x-3)

5: \(9a^2-b^2+4b-4\)

\(=\left(3a\right)^2-\left(b^2-4b+4\right)\)

\(=\left(3a\right)^2-\left(b-2\right)^2\)

=(3a-b+2)(3a+b-2)

6: \(2x^2-x-6\)

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

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

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

7: \(6x^3-15x^2=3x^2\cdot2x-3x^2\cdot5=3x^2\cdot\left(2x-5\right)\)

8: \(9x^2-25=\left(3x\right)^2-5^2=\left(3x-5\right)\left(3x+5\right)\)

9: \(4x^3-4x^2y-x+xy^2\)

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

\(=x\left\lbrack\left(2x-y\right)^2-1\right\rbrack\)

=x(2x-y-1)(2x-y+1)