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HT
1
Y
20 tháng 6 2019
\(ĐK:\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\end{matrix}\right.\)
+ pt \(\Leftrightarrow\frac{x+4}{\left(x-1\right)\left(x-2\right)}-\frac{x+1+2x+5}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{\left(x+4\right)\left(x-3\right)-3\left(x+2\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\left(x^2+x-12\right)-3\left(x^2-4\right)=0\)
\(\Leftrightarrow-2x^2+x=0\) \(\Leftrightarrow-x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\) ( TM )
4 tháng 5 2019
b, \(\frac{1}{x-1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\left(ĐKXĐ:x\ne\pm1;x\ne2\right)\)
\(\Leftrightarrow\)\(\frac{1}{x-1}+\frac{5}{2-x}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow\)\(\frac{\left(x+1\right)\left(2-x\right)+5\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(2-x\right)\left(x-1\right)}=\frac{15\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(2-x\right)}\)
Suy ra:
\(\Leftrightarrow\)(x+1)(2-x)+5(x-1)(x+1) = 15(x-1)
\(\Leftrightarrow\)2x-x2-x+2+5x2-5 = 15x-15
\(\Leftrightarrow\)2x-x2-x+5x2-15x = -15+5-2
\(\Leftrightarrow\)4x2-14x = -12
\(\Leftrightarrow4x^2-14x+12=0\)
\(\Leftrightarrow4x^2-8x-6x+12=0\)
\(\Leftrightarrow\)4x(x-2) - 6(x-2) = 0
\(\Leftrightarrow\left(x-2\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(kotm\right)\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất x = \(\frac{3}{2}\)
24 tháng 1 2022
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
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
LN
9 tháng 5 2020
a,\(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{3+6x}{16x^2-1}\)
ĐKXĐ: x≠1/4, x≠-1/4
⇔\(-\frac{3}{4x-1}=\frac{2}{4x+1}-\frac{3+6x}{16x^2-1}\)
⇔\(\frac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\frac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\frac{3+6x}{16x^2-1}\)
⇒-12x-3=8x-2-3-6x
⇔8x-6x+12x=-3+2+3
⇔14x=2
⇔x=1/7(tmđk)
Vậy phương trình có nghiệm là x=1/7
b, \(\frac{5-x}{4x^2-8x}+\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\) (2)
ĐKXĐ: x≠0, x≠2
(2)⇔\(\frac{2\left(5-x\right)}{2.4x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4.\left(x-1\right)}{4.2x\left(x-2\right)}+\frac{x}{8.x\left(x-2\right)}\)
⇒10-2x+7x-14=4x-4+x
⇔-2x+7x-4x-x=-4-10+14
⇔0x=0
⇔ x∈R
Vậy phương trình có nghiệm là x∈R và x≠0, x≠2
c, \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\) (3)
ĐKXĐ: x≠0
(3)⇒x(x+1)(x2-x+1)-x(x-1)(x2+x+1)=3
⇔x4+x-x4+x=3
⇔2x=3
⇔x=3/2(tmđk)
Vậy phương trình có nghiệm là x=3/2
D
27 tháng 6 2018
\(a,\)
\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}.\frac{4x^2-4x+16}{x^2-4}\right):\frac{16}{x+2}.\frac{x^2+3x+2}{x^2+x+1}\)\(ĐKXĐ:x\ne\pm2\)
\(A=[\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right).4\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x+2\right)}]:\frac{16}{x+2}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=[\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}].\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}.\frac{x+2}{16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{16\left(x+2\right)}{\left(x+2\right)^2.16}.\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}\)
\(A=\frac{-\left(x+1\right)}{x^2+x+1}\)
\(B=\frac{x^2+x-2}{x^3-1}\)\(ĐKXĐ:x\ne1\)
\(B=\frac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(B=\frac{x+2}{x^2+x+1}\)
\(b,\)
Ta có:
\(A+B=\frac{-\left(x+1\right)}{x^2+x+1}+\frac{x+2}{x^2+x+1}\)
\(=\frac{-x-1+x+2}{x^2+x+1}\)
\(=\frac{1}{x^2+x+1}\)
\(\Rightarrow A+B=\frac{1}{x^2+x+1}=\frac{1}{x^2+2.x.\left(\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{1}{\left(x+\frac{1}{2}\right)^2}+\frac{3}{4}\)
Vì:\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\le\frac{1}{\frac{3}{4}}\)
\(\Rightarrow A+B\le\frac{4}{3}\)
\(\Rightarrow GTLN\)của \(A+B=\frac{4}{3}\Leftrightarrow x+\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{-1}{2}\left(TMĐK\right)\)
Vậy........
kq : x=1
\(\frac{4x^x}{3-x}=2\)
\(\Rightarrow4x^x=2\left(3-x\right)\)
\(\Rightarrow4x^x=6-2x\)
\(\Rightarrow x=1\)