a)\(x\ne1;x\ne-1\)

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

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

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

\(A=\frac{1}{x-1}\)

bạn ơi phần a là gì ạ + hay - 1 thế

b) Đặt \(B=7+|2x+3|\)

Vì\(|2x+3|\ge0;\forall x\)

\(\Rightarrow7+|2x+3|\ge7+0;\forall x\)

Hay\(B\ge7;\forall x\)

Dấu "=" xảy ra\(\Leftrightarrow|2x+3|=0\)

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

Vậy MIN B=7\(\Leftrightarrow x=-\frac{3}{2}\)

phần c mình ko rõ lắm bạn ghi lại đề nhé

d) Đặt \(D=|x+2|+|x-3|+2\)

\(=|x+2|+|3-x|+2\ge|x+2+3-x|+2\)

Hay \(D\ge7\)

Dấu "=" xảy ra\(\Leftrightarrow\left(x+2\right)\left(3-x\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x+2\ge0\\3-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x+2< 0\\3-x< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge-2\\x\le3\end{cases}}\)hoặc \(\hept{\begin{cases}x< -2\\x>3\end{cases}\left(loai\right)}\)

\(\Leftrightarrow-2\le x\le3\)

Vậy MIN B =7 \(\Leftrightarrow-2\le x\le3\)

\(b)\) \(\left(2x-1\right)^{2012}=\left(2x-1\right)^{2010}\)

\(\Leftrightarrow\)\(\left(2x-1\right)^{2010}.\left(2x-1\right)^2=\left(2x-1\right)^{2010}\)

\(\Leftrightarrow\)\(\left(2x-1\right)^2=1\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=2\\2x=0\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{2}{2}\\x=\frac{0}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)

Vậy \(x=0\) hoặc \(x=1\)

Chúc bạn học tốt ~ 

2/xy<=1/x^2+1/y^2=1/2

=>xy>=4

Dấu = xảy ra khi x=y=2

(x+y)^2>=4xy>=16

=>x+y>=4

Dấu = xảy ra khi x=y=2

=>x+y+xy+2023>=2023+4+4=2031 

Dấu = xảy ra khi x=y=2

A=(1/x-2 - (2x/(2-x)(2+x) - 1/2+x) ) *(2-x)/x 
=(1/x-2 - x^2+5x-2/(2-x)(2+x))*2-x/x 
=(-x^3-4x^2+12x/(x-2)(2-x)(2+x))*2-x/x 
= - x(x-2)(x+6)(2-x)/x(x-2)(2-x)(2+x) 
= - x+6/x+2

a) \(P=\frac{3x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

\(P=\frac{3\left(x-9\right)}{\left(x-3\right)\left(x-2\right)}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

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

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

\(P=\frac{9-3x-9+x^2-2x^2+4x-x+2}{\left(x-2\right)\left(3-x\right)}\)

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

b) Thay \(x=-\frac{1}{2}\) vào (*) ta có:

\(P=\frac{2-\left(-\frac{1}{2}\right)^2}{\left[\left(-\frac{1}{2}\right)-2\right]\left[3-\left(-\frac{1}{2}\right)\right]}=\frac{2-\frac{1}{4}}{-\frac{5}{2}.\frac{7}{2}}=-\frac{\frac{7}{4}}{\frac{5}{2}.\frac{7}{2}}=-\frac{7}{35}=-\frac{1}{5}\)

c) \(\frac{2-x^2}{\left(x-2\right)\left(3-x\right)}< 0\)

\(\Leftrightarrow2-x^2< 0\)

\(\Leftrightarrow-x^2< -2\)

\(\Leftrightarrow x^2>2\)

\(\Leftrightarrow\hept{\begin{cases}x< -\sqrt{2}\\-\sqrt{2}< x< \sqrt{2}\\x>2\end{cases}}\)

Vậy: ...

a) /x+\(\frac{4}{15}\)/ - / -3,75/ = -2,15

=> \(\orbr{\begin{cases}x+\frac{4}{15}+3,75=-2,15\\x+\frac{4}{15}+3,75=2,15\end{cases}}\)

=> ....v.....v giải ra ( từng th )

bài khác tương tự

ai biet tra loi cho mik voi mik dag rat can gap

chi can lam bai c,d,e la duoc

nhung bai con lai mik tu lam duoc