\(2x-\dfrac{2}{3}=\dfrac{3}{2}-1-7x\)

\(2x+7x=\dfrac{3}{2}-1+\dfrac{2}{3}\)

\(9x=\dfrac{7}{6}\)

\(x=\dfrac{7}{6}:9=\dfrac{7}{54}\)

a) \(A=\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)+6\left(x+1\right)\)

\(=x^2-8x+16-x^2+4+6x+6\)

\(=-2x+26\)

Thay x=-2 ta có:

\(A=-2.\left(-2\right)+26=4+26=30\)

Ta có:

\(\widehat{A}=2\widehat{B}=2.2\widehat{C}=4.3\widehat{D}=12\widehat{D}\)

\(\widehat{B}=2\widehat{C}=2.3\widehat{D}=6\widehat{D}\)

Ta có:

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

\(12\widehat{D}+6\widehat{D}+3\widehat{D}+\widehat{D}=360^o\)

\(22\widehat{D}=360^o\)

\(\widehat{D}=\dfrac{180}{11}^o\)

Đổi \(1\dfrac{1}{2}m=\dfrac{3}{2}m\)

Nửa chu vi là:

\(8:2=4\left(m\right)\)

Chiều dài là:

\(\left(4+\dfrac{3}{2}\right):2=\dfrac{11}{4}\left(m\right)\)

Chiều rộng là:

\(\dfrac{11}{4}-\dfrac{3}{2}=\dfrac{5}{4}\left(m\right)\)

Diện tích là:
\(\dfrac{11}{4}\times\dfrac{5}{4}=\dfrac{55}{16}\left(m^2\right)\)

Vậy ...

\(A=\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{97\times99}\)

\(=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{5}-\dfrac{1}{99}=\dfrac{94}{495}\)

\(2x^2-3x-5=x\left(2x-1\right)-\left(2x-1\right)-6=\left(x-1\right)\left(2x-1\right)-6\)

Để \(2x^2-3x-5⋮2x-1\)

=> \(6⋮2x-1\Rightarrow2x-1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Đến đây e giải các trường hợp ra nha

Chia 3 câu kia ra nha

Câu 1

a) \(x^2+8x+a=x^2+20x+100-12x-120+a+20\)

\(=\left(x+10\right)^2-12\left(x+10\right)+a+20\)

để \(x^2+8x+a⋮x+10\) 

thì a+20=0 => a=-20

b) \(\left(x^3-3x^2+ax+b\right)⋮\left(x^2-1\right)\)

Gọi Q là thương của phép chia \(\left(x^3-3x^2+ax+b\right)cho\left(x^2-1\right)\)

⇒ \(x^3-3x^2+ax+b=Q.\left(x^2-1\right)\)

+) x=1 => a+b=2

+) x=-1 => a-b=-4

=> a=-1; b=3