a) \(ĐKXĐ:\hept{\begin{cases}x\ne0;x\ne2\\x\ne-1\end{cases}}\)

\(Q=1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)

\(\Leftrightarrow Q=1+\left(\frac{x+1}{x^3+1}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right):\frac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)

\(\Leftrightarrow Q=1+\frac{\left(x+1\right)+\left(x+1\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{x\left(x-2\right)}{x^2-x+1}\)

\(\Leftrightarrow Q=1+\frac{x+1+x+1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)

\(\Leftrightarrow Q=1+\frac{-2x^2+4x}{x\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow Q=1+\frac{-2x\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow Q=1+\frac{-2}{x+1}\)

\(\Leftrightarrow Q=\frac{x-1}{x+1}\)

b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(ktm\right)\\x=-\frac{1}{2}\left(tm\right)\end{cases}}\)

Thay \(x=-\frac{1}{2}\)vào Q, ta được :

\(Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}\)

\(\Leftrightarrow Q=\frac{-\frac{3}{2}}{\frac{1}{2}}\)

\(\Leftrightarrow Q=-3\)

c) Để \(Q\inℤ\)

\(\Leftrightarrow x-1⋮x+1\)

\(\Leftrightarrow x+1-2⋮x+1\)

\(\Leftrightarrow2⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)

Vậy để \(Q\inℤ\Leftrightarrow x\in\left\{-2;0;-3;1\right\}\)

Đề không cho biết sớm hơn bao nhiêu phút. Tìm câu hỏi tương tự thì thấy bài giống như vậy ở đây, bạn tham khảo nhé.

Câu hỏi của Hoàng Bảo Ngọc - Toán lớp 8 - Học toán với OnlineMath

Ta có : \(S_{MNP}=S_{ABC}-S_{APC}-S_{CBM}-S_{ABN}\)

\(S_{APC}+S_{PEC}=S_{AEC}=\frac{1}{3}S_{ABC}\)

\(\Rightarrow S_{AEC}=\frac{1}{3}.126=42\left(cm^2\right)\)

Kẻ \(AH\perp CD,EK\perp CD\left(H,K\in CD\right)\)

Ta có : \(\frac{AH.DC}{2}==S_{ADC}=S_{BDC}=3.S_{DEC}=\frac{3}{2}.EK.DC\)

\(\Rightarrow AK=3EK\Rightarrow S_{ADC}=3S_{EPC}\)

\(\Rightarrow S_{EPC}=\frac{1}{4}S_{AEC}=\frac{1}{4}.42=10,5\left(cm^2\right)\)

\(\Rightarrow S_{APC}=42-10,5=31,5\left(cm^2\right)\)

Mà \(S_{CBM}=S_{BCD}-S_{BMD}\)

Tương tự

\(S_{BCD}=\frac{1}{2}.S_{ABC}=\frac{1}{2}.126=63\left(cm^2\right)\)

\(S_{BMC=54cm^2,}S_{ABN}=28cm^2\)

\(\Rightarrow S_{MNP}=126-31,5-54-28=12,5\left(cm^2\right)\)

Bạn ơi vẽ hình hộ mk với

Hơi khó hiểu

cảm ơn

4x^2 + 12x + 9 -3(x^2-16) = x^2 - 4x + 4 +1

4x^2 +12x + 9 -3x^2 + 48 -x^2 + 4x-5 = 0

52 +16x=0

16x = -52

x= -13/4

số bé là 18

số lớn là 40

\(\frac{x-1}{x+1}=\frac{1}{x-1}\)

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

\(\Leftrightarrow x^2-2x+1=x+1\)

\(\Leftrightarrow x^2-3x=0\)

\(\Leftrightarrow x\left(x-3\right)=0\)

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

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

\(\frac{x-1}{x+1}=\frac{1}{x-1}\)

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

\(=>x^2-2x+1=x+1\)

\(=>x^2-3x=0\)

\(=>\hept{\begin{cases}x=0\\x-3=0\end{cases}}\)

\(=>\hept{\begin{cases}x=0\\x=3\end{cases}}\)

Học tốt nhé bạn !!!