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a, \(2+\frac{3}{x-5}=1\Leftrightarrow\frac{3}{x-5}=-1\)
\(\Leftrightarrow x-5=\frac{3}{-1}=-3\Leftrightarrow x=2\)
Vậy .............
b, ....................
\(\Leftrightarrow\frac{x-9}{x^2-3^2}-\frac{2}{x+3}=\frac{1}{x-3}\)
\(\Leftrightarrow\frac{x-9}{\left(x-3\right)\left(x+3\right)}-\frac{2x-6}{\left(x-3\right)\left(x+3\right)}-\frac{x+3}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x-9-2x+6-x+3}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-2x}{\left(x-3\right)\left(x+3\right)}=0\Rightarrow-2x=0\Rightarrow x=0\)
Vậy .............
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
1) (2x - 3)2 = 4x2 - 8
<=> 4x2 - 12x + 9 = 4x2 - 8
<=> 12x + 9 = -8
<=> 12x = -17
<=> x = 17/12
1) (2x - 3)^2 = 4x^2 - 8
<=> 4x^2 - 12x + 9 = 4x^2 - 8
<=> 4x^2 - 12x + 9 - 4x^2 = -8
<=> -12x + 9 = -8
<=> -12x = -8 - 9
<=> -12x = -17
<=> x = 17/12
2) x - (x + 2)(x - 3) = 4 - x^2
<=> x - x^2 + 3x - 2x + 6 = 4 - x^2
<=> 2x - x^2 + 6 = 4 - x^2
<=> 2x - x^2 + 6 + x^2 = 4
<=> 2x + 6 = 4
<=> 2x = 4 + 6
<=> 2x = 10
<=> x = 5
3) 3x - (x - 3)(x + 1) = 6x - x^2
<=> 3x - x^2 - x + 3x + 3 = 6x - x^2
<=> 5x - x^2 + 3 = 6x - x^2
<=> 5x - x^2 + 3 + x^2 = 6x
<=> 5x + 3 = 6x
<=> 3 = 6x - 5x
<=> 3 = x
4) 3x/4 = 6
<=> 3x = 6.4
<=> 3x = 24
<=> x = 8
5) 7 + 5x/3 = x - 2
<=> 21 + 5x = 3x - 6
<=> 5x = 3x - 6 - 21
<=> 5x = 3x - 27
<=> 5x - 3x = -27
<=> 2x = -27
<=> x = -27/2
6) x + 4 = 2/5x - 3
<=> 5x + 20 = 2x - 15
<=> 5x + 20 - 2x = -15
<=> 3x + 20 = -15
<=> 3x = -15 - 20
<=> 3x = -35
<=> x = -35/3
7) 1 + x/9 = 4/3
<=> x/9 = 4/3 - 1
<=> x/9 = 1/3
<=> x = 3
a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)
\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)
\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)
\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0
\(x-1=0\)
\(x=1\)
\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
a)\(2+\frac{3}{x-5}=1\)
\(\Rightarrow\frac{3}{x-5}=-1\)
\(\Rightarrow3=-x+5\)
\(\Leftrightarrow x+3=5\)
\(\Rightarrow x=2\)
Ta thấy \(\left(x-3\right)\left(2x+3\right)=2x^2-3x-9.\)
\(\left(1\right)\Leftrightarrow\frac{x}{x-3}-\frac{2x^2+9}{\left(x-3\right)\left(2x+3\right)}=\frac{1}{2x+3}\)
ĐK: \(x\ne3\)và \(x\ne-\frac{3}{2}\)
\(\Rightarrow x\left(2x+3\right)-2x^2-9=x-3\)
\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\Leftrightarrow2x=6\Leftrightarrow x=2\)
Thỏa mãn ĐK
Các trường hợp khác làm tương tự
5 - 2 x 3 + x + 1 x - 1 3 x - 1 = x + 2 1 - 3 x 9 x - 3 Đ K X Đ : x ≠ 1 3 ⇔ 5 - 2 x 3 x - 1 3 3 x - 1 + 3 x + 1 x - 1 3 3 x - 1 = x + 2 1 - 3 x 3 3 x - 1
⇔ (5 – 2x)(3x – 1) + 3(x + 1)(x – 1) = (x + 2)(1 – 3x)
⇔ 15x – 5 – 6 x 2 + 2x + 3 x 2 – 3 = x – 3 x 2 + 2 – 6x
⇔ - 6 x 2 + 3 x 2 + 3 x 2 + 15x + 2x – x + 6x = 2 + 5 + 3
⇔ 22x = 10 ⇔ x = 5/11 (thỏa mãn)
Vậy phương trình có nghiệm x = 5/11