== bài dễ , cậu giỏi sao giờ nằm viện mà ======== :v

Ta có:

\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)=\left(-4\right)+1+1+1+1\)

\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}=0\)

\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\right)=0\)

Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\ne0\)

\(\Rightarrow x+329=0\)

\(\Rightarrow x=-329\)

Ta có : \(\frac{x+329-327}{327}+\frac{x+329-326}{326}+\frac{x+329-325}{325}+\frac{x+329-324}{324}\)

\(\Rightarrow\frac{x+329}{327}-1+\frac{x+329}{326}-1+\frac{x+329}{325}-1+\frac{x+329}{324}-1=-4\)

\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}=0\)

\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}\right)=0\)

\(\Rightarrow x+329=0\Rightarrow x=-329\)

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0.\)

\(1+\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}-4+\frac{x+349}{5}=0\)

\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

\(\Rightarrow x+329=0\)

\(\Rightarrow x=-329\)

Study well 

\(\Rightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}-4=0\)

\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Mà \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)

Nên \(x+329=0\Rightarrow x=-329\)

Vậy \(x=-329\)

Chúc bạn học tốt !!!

Bài làm

Không viết lại đề nha !

\(=\frac{\left(x+2\right)}{327}+\frac{\left(x+3\right)}{326}+\frac{\left(x+4\right)}{325}+\frac{\left(x+5\right)}{324}+\frac{\left(x+349\right)}{5}=0\)

\(\Leftrightarrow\frac{\left(x+2\right)}{327}+1+\frac{\left(x+3\right)}{326}+1+\frac{\left(x+4\right)}{325}+1+\frac{\left(x+5\right)}{324}+1+\frac{\left(x+349\right)}{5}\)

\(\Leftrightarrow\frac{\left(x+329\right)}{327}+\frac{\left(x+329\right)}{326}+\frac{\left(x+329\right)}{325}+\frac{\left(x+329\right)}{324}+\frac{\left(x+329\right)}{5}-4=0\)

\(\Leftrightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)

Nên:x+329=0=>x=-329

\(\frac{3}{5}.x-x=\frac{-4}{15}\)

<=>\(x.\left(\frac{3}{5}-1\right)=\frac{-4}{15}\)

<=>\(\frac{-2}{5}.x=\frac{-4}{15}\)

<=>\(x=\frac{-4}{15}:\frac{-2}{5}\)

<=>\(x=\frac{2}{3}\)

vậy \(x=\frac{2}{3}\)

Ta có:\(\frac{3}{5}.x-x=-\frac{4}{15}\)

\(\left(\frac{3}{5}-1\right).x=-\frac{4}{15}\)

\(\left(\frac{3}{5}-\frac{5}{5}\right).x=-\frac{4}{15}\)

\(\left(-\frac{2}{5}\right).x=-\frac{4}{15}\)

\(x=-\frac{4}{15}:\left(-\frac{2}{5}\right)\)

\(x=-\frac{4}{15}.\left(-\frac{5}{2}\right)\)

\(x=\frac{\left(-4\right).\left(-5\right)}{15.2}\)

\(x=\frac{20}{30}=\frac{2}{3}\)

a) \(\left(56\times27+56\times35\right)\div62=56\times\left(27+35\right)\div62=56\times62\div62=56\)

b) \(\frac{0,18\times1230+0,9\times4567\times2+3\times5310\times6}{1+4+7+10+....+52+55-514}\)

\(=\frac{0,18\times1230+\left(0,9\times2\right)\times4567+\left(3\times6\right)\times5310}{1+4+5+.....+52+55-514}\)

\(=\frac{0,18\times1230+0,18\times4567+0,18\times5310}{1+4+7+...+52+55-514}\)

\(=\frac{0,18\times\left(1230+4567+5310\right)}{\left(55+1\right)\times55\div2-514}\)

\(=\frac{0,18\times11107}{971}=\frac{1999,26}{971}\)

mình ra cũng giống bạn Forever _ Alone nhé!!!Chẳng qua mình không biết viết phân số

chư học mà

Bài 1:

a) Cách 1: ta có: \(\frac{x}{y}=\frac{3}{5}\Rightarrow\frac{x}{3}=\frac{y}{5}\)

ADTCDTSBN

có: \(\frac{x}{3}=\frac{y}{5}=\frac{x-y}{3-5}=\frac{-6}{-2}=3\)

=> x/3 = 3 => x = 9

y/5 = 3 => y = 15

KL:....

