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f, \(\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\)
\(\Leftrightarrow\) \(\frac{18\left(12x+1\right)}{18\left(11x-4\right)}+\frac{2\left(10x-4\right)\left(11x-4\right)}{18\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\)
\(\Leftrightarrow\) 18(12x + 1) + 2(10x - 4)(11x - 4) = (20x + 17)(11x - 4)
\(\Leftrightarrow\) 216x + 18 + 220x2 − 168x + 32 = 220x2 + 107x − 68
\(\Leftrightarrow\) 216x + 18 + 220x2 − 168x + 32 - 220x2 - 107x + 68 = 0
\(\Leftrightarrow\) −59x + 118 = 0
\(\Leftrightarrow\) -59x = -118
\(\Leftrightarrow\) x = 2
Vậy S = {2}
Chúc bạn học tốt!
a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)
\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)
=>-9/10=-9/10(luôn đúng)
b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)
=>347x+780=1552
=>347x=772
hay x=772/347
a) ĐKXĐ: \(x\notin\pm\frac{1}{3}\)
Ta có: \(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
\(\Leftrightarrow\frac{12x+1}{2\left(3x-1\right)}-\frac{9x-5}{3x+1}=\frac{9\left(12x-4x^2-1\right)}{4\left(9x^2-1\right)}\)
\(\Leftrightarrow\frac{2\left(12x+1\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}-\frac{4\left(9x-5\right)\left(3x-1\right)}{4\left(3x+1\right)\left(3x-1\right)}=\frac{9\left(12x-4x^2-1\right)}{4\left(3x+1\right)\left(3x-1\right)}\)
\(\Leftrightarrow72x^2+30x+2-\left(108x^2-96x+20\right)=108x-36x^2-9\)
\(\Leftrightarrow72x^2+30x+2-108x^2+96x-20-108x+36x^2+9=0\)
\(\Leftrightarrow18x-9=0\)
\(\Leftrightarrow9\left(2x-1\right)=0\)
mà 9≠0
nên 2x-1=0
⇔2x=1
hay \(x=\frac{1}{2}\)(tm)
Vậy: \(x=\frac{1}{2}\)
b)ĐKXĐ: x≠0
Ta có: \(x+\frac{1}{x}=x^2+\frac{1}{x^2}\)
\(\Leftrightarrow x+\frac{1}{x}-x^2-\frac{1}{x^2}=0\)
\(\Leftrightarrow\frac{x^3}{x^2}+\frac{x}{x^2}-\frac{x^4}{x^2}-\frac{1}{x^2}=0\)
\(\Leftrightarrow x^3+x-x^4-1=0\)
\(\Leftrightarrow x^3\left(1-x\right)+\left(x-1\right)=0\)
\(\Leftrightarrow x^3\left(1-x\right)-\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow-\left(x-1\right)^2\cdot\left(x^2+x+1\right)=0\)(1)
Ta có: \(x^2+x+1=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)(2)
Từ (1) và (2) suy ra x-1=0
hay x=1(tm)
Vậy: x=1
c) ĐKXĐ: x≠0
Ta có: \(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+2\right)\)
\(\Leftrightarrow\frac{1}{x}+2-\left(\frac{1}{x}+2\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)\left(2-x^2-2\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)\cdot\left(-x^2\right)=0\)(3)
Ta có: 1≠0
x≠0
Do đó: \(\frac{1}{x}\ne0\)
\(\Leftrightarrow\frac{1}{x}+2\ne0\)(4)
Từ (3) và (4) suy ra x=0(ktm)
Vậy: x∈∅
d) ĐKXĐ: x≠0
Ta có: \(\left(x+1+\frac{1}{x}\right)^2=\left(x-1-\frac{1}{x}\right)^2\)
\(\Leftrightarrow\left(x+1+\frac{1}{x}\right)^2-\left(x-1-\frac{1}{x}\right)^2=0\)
\(\Leftrightarrow\left(x+1+\frac{1}{x}+x-1-\frac{1}{x}\right)\left(x+1+\frac{1}{x}-x+1+\frac{1}{x}\right)=0\)
\(\Leftrightarrow2x\cdot\left(2+\frac{2}{x}\right)=0\)
\(\Leftrightarrow4x\left(1+\frac{1}{x}\right)=0\)
mà 4≠0
và x≠0
nên \(1+\frac{1}{x}=0\)
\(\Leftrightarrow\frac{1}{x}=-1\)
hay x=-1(tm)
Vậy: x=-1