Những câu hỏi liên quan
Đinh Lan Hương
Xem chi tiết
Hoàng Kiệt
Xem chi tiết
Takishima Hotaru
Xem chi tiết
Phương An
30 tháng 1 2017 lúc 20:00

\(P=\frac{\left(\frac{2x-3}{4x^2-12x+5}+\frac{2x-8}{13x-2x^2-20}-\frac{3}{2x-1}\right)}{\left(\frac{21+2x-8x^2}{4x^2+4x-3}\right)}+1\)

\(=\frac{\left(\frac{2x-3}{4x^2-2x-10x+5}+\frac{-2\left(4-x\right)}{8x-2x^2+5x-20}-\frac{3}{2x-1}\right)}{\left(\frac{21-12x+14x-8x^2}{4x^2+6x-2x-3}\right)}+1\)

\(=\frac{\left(\frac{2x-3}{2x\left(2x-1\right)-5\left(2x-1\right)}+\frac{-2\left(4-x\right)}{2x\left(4-x\right)-5\left(4-x\right)}-\frac{3}{2x-1}\right)}{\left(\frac{3\left(7-4x\right)+2x\left(7-4x\right)}{2x\left(2x+3\right)-\left(2x+3\right)}\right)}+1\)

\(=\frac{\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}+\frac{-2\left(4-x\right)}{\left(2x-5\right)\left(4-x\right)}-\frac{3}{2x-1}\right)}{\frac{\left(7-4x\right)\left(3+2x\right)}{\left(2x+3\right)\left(2x-1\right)}}+1\)

\(=\frac{\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}+\frac{-2\left(2x-1\right)}{\left(2x-1\right)\left(2x-5\right)}-\frac{3\left(2x-5\right)}{\left(2x-1\right)\left(2x-5\right)}\right)}{\frac{7-4x}{2x-1}}+1\)

\(=\frac{2x-3-4x+2-6x+15}{\left(2x-1\right)\left(2x-5\right)}\times\frac{2x-1}{7-4x}+1\)

\(=\frac{14-8x}{2x-5}\times\frac{1}{7-4x}+1\)

\(=\frac{2\left(7-4x\right)}{2x-5}\times\frac{1}{7-4x}+\frac{2x-5}{2x-5}\)

\(=\frac{2+2x-5}{2x-5}\)

\(=\frac{2x-3}{2x-5}\)

Bình luận (0)
Hoàng Kiệt
Xem chi tiết
JiYoonMin
Xem chi tiết
Vũ Phương Huyền
Xem chi tiết
Huỳnh Xuân Mai
Xem chi tiết
PRO ANIME
Xem chi tiết
Khánh Ngọc
18 tháng 8 2021 lúc 18:29

a. ĐKXĐ : \(x\ne\frac{1}{2};\frac{5}{2};4;-\frac{3}{2};\frac{1\pm\sqrt{43}}{2}\)

 \(A=\left(\frac{2x-3}{4x^2-12x+5}+\frac{3x-8}{13x-2x^2-20}-\frac{3}{2x-1}\right):\frac{21+2x-2x^2}{4x^2+4x-3}+\)

\(=\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\frac{3x-8}{\left(2x-5\right)\left(x-4\right)}-\frac{3}{2x-1}\right).\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)

\(=\frac{\left(2x-3\right)\left(x-4\right)-\left(3x-8\right)\left(2x-1\right)-3\left(2x-5\right)\left(x-4\right)}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}.\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)

\(=\frac{-10x^2+47x-56}{\left(2x-5\right)\left(x-4\right)}.\frac{2x+3}{-2x^2+2x+21}+1\) số to wa

Bình luận (0)
 Khách vãng lai đã xóa
việt anh ngô
Xem chi tiết
Nguyễn Lê Phước Thịnh
4 tháng 2 2020 lúc 21:45

Bài 1:

d)ĐKXĐ: \(x\ne8\)

Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)

\(\Leftrightarrow\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3x-24}=0\)

\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)

MTC=24(x-8)

\(\Leftrightarrow\frac{36}{24\left(x-8\right)}+\frac{72x-480}{24\left(x-8\right)}+\frac{3x-24}{24\left(x-8\right)}-\frac{104x-816}{24\left(x-8\right)}=0\)

\(\Leftrightarrow36+72x-480+3x-24-104x+816=0\)

\(\Leftrightarrow348-29x=0\)

\(\Leftrightarrow-29x+348=0\)

\(\Leftrightarrow x=\frac{-348}{-29}=12\)

Vậy: x=12

e) ĐKXĐ: \(x\ne\pm1\)

Ta có: \(\frac{6}{x^2-1}+5=\frac{8x-1}{4x+4}-\frac{12x-1}{4-4x}\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4x+4}+\frac{12x-1}{4-4x}=0\)

\(\Leftrightarrow\frac{6}{\left(x-1\right)\left(x+1\right)}+5-\frac{8x-1}{4\left(x+1\right)}+\frac{12x-1}{4\left(1-x\right)}=0\)

MTC=4(x+1)(x-1)

\(\Leftrightarrow\frac{24}{4\left(x-1\right)\left(x+1\right)}+\frac{20x^2-20}{4\left(x-1\right)\left(x+1\right)}-\frac{8x^2-9x+1}{4\left(x-1\right)\left(x+1\right)}-\frac{12x^2-11x-1}{4\left(x-1\right)\left(x+1\right)}=0\)

\(\Leftrightarrow24+20x^2-20-8x^2+9x-1-12x^2+11x+1=0\)

\(\Leftrightarrow20x+4=0\)

\(\Leftrightarrow20x=-4\)

\(\Leftrightarrow x=-\frac{4}{20}=-0,2\)(loại)

Vậy: x không có giá trị

g) Ta có: \(\frac{\frac{x+1}{x-1}-\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\frac{1}{2}\)

\(\Leftrightarrow\frac{\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}}{\frac{x-1}{x-1}+\frac{x+1}{x-1}}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}}{\frac{2x}{x-1}}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-1}{2x}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{4x\cdot\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\cdot2x}-\frac{1}{2}=0\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{2}=0\)

MTC=2(x+1)

\(\Leftrightarrow\frac{2}{2\left(x+1\right)}-\frac{x+1}{2\left(x+1\right)}=0\)

\(\Leftrightarrow2-x+1=0\)

\(\Leftrightarrow1-x=0\)

\(\Leftrightarrow x=1\)(loại vì không thỏa mãn ĐKXĐ)

Vậy: x không có giá trị

Bình luận (0)
 Khách vãng lai đã xóa