\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)

\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)=0\)

Vì \(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)

\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)

\(\Leftrightarrow\left(x-1\right)\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\right)=0\)

\(\Leftrightarrow x=1\)

\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)

\(< =>\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)

\(< =>\left(x-1\right)\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)=0\)

Do \(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\)

\(< =>x-1=0\)

\(< =>x=1\)

By Titu's Lemma we easy have:

\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)

\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

1) có \(2y\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)

\(\Rightarrow\left(\sqrt{xy}+\frac{1}{4\sqrt{xy}}\right)^2+\frac{15}{16xy}+\frac{1}{2}\ge\frac{15}{16}\cdot4+\frac{1}{2}=\frac{17}{4}\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)

A (min) khi

\(\frac{4}{x}=\frac{1}{4y}=>x=16y\)

\(y=\frac{5}{4.17};x=\frac{5.16}{4.17}\)\(x.y=\frac{5.5}{17.17}\)

A(min)=2.\(2\sqrt{\frac{1}{xy}}=2.\frac{17}{5}=\frac{34}{5}\)

Bạn có thể giải thích rõ hơn cho mình dc ko?? Mình ko hiểu cho lắm!

Oh ok mình hiểu r!! Thanks

​

O

\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)

\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)

\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)

=> x - 29 = 0

=> x = 29.

1) \(x^2y+x\left(2y-1\right)=7\)

\(\Leftrightarrow x^2y+2xy-x=7\)

\(\Leftrightarrow xy\left(x+2\right)-x-2=7-2\)

\(\Leftrightarrow xy\left(x+2\right)-\left(x+2\right)=5\)

\(\Leftrightarrow\left(xy-1\right)\left(x+2\right)=5\)

\(\Rightarrow\)xy - 1 và x + 2 là ước của 5 là \(\pm1;\pm5\)

đến đây tự lm đc

2 ) \(B=\frac{255}{1}+\frac{254}{2}+\frac{253}{3}+....+\frac{3}{253}+\frac{2}{254}+\frac{1}{255}\)

\(=\left(\frac{254}{2}+1\right)+\left(\frac{253}{3}+1\right)+....+\left(\frac{2}{254}+1\right)+\left(\frac{1}{255}+1\right)+1\)

\(=\frac{256}{2}+\frac{256}{3}+....+\frac{256}{255}+\frac{256}{256}\)

\(=256\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{255}+\frac{1}{256}\right)=256A\)

\(\Rightarrow\frac{B}{A}=256=16^2\) Là số CP (đpcm)

ừ, z câu 2 đâu?

\(x+3,5-\frac{4}{7}=\frac{3}{2}-5\frac{1}{8}\)

\(x+3,5-\frac{4}{7}=-\frac{29}{8}\)

\(x+3,5=-\frac{29}{8}+\frac{4}{7}\)

\(x+3,5=-\frac{117}{56}\)

      \(\Leftrightarrow x=-\frac{367}{56}\)

vì x + 2 = y + 1 = z + 3 => x = y - 1 = z + 1 ; y = x + 1 = z + 2; z = x + 1 = y - 2  và z < x < y

ta có (x-1/3).(y-1/2).(z-5)=0 => ta có 3 TH

TH₁ z - 5 = 0 => z = 5 ; y = 7 ; x = 4

TH₂ x - 1/3 = 0 => x = 1/3 ; y = 4/3 ; z = -2/3

TH₃ y - 1/2 = 0 => y = 1/2 ; x = -1/2 ; z = -3/2

nhớ cho mik nha 

Ta có:

\(\left(x-\frac{1}{2}\right).\left(y-\frac{1}{2}\right).\left(z-5\right)=0\)

\(\Rightarrow x-\frac{1}{2}=0;y-\frac{1}{2}=0\)hoặc \(z-5=0\)

Với \(x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)\(\Rightarrow\)\(x+2=\frac{1}{3}+2=\frac{7}{3}=y+1=z+3\)\(\Rightarrow y=...;z=...\)

Với \(y-\frac{1}{2}=0\Rightarrow y=\frac{1}{2}\)\(\Rightarrow....\)

Với \(z-5=0\)\(\Rightarrow.....\)

B tự làm nốt nhé

nhớ k nha