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a) \(\left(x-11\right)+\frac{3x}{x-11}=3+\frac{33}{x-11}\)
\(\Leftrightarrow x+\frac{3x}{x-11}-\frac{33}{x-11}=14\)
\(\Leftrightarrow x^2-11x+3x-33=14x-154\)
\(\Leftrightarrow x^2-22x+121=0\)
\(\Leftrightarrow\left(x-11\right)^2=0\Leftrightarrow x=11\)
Vậy .......
b) \(\frac{7-2x}{x-1}=\frac{1-4x}{x+2}\Leftrightarrow\left(7-2x\right)\left(x+2\right)=\left(1-4x\right)\left(x-1\right)\)
\(\Leftrightarrow7x-2x^2+14-4x=x-4x^2-1+4x\)
\(\Leftrightarrow2x^2=-15\)(vô lí)
Vậy pt vô nghiệm
c) \(\frac{3-2x}{x+1}=2+\frac{1-4x}{x-2}\)
\(\Leftrightarrow\left(3-2x\right)\left(x-2\right)=2\left(x+1\right)\left(x-2\right)+\left(1-4x\right)\left(x+1\right)\)
\(\Leftrightarrow3x-2x^2-6x+4x=2x^2+2x-4x-4+x-4x^2+1-4x\)
\(\Leftrightarrow6x=-3\Leftrightarrow x=-\frac{1}{2}\)
Vậy.........
(gửi trước 3 câu)
d) \(\frac{109x-4}{111x+1}-1=0\Leftrightarrow109x-4=111x+1\Leftrightarrow2x=-5\Leftrightarrow x=-\frac{5}{2}\)
Vậy x=-5/2
e) \(\frac{x^2-7}{x}=x-\frac{1}{2}\Leftrightarrow\frac{x^2-7}{x}-\frac{x^2}{x}=-\frac{1}{2}\Leftrightarrow-\frac{7}{x}=\frac{1}{2}\Leftrightarrow x=-14\)
f) \(\frac{x+1}{x+2}=3\Leftrightarrow x+1=3x+6\Leftrightarrow2x=7\Leftrightarrow x=\frac{7}{2}\)
Câu A
X + (X+1) + (X+3) +...+ (X+2003) = 2004
Số số hạng trong tổng 1 + 3 + ... + 2003 là
(2003 - 1) : 2 + 1 = 1002
Tổng dãy 1 + 3 + ... + 2003 là:
(1 + 2003) * 1002 : 2 = 1004004
=> (1003.X) + 1004004 = 2004
=> (1003.X)= 2004 - 1004004
=> 1003.X = - 1002000
X = - 1002000/1003
E chỉ giải đc đến đây thui!!!!!!!!!!!!!!! :)))
x + ( x + 1) + (x + 3) ... + (x + 2003) = 2004
x + x + x + ... + x (có 1003 x) + 1 + 3 + 5 + ... + 2003 = 2004
x . 1003 + 1004004 = 2004
x . 1003 = 2004 - 1004004
x . 1003 = -1002000
x = -1002000 : 1003
x = -999,00299 = ~-999
\(\frac{x-3}{11}+\frac{x+1}{3}=\frac{x+7}{9}-1\)
\(\Leftrightarrow\frac{9\left(x-3\right)}{99}+\frac{33\left(x+1\right)}{99}=\frac{11\left(x+7\right)}{99}-\frac{99}{99}\)
\(\Leftrightarrow\frac{9\left(x-3\right)+33\left(x+1\right)}{99}=\frac{11\left(x+7\right)-99}{99}\)
\(\Leftrightarrow9\left(x-3\right)+33\left(x+1\right)=11\left(x+7\right)-99\)
\(\Leftrightarrow9x-27+33x+33=11x+77-99\)
\(\Leftrightarrow42x+6=11x-22\Leftrightarrow42x-11x=-6-22\)
\(\Leftrightarrow31x=-28\Leftrightarrow x=-\frac{28}{31}\)
Vậy phương trình có tập nghiệm S={-28/31}
\(ĐKXĐ:x\ne3;x\ne5;x\ne4;x\ne6\)
\(\frac{x}{x-3}-\frac{x}{x-5}=\frac{x}{x-4}-\frac{x}{x-6}\)
\(\Rightarrow\frac{x}{x-3}-\frac{x}{x-5}-\frac{x}{x-4}+\frac{x}{x-6}=0\)
