Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Cái bài đầu giải BPT bn ghi cái dj ak ,mik cx k hỉu nữa
V mik giải bài 2 nghen, sửa lại đề bài đầu rồi mik giải cho
\(3x-3=|2x+1|\)
Điều kiện: \(3x-3\ge0\Leftrightarrow3x\ge3\Leftrightarrow x\ge1\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=3x-3\\2x+1=-3x+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1-3\\2x+3x=-1+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=-3\\5x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\left(n\right)\\x=\frac{2}{5}\left(l\right)\end{cases}}}\)
Vậy S={3}
Cài đề câu b ,bn xem lại nhé!
\(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}>\frac{x^2}{7}-\frac{2x-3}{5}\)
\(\Leftrightarrow\frac{2x-3}{35}+\frac{5x\left(x-2\right)}{35}-\frac{5x^2}{35}+\frac{7\left(2x-3\right)}{35}>0\)
\(\Leftrightarrow2x-3+5x\left(x-2\right)-5x^2+7\left(2x-3\right)>0\)
\(\Leftrightarrow2x-3+5x^2-10x-5x^2+14x-21>0\)
\(\Leftrightarrow6x-24>0\)
\(\Leftrightarrow x>4\)
VẬY TẬP NGHIỆM CỦA BẤT PHƯƠNG TRÌNH LÀ : S = { \(x\text{\x}>4\)}
\(\frac{6x+1}{18}+\frac{x+3}{12}\le\frac{5x+3}{6}+\frac{12-5x}{9}\)
\(\Leftrightarrow\frac{6\left(6x+1\right)}{108}+\frac{9\left(x+3\right)}{108}\le\frac{18\left(5x+3\right)}{108}+\frac{12\left(12-5x\right)}{108}\)
\(\Leftrightarrow36x+6+9x+27\le90x+54+144-60x\)
\(\Leftrightarrow36x+6+9x+27-90x-54-144+60x\le0\)
\(\Leftrightarrow15x-165\le0\)
\(\Leftrightarrow x\le11\)
VẬY TẬP NGHIỆM CỦA BẤT PHƯƠNG trình ..........
tk mk nka !!! chúc bạn học tốt !!!
b) \(\frac{x-3}{x-2}+\frac{x+2}{x-4}=-1\)
\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)+x^2-4}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{x^2-7x+12+x^2-4}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{2x^2-7x+8}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{2x^2-7x+8}{\left(x-2\right)\left(x-4\right)}=-1\)
.................
a) \(\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
\(\Rightarrow\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x+3\right)\left(x-1\right)}{\left(x+1\right)\left(x^2+x+1\right)}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
\(\Rightarrow\frac{2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)}{x^3-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
\(\Rightarrow\left(x^3-1\right)\left[2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)\right]=\left(x^3-1\right)\left(2x-1\right)\left(2x+1\right)\)
\(\Rightarrow2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)=\left(2x-1\right)\left(2x+1\right)\)
\(\Rightarrow2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)-\left(2x-1\right)\left(2x+1\right)=0\)
\(\Rightarrow2x^2+2x+2+2x^2-2x+3x-3-\left(4x^2-1\right)=0\)
\(\Rightarrow2x^2+2x+2+2x^2-2x+3x-3-4x^2+1=0\)
\(\Rightarrow3x=0\)
\(\Rightarrow luon-dung-voi-moi-x\)
\(a,ĐKXĐ:x\ne\pm\frac{1}{2}\)
Ta có: \(\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)
\(\Leftrightarrow2\left(2x-1\right)-3\left(2x+1\right)=4\)
\(\Leftrightarrow4x-2-6x-3=4\)
\(\Leftrightarrow-2x=9\)
\(\Leftrightarrow x=-\frac{9}{2}\)(Tm ĐKXĐ)
Vậy pt có nghiệm duy nhất \(x=-\frac{9}{2}\)
\(b,ĐKXĐ:x\ne\pm1;-3\)
Ta có: \(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow2x\left(x^2+2x-3\right)+18x+18=\left(2x-5\right)\left(x^2-1\right)\)
\(\Leftrightarrow2x^3+4x^2-6x+18x+18=2x^3-2x-5x^2+5\)
\(\Leftrightarrow9x^2+14x+13=0\)
\(\Leftrightarrow\left(9x^2+14x+\frac{49}{9}\right)+\frac{68}{9}=0\)
\(\Leftrightarrow\left(3x+\frac{7}{3}\right)^2+\frac{68}{9}=0\)
Pt vô nghiệm
