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a) 0,5(2y-1) - (0,5-0,2y) + 1 = 0
\(\Leftrightarrow\) y - 0,5 - 0,5 + 0,2y + 1 = 0
\(\Leftrightarrow\) 1,2y - 1 + 1 = 0
\(\Leftrightarrow\) 1,2y = 0
\(\Leftrightarrow\) y = \(\frac{0}{1,2}\)= 0
Vậy y = 0
b) 3(3x-1) + 2 = 5(1-2x) - 1
\(\Leftrightarrow\) 9x - 3 + 2 = 5 - 10x - 1
\(\Leftrightarrow\) 9x + 10x = 5 -1 + 3 -2
\(\Leftrightarrow\) 19x = 5
\(\Leftrightarrow\) x = \(\frac{5}{19}\)
Vậy x = \(\frac{5}{19}\)
c) \(\frac{3x-1}{24}\)- \(\frac{2x+6}{36}\)- 1 = 0
\(\Leftrightarrow\) 3(3x-1) - 2(2x+6) -1.72 = 0
\(\Leftrightarrow\) 9x - 3 - 4x - 12 - 72 = 0
\(\Leftrightarrow\) 5x - 87 = 0
\(\Leftrightarrow\) 5x = 87
\(\Leftrightarrow\) x = \(\frac{87}{5}\)
Vậy x = \(\frac{87}{5}\)
d) \(\frac{11a-4}{7}\)- \(\frac{a-9}{2}\)= 5
\(\Leftrightarrow\) 2(11a-4) - 7(a-9) = 5
\(\Leftrightarrow\) 22a -8 -7a +63 = 5
\(\Leftrightarrow\) 15a + 55 = 5
\(\Leftrightarrow\) 15a = 5 - 55 = -50
\(\Leftrightarrow\) a = \(\frac{-50}{15}\)= \(\frac{-10}{3}\)
Vậy a = \(\frac{-10}{3}\)có vẻ như bạn viết sai đề rồi !
a )
c) (3x−1)/24 ) - 2x+6/36 - 1 = 0
<=> 9x - 3-4x-12-72 =0
<=> 9x-4x = 3+12+72
<=> 5x = 87 => x = 17,4
d ) (11a-4)/7 -( a-9)/2=5
<=> (11a-4)2 - (a-9)7=5.14
<=> 22a-8 - 7a +63 = 70
<=> 22a-7a = 8-63+70
<=> 15a = 15
<=> a = 1
Bạn làm bài kiểm tra hả sao nhiều bài tek. Mk làm mất khá nhiều tg luôn đó 
Có một số câu thì mình không làm được. Mong bạn thông cảm!!!
bài 1:
\(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}\)
<=>\(\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}+-1\right)+\left(\dfrac{x-6}{1998}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}\)
<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}=0\)
<=>(x-2004)\(\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}\right)\)
vì 1/1994+1/1996+1/1998-1/2-1/4-1/6 khác 0
nên x-2004=0=>x=2004
vyaj.......
bài 2:
\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)
<=>\(\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)
<=>\(\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\)
<=>\(\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)
vì 1/15+1/13+1/11+1/9 khác 0
=>x-100=0<=>x=100
a) dk: \(\left\{{}\begin{matrix}a,d\ne0\\5a\ne3b\\5c\ne3d\end{matrix}\right.\) \(VT=\dfrac{5a+3b}{5a-3b}=\dfrac{5.\dfrac{a}{b}+3}{5\dfrac{a}{b}-3}=\dfrac{5.\dfrac{c}{d}+3}{5\dfrac{c}{d}-3}=\dfrac{\dfrac{5c+3d}{d}}{\dfrac{5c-3d}{d}}=\dfrac{5c+3d}{d}.\dfrac{d}{5c-3d}=\dfrac{5c+3d}{5c-3d}=VP\)
b)
\(\left\{{}\begin{matrix}b,d\ne0\\11a^2\ne8b^2\\11c^2\ne8d^2\end{matrix}\right.\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\right)\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7.\dfrac{a^2}{b^2}+3\dfrac{a}{b}}{11\dfrac{.a^2}{b^2}-8}=\dfrac{7.\dfrac{c^2}{d^2}+3\dfrac{c}{d}}{11\dfrac{.c^2}{d^2}-8}=\dfrac{7c^2+3cd}{11c^2-8d^2}=VP\)
\(ĐKXĐ:\hept{\begin{cases}2x+y+1\ge0\\x+y\ge0\end{cases}}\)
Hệ \(\hept{\begin{cases}\sqrt{2x+y+1}-\sqrt{x+y}=1\\3x+2y=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x+y+1}=1+\sqrt{x+y}\\y=\frac{4-3x}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+y+1=1+2\sqrt{x+y}+x+y\\y=\frac{4-3x}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2\sqrt{x+y}\\y=\frac{4-3x}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2=4x+4y\\y=\frac{4-3x}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge0\left(1\right)\\x^2=4x+4.\frac{4-3x}{2}\left(2\right)\\y=\frac{4-3x}{2}\left(3\right)\end{cases}}\)
Giải (2) :
\(\left(2\right)\Leftrightarrow x^2=4x+8-6x\)
\(\Leftrightarrow x^2+2x-8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)
Kết hợp với (1) được x = 2
Thay x = 2 vào (3) được y = -1
Thấy x = 2 ; y = -1 t/m ĐKXĐ
Vậy hệ có nghiệm \(\hept{\begin{cases}x=2\\y=-1\end{cases}}\)