Thực hiện phép tính \(\left(\frac{1}{3+2x}+\frac{1}{3-2x}\right)\div\frac{1}{3+2x}\) giúp mình vs,minh cần gấp
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right).\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\) thực hiện phép tính
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right)\times\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(=\left[\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right]\times\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x^2-x+1\right)-3+3\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\times\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2-x+1-3+3x+3}{x+1}\times\frac{3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{x+1}\times\frac{3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{3\left(x+1\right)^2}{\left(x+1\right)\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{3x}{x\left(x+2\right)}-\frac{2x-2}{x\left(x+2\right)}\)
\(=\frac{3x-2x+2}{x\left(x+2\right)}\)
\(=\frac{x+2}{x\left(x+2\right)}\)
\(=\frac{1}{x}\)
Thực hiện phép tính :
\(4\left(\frac{3}{4}x-1\right)+\left(12x^2-3x\right)\div\left(-3x\right)-\left(2x+1\right)\)
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right).\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\) Thực hiện phép tính
thực hiện phép tính: \(\frac{4,5\left[47,375-\left(26\frac{1}{3}-18.0,75\right).0,24:0,88\right]}{17,81:1,37-23\frac{2}{3}:1\frac{5}{6}}\)
tìm các giá trị của x,y thỏa mãn : giá trị tuyệt đối của( \(2x-27^{2007}\) ) \(+\left(3Y+10\right)^{2008}=0\)
GIÚP MK VS. MK CẦN GẤP
thực hiện phép tính bằng -4,8172906
còn lại dễ bạn tự làm nhé!!!
hỏi luffy toán học ấy
b) Ta có: \(\hept{\begin{cases}\left|2x-27\right|^{2007}\ge0\left(\forall x\right)\\\left(3y+10\right)^{2008}\ge0\left(\forall y\right)\end{cases}}\)
Để \(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2008}=0\)
\(\Rightarrow\hept{\begin{cases}\left|2x-27\right|^{2017}=0\\\left(3y+10\right)^{2008}=0\end{cases}}\Rightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{27}{2}=13,5\\y=-\frac{10}{3}\end{cases}}\)
thực hiện phép tính sau:
a)\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
b)\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
\(a,\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(x\left(x+1\right)+x\left(x-3\right)=4x\)
\(x^2+x+x^2-3x=4x\)
\(2x^2-2x=4x\)
\(2x^2-2x-4x=0\)
\(2x\left(x-3\right)=0\)
\(2x=0\Leftrightarrow x=0\)
hoặc
\(x-3=0\Leftrightarrow x=3\)
b) \(ĐKXĐ:x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)
\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)
\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)
\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )
Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)
giúp mình hai câu này với mình đang cần gấp ạ mai học rôi <3
1) thực hiện phép tính
a) \(\frac{x-x}{x+1}-\frac{x+1}{x-1}+\frac{4}{x^2-1}\)
b) \(\frac{x^3y+xy^3}{x^4y}:\left(x^2+y^2\right)\)
c) ( rút gọn nha )
\(\frac{4x-1}{2x^2-2}\)
giúp mình với
Bài 1:
a: \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{-4}{x+1}\)
b: \(=\dfrac{xy\left(x^2+y^2\right)}{x^4y}\cdot\dfrac{1}{x^2+y^2}=\dfrac{x}{x^4}=\dfrac{1}{x^3}\)
c: Đề thiếu rồi bạn
thực hiện phép tính:
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(ĐKXĐ:x\ne3;x\ne-1\)
Nếu x=0 là nghiệm của phương trình
Nếu x khác 0 ta có:
\(\frac{1}{2\left(x-3\right)}+\frac{1}{2\left(x-1\right)}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{x-1+x-3}{\left(x-1\right)\left(x-3\right)}=\frac{4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{2x-4}{\left(x-1\right)\left(x-3\right)}=\frac{4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow2x-4=4\)
\(\Leftrightarrow x=4\)
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne-1;x\ne3\right)\)
<=> \(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
<=> \(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\frac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}=0\)
<=> \(\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
=> 2x2-6x=0
<=> 2x(x-3)=0
<=> \(\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
ĐCĐK x khác -1 và x khác 3 => x=0
Vậy x=0 là nghiệm của phương trình
Bài của @shibo@ chép sai đề ở vế phải
Bài 1 : thực hiện phép tính
a) \(\frac{3x+5}{2.\left(x-1\right)}+\frac{4}{x-2}\)
b) \(\frac{2x^2+1}{4x^2-2x}+\frac{3-3x}{1-2x}+\frac{3}{2x}\)
a) \(\frac{3x+5}{2\left(x-1\right)}+\frac{4}{x-2}=\frac{\left(3x+5\right)\left(x-2\right)+4\cdot2\left(x-1\right)}{2\left(x-1\right)\left(x-2\right)}=\frac{3x^2-6x+5x-10+8x-8}{2\left(x-1\right)\left(x-2\right)}\)
\(=\frac{3x^2+7x-18}{2\left(x-1\right)\left(x-2\right)}\)
b) \(\frac{2x^2+1}{4x^2-2x}+\frac{3-3x}{1-2x}+\frac{3}{2x}=\frac{2x^2+1+4x\left(3-3x\right)+2\cdot3\left(1-2x\right)}{4x\left(1-2x\right)}=\frac{2x^2+1+12-12x+6-12x}{4x\left(1-2x\right)}\)\(=\frac{2x^2-24x+19}{4x\left(1-2x\right)}\)
Đề này... bạn xem lại đi. Chứ thế này thì dùng máy tính cũng không làm nổi T-T
Thực hiện phép tính sau:
\(\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x\left(1-x\right)}{9-x^2}\)
GIÚP MK NAH~ THANKS!!!
= (x+1).(x+3)-(1-x).(x-3)+2x.(1-x)/(x-3).(x+3)
= x^2+4x+3+x^2-4x+3+2x-2x^2/(x+3).(x-3)
= 2x+6/(x+3).(x-3) = 2.(x+3)/(x+3).(x-3) = 2/x-3
k mk nha
\(\frac{x+1}{x-3}\)\(-\)\(\frac{1-x}{x+3}\)\(-\)\(\frac{2x\left(1-x\right)}{9-x^2}\)
\(=\)\(\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)\(-\)\(\frac{\left(1-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)\(-\)\(\frac{2x\left(1-x\right)}{9-x^2}\)
\(=\)\(\frac{x^2+4x+3}{x^2-9}\)\(-\)\(\frac{4x-x^2-3}{x^2-9}\)\(+\)\(\frac{2x-2x^2}{x^2-9}\)
\(=\)\(\frac{x^2+4x+3-4x+x^2+3+2x-2x^2}{x^2-9}\)\(=\)\(\frac{6+2x}{\left(x-3\right)\left(x+3\right)}\)\(=\)\(\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\)\(\frac{2}{x-3}\)