\(c,ĐKXĐ:x\ne1\)
Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow x^2+x+1+2x^2-5=x-1\)
\(\Leftrightarrow3x^2=3\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=\pm1\)
Kết hợp vs ĐKXĐ được x = -1
Vậy pt có nghiệm duy nhất x = -1
làm lần lượt nha(bài nào k bt bỏ qua)
\(a,\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow\frac{2\left(2x-1\right)-3\left(2x+1\right)}{4x^2-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow-2x-5=4\)
\(\Rightarrow-2x=9\)
\(\Rightarrow x=\frac{9}{-2}\)
\(ĐKXĐ:x\ne-1\)
Từ phương trình suy ra \(\frac{x^2-x+1}{x^3+1}+\frac{2x^2+1}{x^3+1}+\frac{2x^2\left(x-1\right)\left(x+1\right)}{x^3+1}=2x\)
\(\Leftrightarrow\frac{2x^4+x^2-x+2}{x^3+1}=2x\)
\(\Leftrightarrow2x^4+x^2-x+2=2x^4+2x\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\left(tmđk\right)\)
Vậy tập nghiệm của phương trình \(S=\left\{1;2\right\}\)
Đặt x2 + 2x = a ta có
\(\frac{1}{a-3}\)+ \(\frac{18}{a+2}\)= \(\frac{18}{a+1}\)
<=> a2 - 15a + 56 = 0
<=> a = (7;8)
Thế vô tìm được nghiệm
\(\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
Quy đồng mẫu chung :
\(\frac{2.\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(4x^2-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
Sau đó ta khử mẫu:
\(\Rightarrow\)\(2x^2+2x+2+2x^2+x-3=4x^2-1\)
\(\Rightarrow\)\(2x^2+2x+2x^2+x-4x^2=-1-2+3\)
\(\Rightarrow\)\(3x=0\)
\(\Rightarrow\)\(x=0\)
Vậy bạn tự kết luận
ĐKXĐ: \(x\ne1\)
\(\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
\(\Leftrightarrow\)\(\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{4x^2-1}{x^3-1}\)
\(\Leftrightarrow\)\(\frac{2x^2+2x+2}{x^3-1}+\frac{2x^2+x-3}{x^3-1}=\frac{4x^2-1}{x^3-1}\)
\(\Rightarrow\)\(2x^2+2x+2+2x^2+x-3=4x^2-1\)
\(\Leftrightarrow\)\(4x^2+3x-1=4x^2-1\)
\(\Leftrightarrow\)\(3x=0\)
\(\Leftrightarrow\)\(x=0\) (thỏa mãn)
Vậy....
a) \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\left(x\ne1\right)\)
\(\Leftrightarrow\frac{1}{x-1}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4}{x^2+x+1}=0\)
\(\Leftrightarrow\frac{1\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4x-4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+1+2x^2-5-4x+4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{3x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{3x}{x^2+x+1}=0\)
=> 3x=0
<=> x=0 (tmđk)
a) \(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}\)
Đặt \(x^2-2x+3=t\left(t\ge2\right)\), khi đó phương trình trở thành:
\(\frac{1}{t-1}+\frac{2}{t}=\frac{6}{t+1}\)
\(\Leftrightarrow\frac{t\left(t+1\right)+t^2-1}{\left(t-1\right)t\left(t+1\right)}=\frac{6t\left(t-1\right)}{\left(t-1\right)t\left(t+1\right)}\)
\(\Leftrightarrow t\left(t+1\right)+t^2-1=6t\left(t-1\right)\)
\(\Leftrightarrow2t^2+t-1=6t^2-6t\)
\(\Leftrightarrow-4t^2+7t-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=\frac{7+\sqrt{33}}{8}\\t=\frac{7-\sqrt{33}}{8}\end{cases}}\left(ktmđk\right)\)
Vậy phương trình vô nghiệm.
\(\frac{2}{x^2-x+1}-\frac{1}{x+1}=\frac{2x+2-x^2+x-1}{x^3+1}=\frac{3x+1-x^2}{x^3+1}\\ \)
\(\Rightarrow3x+1-x^2=2x-1\Rightarrow\left(x+1\right)\left(x-2\right)=0\Rightarrow x=2\)
\(\frac{2}{x^2-x+1}=\frac{1}{x+1}+\frac{2x-1}{x^3+1}\)
\(\frac{2}{x^2-x+1}=\frac{1}{x+1}+\frac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
ĐKXĐ : x\(\ne\)-1
MTC : (x+1)(x^2-x+1)
\(\frac{2x+2}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^2-x+1+2x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
2x+2=x^2-x+1+2x-1
-x^2+2x-2x+x=-2+1-1
x-x^2=-2
x(1-x)=-2
.....................