\(\frac{5x-1}{10}+\frac{2x+3}{6}=\frac{x-8}{15}-\frac{x}{30}\)

\(\Leftrightarrow3\left(5x-1\right)+5\left(2x+3\right)=2\left(x-8\right)-x\)

\(\Leftrightarrow15x-3+10x+15=2x-16-x\)

\(\Leftrightarrow15x+10x-2x+x=-16+3-15\)

\(\Leftrightarrow24x=-28\)

\(\Leftrightarrow x=\frac{-28}{24}=\frac{-7}{6}\)

Vậy ... 

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

<=>x²+5x-3x-15=x²-2x+3x-6

<=>x²+2x-15=x²+x-6

<=>x²+2x-15-(x²+x-6)=0

<=>x²+2x-15-x²-x+6=0

<=>x-9=0

<=>x=9

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

\(x^2+5x-3x-15=x^2-2x+3x-6\)

\(x^2+5x-3x-x^2+2x-3x=15-6\)

\(x=9\)

=(a²+ac+ba+bc)(a²+ab+ac)+b²c² 
=(a²+ab+ac)²+bc(a²+ab+ac)+b²c²      (2)
Đặt m=a²+ab+ac và n=bc
 (2) viết lại thành : m²+mn+n² = m²+2*m*n/2+n²/4+n²*3/4
                                          \(=\left(m+\frac{n}{2}\right)^2+\frac{3n^2}{4}\ge0\left(đpcm\right)\)

1/ Sửa đề a+b=1

\(M=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

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

Thay a+b=1 vào M ta được:

\(M=1-3ab+3ab\left[1-2ab\right]+6a^2b^2\)

\(=1-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2/ Đặt \(A=\frac{2n^2+7n-2}{2n-1}=\frac{\left(2n^2-n\right)+\left(8n-4\right)+2}{2n-1}=\frac{n\left(2n-1\right)+4\left(2n-1\right)+2}{2n-1}=n+4+\frac{2}{2n-1}\)

Để \(A\in Z\Leftrightarrow2n-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Ta có bảng:

Vậy n={1;0}

câu 4c phải là x-1 mới đúng chứ

Vẽ đường chéo AC.

Hình thang ABCD có: AB//CD

                               => \(\widehat{BAC}=\widehat{ACD}\)

Xét \(\Delta ABC\)và \(\Delta ACD\)có:

- \(AB=CD\)

- \(\widehat{BAC}=\widehat{ACD}\)

- \(AC\) là cạnh chung

Do đó:  \(\Delta ABC=\Delta ACD\left(c-g-c\right)\)

         \(=>\widehat{DAC}=\widehat{BCA}\)(2 góc tương ứng nằm ở vị trí so le trong)

           \(=>AD//BC\)

Hình vẽ:

Vẽ đường chéo AC

Hình thang ABCD có: AB//CD

                               \(\Rightarrow\widehat{BAC}=\widehat{ACD}\)

Xét \(\Delta ABC\)và \(\Delta ACD\)có:

\(-AB=CD\)

\(-\widehat{BAC}=\widehat{ACD}\)

\(-AC\)là cạnh chung

Do đó: \(\Delta ABC=\Delta ACD\left(c-g-c\right)\)

          \(\Rightarrow\widehat{DAC}=\widehat{BCA}\)(2 góc tương ứng nằm ở vị trí so le trong)

          \(\Rightarrow AD//BC\)

\(1.x^4+x^3-x^2-x\)

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

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

\(2.x^5-x^3+x^2-1\)

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

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

\(3.x^4+2x^3y^2+y^2\)( ko biết làm )

\(4.x+y\left(x-1\right)-1\)

\(=\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(1+y\right)\)

\(5.a^3+a^2b-a^2c-abc\)

\(=\left(a^3-a^2c\right)+\left(a^2b-abc\right)=a^2\left(a-c\right)+ab\left(a-c\right)\)

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

\(6.a^3-b^3+\left(a-b\right)^2\)

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

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