b

Chọn B

\(2,\\ a,=2x^2+4x-3x-6-2x^2-4x-2=-3x-8\\ b,=\left[x-2+2\left(x+1\right)\right]^2=\left(x-2+2x+2\right)^2=9x^2\)

Phần BA là phần nào á b mình zoom ảnh thấy phần C

Xét tứ giác GHKI có 

GH//KI

GH=KI

Do đó: GHKI là hình bình hành

Suy ra: GI=HK

\(\dfrac{5}{8}+\dfrac{3}{8}x=1\dfrac{1}{8}\)

\(\Rightarrow\dfrac{5}{8}+\dfrac{3}{8}x=\dfrac{9}{8}\)

\(\Rightarrow\dfrac{3}{8}x=\dfrac{9}{8}-\dfrac{5}{8}\)

\(\Rightarrow\dfrac{3}{8}x=\dfrac{4}{8}\)

\(\Rightarrow x=\dfrac{4}{8}:\dfrac{3}{8}\)

\(\Rightarrow x=\dfrac{4}{8}\cdot\dfrac{8}{3}=\dfrac{4}{3}\)

\(---\)

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

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

\(\Rightarrow\dfrac{1}{3}x-2=-\dfrac{1}{6}\)

\(\Rightarrow\dfrac{1}{3}x=-\dfrac{1}{6}+2\)

\(\Rightarrow\dfrac{1}{3}x=\dfrac{11}{6}\)

\(\Rightarrow x=\dfrac{11}{6}:\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{11}{6}\cdot3=\dfrac{11}{2}\)

\(---\)

\(\left(3x+4\right)\left(2x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x+4=0\\2x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=-4\\2x=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(--\)

\(-4+4x=9x-14\)

\(\Rightarrow4x-9x=-14+4\)

\(\Rightarrow-5x=-10\)

\(\Rightarrow x=\left(-10\right):\left(-5\right)=2\)

\(---\)

\(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)

\(=\left[\left(-\dfrac{14}{25}\right)^2\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}-\dfrac{1}{6}\)

\(=\left(\dfrac{196}{625}\cdot\dfrac{125}{49}\right)\cdot\dfrac{5}{6}-\dfrac{1}{6}\)

\(=\dfrac{4}{5}\cdot\dfrac{5}{6}-\dfrac{1}{6}\)

\(=\dfrac{4}{6}-\dfrac{1}{6}\)

\(=\dfrac{3}{6}\)

\(=\dfrac{1}{2}\)

\(Toru\)

2.1

ĐKXĐ: \(x\ge-\dfrac{1}{16}\)

\(x^2-x-20-2\left(\sqrt{16x+1}-9\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)-\dfrac{32\left(x-5\right)}{\sqrt{16x+1}+9}=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4-\dfrac{32}{\sqrt{16x+1}+9}\right)=0\) (1)

Do \(x\ge-\dfrac{1}{16}\Rightarrow\left\{{}\begin{matrix}\dfrac{32}{\sqrt{16x+1}+9}< \dfrac{32}{9}\\x+4\ge-\dfrac{1}{16}+4=\dfrac{63}{16}>\dfrac{32}{9}\end{matrix}\right.\)

\(\Rightarrow x+4-\dfrac{32}{\sqrt{16x+1}+9}>0\)

Nên (1) tương đương:

\(x-5=0\)

\(\Leftrightarrow x=5\)

Câu 2.2, 2.3 đề lỗi không dịch được

dài quá bạn ơi, tách ra làm 3 đến 4 phần đăng riêng nhé