Vì 6n+7 chia hết cho 2n-1

=> (6n+7):(2n-1)=1

     6n+7=1.(2n-1)=2n-1

     6n+7+1=2n

     6n+8=2n

     8=2n-6n=(-4)n

     n=8:(-4)=-2

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

Viết đoạn mở bài cho bài văn tả con vật em vừa làm trong tiết tập làm văn trước theo cách mở bài gián tiếp :

\(\text{a) }A=x^2-10x+25\\ A=x^2-2\cdot x\cdot5+5^2\\ A=\left(x-5\right)^2\\ Do\text{ }\left(x-5\right)^2\ge0\forall x\\ \Leftrightarrow A\ge0\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-5\right)^2=0\\ \Leftrightarrow x-5=0\\ \Leftrightarrow x=5\\ \text{Vậy }A_{\left(Min\right)}=0\text{ }khi\text{ }x=5\)

\(\text{b) }B=x^2+y^2-x+6y+10\\ B=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\\ B=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\\ Do\text{ }\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\ \left(y+3\right)^2\ge0\forall y\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x;y\\ \text{Dấu "=" xảy ra khi: }\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\\\left(y+3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\\ \text{ Vậy }B_{\left(Min\right)}=\dfrac{3}{4}\text{ }khi\text{ }x=\dfrac{1}{2};y=-3\)

\(\text{c) }C=2x^2-6x+10\\ C=\left(2x^2-6x+\dfrac{9}{2}\right)+\dfrac{11}{2}\\ C=2\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{2}\\ C=2\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{11}{2}\\ C=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{2}\\ Do\text{ }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{2}\ge\dfrac{11}{2}\\ \text{Dấu "=" xảy ra khi: }\\ \left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{Vậy }C_{\left(Min\right)}=\dfrac{11}{2}khi\text{ }x=\dfrac{3}{2}\)

\(\)

\(\text{a) }x^4-4x^2+4x-1\\ \\=x^4-\left(4x^2-4x+1\right)\\ \\ =\left(x^2\right)^2-\left(2x-1\right)^2\\ \\=\left(x^2-2x+1\right)\left(x^2+2x-1\right)\\ \\=\left(x-1\right)^2\left(x^2+2x-1\right)\)

\(\text{b) }4x^2-y^2+4x+1\\ \\=\left(4x^2+4x+1\right)-y^2\\ \\=\left(2x+1\right)^2-y^2\\ \\=\left(2x+1+y\right)\left(2x+1-y\right)\)

\(\text{Ta có : }A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ A=\dfrac{2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\\ A=\dfrac{3^{32}-1}{2}\\ Do\text{ }\dfrac{3^{32}-1}{2}< 3^{32}-1\\ nên\text{ }\Rightarrow A< B\)

Vậy \(A< B\)

100 lít dầu thj động cơ đó chạy đc trong số giờ là

100:0,8=800[ giờ]

Vay.....