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/5x-4/=/x+2/
5x-4=+-(x+2)
5x-4=x+2
5x-x=2+4
4x=6
x=3/2
+5x-4=-x-2
5x+x=-2+4
6x=2
x=1/3
Mình không chắc đâu
/5x-4/=/x+2/
TH1:5x-4=x+2
4+2=5x-x
6=4x
=>x=1,5
TH2:5x-4=-x+2
4+2=5x+x
6=6x
=>x=1
TH3:-5x-4=x+2
4+2=-5x-x
6=-6x
=>x=-1
Vậy x\(\in\)\(\left\{-1;1;1,5\right\}\)
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\(1-\frac{5x}{2}=\frac{13}{36}-x+\frac{1}{3}\)
\(\Leftrightarrow\frac{2-5x}{2}=\frac{13}{36}-\frac{36x}{36}+\frac{12}{36}\)
\(\Leftrightarrow\frac{36-60x}{36}=\frac{13-36x+12}{36}\Leftrightarrow36-60x=13-36x+12\)
\(\Leftrightarrow11-24x=0\Leftrightarrow x=\frac{11}{24}\)
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Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)
Do đó: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
b: \(\left(\dfrac{a-b}{c-d}\right)^4=\left(\dfrac{bk-b}{dk-d}\right)^4=\left(\dfrac{b}{d}\right)^4\)
\(\dfrac{a^4+b^4}{c^4+d^4}=\dfrac{b^4k^4+b^4}{d^4k^4+d^4}=\dfrac{b^4}{d^4}\)
Do đó: \(\left(\dfrac{a-b}{c-d}\right)^4=\dfrac{a^4+b^4}{c^4+d^4}\)
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A=1+5+52+53+...+550
=> 5A=(5+52+53+...+550)+551
=> 5A-A=551-1
4A=551-1 => \(A=\frac{5^{51}-1}{4}\)
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đề phải tìm x ak
13-2(x+1)=4x
13=4x+2(x +1)
13=13+0=12+1=11+2=10+3=9+4=8+5=7+6
ma 4xla mot so hang vay x=2 vi 8:4=2