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\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)
\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)
\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)
\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)
\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)
\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)
\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)
\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)
\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)
\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)
\(< =>5x-35=32x-8< =>32x-5x=-35+8\)
\(< =>27x=-27< =>x=-1\)
\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)
\(< =>12=6x-3< =>6x=12+3\)
\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)
\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)
\(< =>8x-4=9x+3< =>9x-8x=-4-3\)
\(< =>9x-8x=-7< =>x=-7\)
\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)
\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)
\(< =>12x-7x=14-4< =>5x=10\)
\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)
\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)
\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)
\(< =>6x-4x=4+6< =>2x=10\)
\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)
\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)
\(< =>\left(x+1\right)\left(x+1\right)=3.3\)
\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)
\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)
\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)
\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)
Bài 1 :Bỏ dấu ngoặc
2007-(7-3+4)
= 2007 -7+3-4
= 1999
6+[(-5) + 4 - 1 ]
= 6-5+4-1
=4
5-[(-6+8-2]
= 5+6-8+2
=5
-10+(7-3+1)
= -10 +7-3+1
= -5
Bài 3 Tìm x
\(\dfrac{1}{3} = \dfrac{x}{6}\)
\(<=> x= \dfrac{1.6}{3}\)
\(<=> x=2\)
\(\frac{3+x}{5+y}=\frac{3}{5}\)
=> (3 + x).5 = (5 + y).3
=> 15 + 5x = 15 + 3y
=> 5x = 3y
=> \(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{16}{8}=2\)
=> x = 2.3 = 6; y = 2.5 = 10
\(\frac{x-7}{y-6}=\frac{7}{6}\)
=> (x - 7).6 = (y - 6).7
=> 6x - 42 = 7y - 42
=> 6x = 7y
=> \(\frac{x}{7}=\frac{y}{6}=\frac{x-y}{7-6}=\frac{-4}{1}=-4\)
=> x = -4.7 = -28; y = -4.6 = -24
a) \(\frac{2}{3x}=\frac{5}{2}\Leftrightarrow3x=\frac{4}{5}\Rightarrow x=\frac{4}{15}\)
b) \(\frac{x-3}{4}=\frac{1}{2}\Rightarrow x-3=2\Rightarrow x=5\)
c) \(\frac{5}{24+x}=\frac{7}{12}\Rightarrow24+x=\frac{60}{7}\Rightarrow x=\frac{-108}{7}\)
d) \(-6x=18\Rightarrow x=-3\)
Lời giải:
$\frac{4}{x}+\frac{x}{3}=\frac{5}{6}$
$\frac{12+x^2}{3x}=\frac{5}{6}$
$12+x^2=\frac{5}{2}x$
$2x^2-5x+24=0$
$x^2+(x-2,5)^2=-17,75<0$ (vô lý)
Do đó không tồn tại $x$ thỏa mãn.