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\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)
\(-18x^3+51x^2+9x-60=0\)
\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)
\(\Rightarrow\frac{x-1}{65}-1+\frac{x-3}{63}-1=\frac{x-5}{61}-1+\frac{x-7}{59}-1\)
\(\Rightarrow\frac{x-66}{65}+\frac{x-66}{63}=\frac{x-66}{61}+\frac{x-66}{59}\)
\(\Rightarrow\frac{x-66}{65}+\frac{x-66}{63}-\frac{x-66}{61}-\frac{x-66}{59}=0\)
\(\Rightarrow x-66=0\).Do\(\frac{x-66}{65}+\frac{x-66}{63}-\frac{x-66}{61}-\frac{x-66}{59}\ne0\)
\(\Rightarrow x=66\)
a) 3x - 2(5 + 2x) =45 - 2x
=> 3x - 10 - 4x = 45 - 2x
=> 3x - 4x + 2x = 45 + 10
=> x = 55
b) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
=> \(\frac{x-3}{5}=\frac{2x+17}{3}\)
=> 5(2x + 17) = 3(x - 3)
=> 10x + 85 = 3x - 9
=> 7x = -94
=> x = -94/7
c) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)
=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x-33}{7}\)
=> \(\frac{10x-6}{12}-\frac{21x-3}{12}=\frac{4x-33}{7}\)
=> \(\frac{-11x-3}{12}=\frac{4x-33}{7}\)
=> (-11x - 3).7 = (4x - 33).12
= -77x - 21 = 48x - 396
=> x = 3
d) (x - 1)(5x + 3) = (3x - 8)(x - 1)
=> (x - 1)(5x + 3) - (3x - 8)(x -1) = 0
=> (x - 1)(2x + 11) = 0
=> \(\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5,5\end{cases}}\)
e) (x - 1)(x2 + 5x - 2) - (x3 - 1) = 0
=> (x - 1)(x2 + 5x - 2) - (x - 1)(x2 + x + 1) = 0
=> (x - 1)(4x - 3) = 0
=> \(\orbr{\begin{cases}x-1=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0,75\end{cases}}\)
f) \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
=> \(\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)
=> \(\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
=> \(\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
=> x - 50 = 0 (Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\))
=> x = 50
b, \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
\(\Leftrightarrow\frac{x-3}{5}=\frac{17+2x}{3}\Leftrightarrow3x-9=85+10x\)
\(\Leftrightarrow-7x=94\Leftrightarrow x=-\frac{94}{7}\)
f, sửa : \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)=0\)
\(\Leftrightarrow x=-66\)
a) \(\left|2-\frac{3}{2}x\right|-4=x+2\)
=> \(\left|2-\frac{3}{2}x\right|=x+2+4\)
=> \(\left|2-\frac{3}{2}x\right|=x+6\)
ĐKXĐ : \(x+6\ge0\) => \(x\ge-6\)
Ta có: \(\left|2-\frac{3}{2}x\right|=x+6\)
=> \(\orbr{\begin{cases}2-\frac{3}{2}x=x+6\\2-\frac{3}{2}x=-x-6\end{cases}}\)
=> \(\orbr{\begin{cases}2-6=x+\frac{3}{2}x\\2+6=-x+\frac{3}{2}x\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{2}x=-4\\\frac{1}{2}x=8\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{8}{5}\\x=16\end{cases}}\) (tm)
b) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)
=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
=> \(\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\)
=> x = 1/4
hoặc x = 0 hoặc x = 1/2
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{\left(x+1\right)}{65}-1+\frac{\left(x+3\right)}{63}-1=\frac{\left(x+5\right)}{61}-1+\frac{\left(x+7\right)}{59}\)
\(\Leftrightarrow\left(x-66\right).\left(\frac{1}{65}+\frac{1}{63}\right)=\left(x-66\right).\left(\frac{1}{61}+\frac{1}{59}\right)\)
\(\Leftrightarrow\left(x-66\right).\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
\(\Rightarrow x=66\)
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
=> \(\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)=\left(\frac{x+5}{61}+1\right)+\left(\frac{x+7}{59}+1\right)\)
=> \(\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
=> \(\left(x+66\right).\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
=> x + 66 = 0
=> x = 0 - 66
=> x = -66
a, \(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)
\(\Leftrightarrow1+\frac{x+16}{49}+1+\frac{x+18}{47}=\frac{x+20}{45}-1+2\)
\(\Leftrightarrow\frac{x+16+49}{49}+\frac{x+18+47}{47}=\frac{x+20+45}{45}\)
\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\)
\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)
Ta có: \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\)>0
\(\Rightarrow x+65=0\)
\(\Leftrightarrow x=-65\)
Vậy x = -65
b, \(\frac{x-69}{30}+\frac{x-67}{32}+\frac{x-65}{34}=\frac{x-63}{36}+\frac{x-61}{38}+\frac{x-59}{40}\)
\(\Leftrightarrow\frac{x-69}{30}-1+\frac{x-67}{32}-1+\frac{x-65}{34}-1+\frac{x-63}{36}-1+\frac{x-61}{38}-1+\frac{x-59}{40}-1\)
\(\Leftrightarrow\frac{x-99}{30}+\frac{x-99}{32}+\frac{x-99}{34}-\frac{x-99}{36}-\frac{x-99}{38}-\frac{x-99}{40}=0\)
\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\right)=0\)
Vì \(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\)>0
\(\Rightarrow x-99=0\)
\(\Leftrightarrow x=99\)
Vậy x =99
\(|3x-5|=|\frac{1}{2}+3|\)
\(\Rightarrow|3x-5|=|\frac{7}{2}|=\frac{7}{2}\)
\(\orbr{\begin{cases}3x-5=\frac{7}{2}\\3x-5=-\frac{7}{2}\end{cases}\Rightarrow}\orbr{\begin{cases}3x=\frac{17}{2}\\3x=-\frac{3}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{17}{6}\\x=-\frac{1}{2}\end{cases}}}\)