\(\Rightarrow x\left(\frac{1}{x-3}-\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(tm\right)\\\frac{1}{x-3}-\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}=0\left(1\right)\end{cases}}\)
\(\left(1\right)\Rightarrow\frac{1}{x-3}+\frac{1}{x-6}=\frac{1}{x-5}+\frac{1}{x-4}\)
\(\Rightarrow\frac{2x-9}{\left(x-3\right)\left(x-6\right)}=\frac{2x-9}{\left(x-5\right)\left(x-4\right)}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\left(tm\right)\\\left(x-3\right)\left(x-6\right)=\left(x-5\right)\left(x-4\right)\left(2\right)\end{cases}}\)
\(\left(2\right)\Leftrightarrow x^2-9x+18=x^2-9x+20\)
\(\Leftrightarrow0=2\left(L\right)\)
Vậy pt có 2 nghiệm \(\left\{0;\frac{9}{2}\right\}\)
a, 2(4x - 7 ) = 3(x + 1) + 18
⇌ 8x -14 = 3x + 3 + 18
⇌ 5x = 35 ⇌ x = 7
→ S = \(\left\{7\right\}\)
b, ( 2x - 1 )2 - 4x ( x - 3 ) = -11
⇌ 4x2 - 2x + 1 - 4x2 + 12 = -11
⇌ 10x = -12
⇌ x = \(-\frac{12}{10}\)
→ S = \(\left\{-\frac{12}{10}\right\}\)
c, ( 2x - 5 )2 - ( x + 2 )2 = 0
⇌ ( 2x - 5 -x + 2 )2 = 0
⇌ ( x - 3 )2 = 0
⇌ x - 3 = 0 ⇌ x = 3
→ S = \(\left\{3\right\}\)
d, ( x - 6 ) ( x + 1 ) = 2(x + 1)
⇌ ( x - 6 - 2 ) ( x+ 1) = 0
⇌ x2 - 7x - 8 =0
⇌ ( x - 8 ) ( x + 1 ) = 0
⇒\(\left\{{}\begin{matrix}x-8=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)
→ S = \(\left\{8;-1\right\}\)
e, \(\frac{x-3}{2}=2-\frac{1-2x}{5}\)
⇌ 5( x - 3) = 20 - 2(1 - 2x)
⇌ 5x - 4x = 15 + 20 + 2
⇌ x = 37
→ S = \(\left\{37\right\}\)
g, \(\frac{3x+2}{2}+\frac{5-2x}{3}=\frac{11}{6}\)
⇌ 3(3x + 2) + 2(5 - 2x) = 11
⇌ 6x + 6 + 10 - 4x = 11
⇌ 2x = -5
⇌ x = \(-\frac{5}{2}\)
→ S = \(\left\{-\frac{5}{2}\right\}\)
h, \(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{9x-66}{x^2-4}\)
⇌ (x - 2)2 - 3(x - 2) = 9x - 66
⇌ x2 - 4x + 4 - 3x - 6 = 9x - 66
⇌ x2 -16 + 64 = 0
⇌ (x - 8)2 = 0
⇌ x - 8 = 0
⇌ x = 8
→ S = \(\left\{8\right\}\)
ko ghi lại đề TL:
a,\(\frac{6x+30}{12}+\frac{9-6x}{12}=\frac{12x}{12}-\frac{14+2x}{12}\)
6x+30+9-6x=12x-14-2x
39=10x-14
10x=39+14=53
x=5,3
vậy.........
câu B ,c làm tương tự
học tốt
a)
\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)
\(\Leftrightarrow\frac{49-13x}{12}=0\)
\(\Rightarrow49-13x=0\)
\(\Rightarrow x=\frac{-49}{13}\)
b)
\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)
\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)
\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)
\(\Leftrightarrow\frac{-3x}{4}=0\)
\(\Rightarrow-3x=0\)
\(\Rightarrow x=0\)
\(\frac{x-3}{11}+\frac{x+1}{3}=\frac{x+7}{9}-11\)
\(9\left(x-3\right)+33\left(x+1\right)=11\left(x+7\right)-1089\)
\(9x-27+33x+3=11x+77-1089\)
\(42x+6=11x+77-1089\)
\(42x+6=11x-1012\)
\(42x-11x=-1012-6\)
\(31x=-1018\Leftrightarrow x=-\frac{1018}{31}\)