Đk:\(x\ne0;1;2;3;4\)

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

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

\(\Leftrightarrow\frac{1}{x-4}-\frac{1}{x}=2-\frac{1}{4-x}\)\(\Leftrightarrow\frac{4}{x\left(x-4\right)}=\frac{2x-7}{x-4}\)

Dễ thấy \(x\ne4\) nên nhân 2 vế của pt vừa biến đổi với \(x-4\) ta dc:

\(\Leftrightarrow\frac{4}{x}=2x-7\Leftrightarrow x\left(2x-7\right)=4\)

\(\Leftrightarrow2x^2-7x=4\Leftrightarrow2x^2-7x-4=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x+1\right)=0\)\(\Leftrightarrow x=-\frac{1}{2}\left(x\ne4\right)\)

d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

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

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

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

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

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

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

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)

1/ Ta có : \(\frac{\left(x+2\right)+\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\frac{1}{x-2}\)

=> \(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}=\frac{1}{x-2}\)

=>  \(\left(2x+1\right)\left(x-2\right)=\left(x-1\right)\left(x+2\right)\)

=>   \(2x^2-3x-2=x^2+x-2\)

=>    \(x^2-4x=0\)

=>    \(x\left(x-4\right)=0\)

=>    \(\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

2/ Ta có:   \(\frac{x+3+2\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}=\frac{3}{x+2}\)

=>    \(\frac{x+3+2x+2}{\left(x+1\right)\left(x+3\right)}=\frac{3}{x+2}\)

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

=>    \(\left(x+1\right)\left(x+3\right).3=\left(3x+5\right)\left(x+2\right)\)

=>     \(3x^2+12x+9=3x^2+11x+10\)

=>     \(x=1\)

a)

\(2^x\left(1+2+2^2+2^3\right)=480\)

\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)

Chính Xác 100% là X=5 

k cho mink nhé các pạn

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

\(\Rightarrow\frac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x-1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}\)

\(\Rightarrow\)\(x^2+3x+2+x^2-2x-3=2+x^2+4x+3\)

\(\Rightarrow x^2-3x-6=0\)

.....

\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)

\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)

.....

\(3,\)\(-2x^2+4x+3\)

\(=-2\left(x^2-2x-\frac{3}{2}\right)\)

\(=-2\left[\left(x^2-2x+1\right)-\frac{5}{2}\right]\)

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

Đa thức này lớn nhất =5 khi và chỉ khi \(\left(x-1\right)^2\)nhỏ nhất 

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

chứng minh hay tìm bạn

Tìm x bạn

1/ <=> x² - x -(x² - x)/x³ = 0

<=> (x² - x)(1 - 1/x³) = 0 

Phần còn lại bạn làm tiếp nha điều kiện x#0

Câu 1:
\(F=\frac{\frac{x^3-x}{x+1}+\frac{2x-2}{1+\frac{x}{2}}}{\frac{x^3-3x^2}{x-3}-\frac{2x^2+8}{x+2}}\left(ĐKXĐ:x\ne3;-2;-1\right)\)

\(F=\frac{\frac{x\left(x-1\right)\left(x+1\right)}{x+1}+\frac{2x-2}{1+\frac{x}{2}}}{\frac{x^2\left(x-3\right)}{x-3}-\frac{2x^2+8}{x+2}}\)

\(F=\frac{\frac{\left(x^2-x\right)\left(1+\frac{x}{2}\right)+2x-2}{1+\frac{x}{2}}}{\frac{x^2\left(x+2\right)-2x^2-8}{x+2}}\)

\(F=\frac{\frac{x^2+\frac{x^3}{2}-x-\frac{x^2}{2}+2x-2}{1+\frac{x}{2}}}{\frac{x^3-8}{x+2}}\)

\(F=\frac{\frac{x^2}{2}+\frac{x^3}{2}+x-2}{1+\frac{x}{2}}.\frac{x+2}{x^3-8}\)

Câu 2:

\(G=\frac{\frac{x^4+1}{x^3-1}-x}{\frac{x}{x^2+x+1}-\frac{2}{x-1}}\left(ĐKXĐ:x\ne1\right)\)

\(G=\frac{\frac{x^4+1-x\left(x^3-1\right)}{x^3-1}}{\frac{x\left(x-1\right)-2\left(x^2+x+1\right)}{x^3-1}}\)

\(G=\frac{x+1}{x^3-1}:\frac{x^2-x-2x^2-2x-2}{x^3+1}\)

\(G=\frac{x+1}{-x^2-3x-2}\)

\(G=\frac{x+1}{-\left(x+2\right)\left(x+1\right)}\)

\(G=-\frac{1}{x+2}\)Tại x=2017 ta đc:\(G=-\frac{1}{2+2017}=-\frac{1}{2019}\)

ta có

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

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

\(x+\frac{1}{\frac{5}{3}}=\frac{1}{2+\frac{1}{\frac{12}{5}}}\)

\(x+\frac{3}{5}=\frac{1}{\frac{29}{12}}=\frac{12}{29}\)

\(x=\frac{12}{29}-\frac{3}{5}=-\frac{27}{145}\)

nếu thấy đúng thì tick nha

\(-\frac{17}{95}\)

\(x+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}=\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}\)

\(=>x+\frac{1}{1+\frac{1}{\frac{3}{2}}}=\frac{1}{2+\frac{1}{2+\frac{1}{\frac{3}{2}}}}\)

\(=>x+\frac{1}{1+\frac{2}{3}}=\frac{1}{2+\frac{1}{2+\frac{2}{3}}}\)

\(=>x+\frac{1}{\frac{5}{3}}=\frac{1}{2+\frac{1}{\frac{8}{3}}}\)

\(=>x+\frac{3}{5}=\frac{1}{2+\frac{8}{3}}\)

\(=>x+\frac{3}{5}=\frac{1}{\frac{19}{8}}\)

\(=>x+\frac{3}{5}=\frac{8}{19}\)

\(=>x=\frac{8}{19}-\frac{3}{5}=-\frac{17}{95}\)

\(P=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x^2-x}\right)\)

     \(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)

      \(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{x+1}{x\left(x-1\right)}\)

        \(=\frac{x^2}{x-1}\)

Vì \(P=-\frac{1}{2}\)

\(\Leftrightarrow\frac{x^2}{x-1}=-\frac{1}{2}\)ĐKXĐ:\(x\ne1\)

\(\Rightarrow2x^2=-x+1\)

\(\Rightarrow x^2+x^2+x-1=0\)

\(\Rightarrow x\left(x+1\right)+\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow x\left(x+1\right)\left(x-1\right)=0\)

\(\Rightarrow\)\(x=0\)           \(\Rightarrow\)\(x=0\)(TM)

         \(x+1=0\)          \(x=-1\)(TM)

        \(x-1=0\)           \(x=1\)(KTM)

Vậy để \(P=-\frac{1}{2}\)thì x=0 hoặc x=-1