Cách 2:

ta có: \(\frac{x}{y}=\frac{3}{5}\Rightarrow\frac{x}{3}=\frac{y}{5}=k\Rightarrow\hept{\begin{cases}x=3k\\y=5k\end{cases}}\)

mà x -y = -6 => 3k - 5k = -6 => -2k = 6 => k = 3

=> x = 3k =>...

...

b) ta có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{2y}{6}\)

ADTCDTSBN

có: \(\frac{x}{2}=\frac{2y}{6}=\frac{z}{5}=\frac{x+2y+z}{2+6+5}=\frac{26}{13}=2\)

=> x/2 = 2 => x = 4

y/3 = 2 => y = 6

z/5 = 2 => z = 10

KL:...

cách 2 bn cx lm như cách kia nha

a,C1: \(\frac{x}{y}=\frac{3}{5}\Rightarrow\frac{x}{3}=\frac{y}{5}=\frac{x-y}{3-5}=\frac{-6}{-2}=3\)

=>x=9,y=15

C2: Đặt x/3=y/5=k => x=3k,y=5k

Ta có: x - y = 3k - 5k = -2k = -6 =>k=3

=>x=9,y=15

b, tương tự a

2/

C1: \(\frac{3x-5}{4}=\frac{x-2}{3}\Rightarrow3\left(3x-5\right)=4\left(x-2\right)\Rightarrow9x-15=4x-8\Rightarrow5x=7\Rightarrow x=\frac{7}{5}\) 

C2: \(\frac{3x-5}{4}=\frac{x-2}{3}\Rightarrow3x-5=\frac{x-2}{3}\cdot4\Rightarrow3x-5=\frac{4x-8}{3}\Rightarrow9x-15=4x-8\Rightarrow5x=7\Rightarrow x=\frac{7}{5}\)

Bài 2:

\(\frac{3x-5}{4}=\frac{x-2}{3}\)

\(\Rightarrow9x-15=4x-8\)

\(\Rightarrow9x-4x=-8+15\)

\(5x=7\Rightarrow x=\frac{7}{5}\)

\(\frac{x+2}{327}+\frac{x+3}{326}+....+\frac{x+349}{5}=\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}+4=\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\frac{x+329}{5}=\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+....+\frac{1}{324}+\frac{1}{5}\right)=0\Rightarrow x+329=0\Leftrightarrow x=-329\)

Bạn tham khảo tại đây nhé: Câu hỏi của trần như.

Chúc bạn học tốt!

Ta có : \(\frac{x+6}{2010}+\frac{x+5}{2009}=\frac{x+4}{2008}+\frac{x+3}{2007}\)

\(\Leftrightarrow\frac{x+6}{2010}-1+\frac{x+5}{2009}-1=\frac{x+4}{2008}-1+\frac{x+3}{2007}-1\)

\(\Leftrightarrow\frac{x-2004}{2010}+\frac{x-2004}{2009}=\frac{x-2004}{2008}+\frac{x-2004}{2007}\)

\(\Leftrightarrow\frac{x-2004}{2010}+\frac{x-2004}{2009}-\frac{x-2004}{2008}-\frac{x-2004}{2007}=0\)

\(\Leftrightarrow\left(x-2004\right)\left(\frac{1}{2010}+\frac{1}{2009}-\frac{1}{2008}-\frac{1}{2007}\right)=0\)

Mà : \(\frac{1}{2010}+\frac{1}{2009}-\frac{1}{2008}-\frac{1}{2007}\ne0\)

Nên : x - 2004 = 0 

=> x = 2004

Câu b

Ta có :x + 3 /1.3  +3/3.5 + 3/5.7+...+3/13.15=2 1/5

X + 2/3.(1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15)1=11/5

X+2/3.(1-1/15)=11/5

X+ 2/3.14/15=11/5

X + 28/45=11/5

X = 11/5 -28/45

X=71/45

Câu a  gợi ý 

1/2-1/3/1/6=0 

1/2- 1/3 - 1/6 ) x (1/2 + 2/3 + 3/4 +4/5 + .......+ 2019 /2020 ) =0 

3/4:x=9/10

X = 3/4:9/10

X = 5/6

Câu b

Ta có : x + 3 /1.3  +3/3.5 + 3/5.7+...+3/13.15 = 2 1/5

X + 2/3.(1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15)1 = 11/5

X + 2/3.(1-1/15) = 11/5

X + 2/3.14/15 = 11/5

X + 28/45 = 11/5

X = 11/5 - 28/45

X = 71/45

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