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4 tháng 2 2020
Đề nghỉ ghi cái đề? @@ Rút gọn đúng ko?
\(Đkxđ:\hept{\begin{cases}x\ne\pm2\\x\ne-\frac{6}{13}\end{cases}}\)
\(A=\left[\frac{\left(1+2x\right)\left(x-2\right).3-2x\left(x+2\right)-4x^2}{6\left(x^2-4\right)}\right].\frac{12\left(2-x\right)}{6.13x}\)
\(=\left[\frac{3x-6+6x^2-12x-2x^2-4x-4x^2}{6\left(x^2-4\right)}\right].\frac{12\left(2-x\right)}{6+13x}\)
\(=\frac{13x+6}{6\left(x+2\right)\left(2-x\right)}.\frac{12\left(2-x\right)}{6+13x}\)
\(=\frac{2}{x+2}\)
4 tháng 2 2020
\(A=\left(\frac{1+2x}{4+2x}-\frac{x}{3x-6}+\frac{2x^2}{12-3x^2}\right)\times\frac{24-12x}{6+13x}\)
\(=\left(\frac{1+2x}{2\left(2+x\right)}+\frac{x}{3\left(2-x\right)}+\frac{2x^2}{3\left(4-x^2\right)}\right)\times\frac{2.\left(12-x\right)}{6+13x}\)
\(=\left(\frac{\left(1+2x\right).3.\left(2-x\right)}{2.3.\left(2+x\right)\left(2-x\right)}+\frac{2x\left(2+x\right)}{2.3.\left(2-x\right)\left(2+x\right)}+\frac{2.2x^2}{2.3.\left(2-x\right)\left(2+x\right)}\right)\times\frac{2.\left(12-x\right)}{6+13x}\)
\(=\left(\frac{6+12x-3x-6x^2+4x+2x^2+4x^2}{6\left(2-x\right)\left(2+x\right)}\right)\times\frac{2\left(12-x\right)}{6+13x}\)
\(=\frac{6+13x}{6\left(2-x\right)\left(2+x\right)}\times\frac{2\left(12-x\right)}{6+13x}\)
\(=\frac{12-x}{\left(2-x\right)\left(2+x\right)}=\frac{12-x}{4-x^2}\)
27 tháng 4 2020
a, \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\frac{8x-3}{4}-\frac{6x-4}{4}=\frac{4x-2}{4}+\frac{x+3}{4}\)
\(8x-3-6x-4=4x-2+x+2\)
\(2x-7=5x\Leftrightarrow2x-5x+7=0\Leftrightarrow-3x=-7\Leftrightarrow x=\frac{7}{3}\)
27 tháng 4 2020
b, \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
\(\frac{3x+9}{6}-\frac{2x-2}{6}=\frac{x+5}{6}+\frac{6}{6}\)
\(3x+9-2x-2=x+5+6\)
\(x+7=x+11\Leftrightarrow-4\ne0\)
9 tháng 6 2020
a, 2(4x - 7 ) = 3(x + 1) + 18
⇌ 8x -14 = 3x + 3 + 18
⇌ 5x = 35 ⇌ x = 7
→ S = \(\left\{7\right\}\)
b, ( 2x - 1 )2 - 4x ( x - 3 ) = -11
⇌ 4x2 - 2x + 1 - 4x2 + 12 = -11
⇌ 10x = -12
⇌ x = \(-\frac{12}{10}\)
→ S = \(\left\{-\frac{12}{10}\right\}\)
c, ( 2x - 5 )2 - ( x + 2 )2 = 0
⇌ ( 2x - 5 -x + 2 )2 = 0
⇌ ( x - 3 )2 = 0
⇌ x - 3 = 0 ⇌ x = 3
→ S = \(\left\{3\right\}\)
d, ( x - 6 ) ( x + 1 ) = 2(x + 1)
⇌ ( x - 6 - 2 ) ( x+ 1) = 0
⇌ x2 - 7x - 8 =0
⇌ ( x - 8 ) ( x + 1 ) = 0
⇒\(\left\{{}\begin{matrix}x-8=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
→ S = \(\left\{8;-1\right\}\)
e, \(\frac{x-3}{2}=2-\frac{1-2x}{5}\)
⇌ 5( x - 3) = 20 - 2(1 - 2x)
⇌ 5x - 4x = 15 + 20 + 2
⇌ x = 37
→ S = \(\left\{37\right\}\)
g, \(\frac{3x+2}{2}+\frac{5-2x}{3}=\frac{11}{6}\)
⇌ 3(3x + 2) + 2(5 - 2x) = 11
⇌ 6x + 6 + 10 - 4x = 11
⇌ 2x = -5
⇌ x = \(-\frac{5}{2}\)
→ S = \(\left\{-\frac{5}{2}\right\}\)
h, \(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{9x-66}{x^2-4}\)
⇌ (x - 2)2 - 3(x - 2) = 9x - 66
⇌ x2 - 4x + 4 - 3x - 6 = 9x - 66
⇌ x2 -16 + 64 = 0
⇌ (x - 8)2 = 0
⇌ x - 8 = 0
⇌ x = 8
→ S = \(\left\{8\right\}\)
4 tháng 2 2020
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
HH
3 tháng 7 2020
a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)
\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)
Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\)
Ta được :
\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)
Từ đó ta lắp vào ( 1 ) tính được x
\(\frac{2x^2-x-1}{x^2-13x+1}=\frac{2x^2-2x+x-1}{x^2-13x+1}=\frac{2x\left(x-1\right)+x-1}{x^2-13x+1}=\frac{\left(2x+1\right)\left(x-1\right)}{x^2-13x+1}\)
Đề hơi lạ
Khôi Bùi , hơ hơ