Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x-1\right)\left(x+1\right)-2\left(2x+3\right)\le\left(x-2\right)^2+x\)
\(\Leftrightarrow x^2-1-4x-6\le x^2-4x+4+x\)
\(\Leftrightarrow x^2-4x-7\le x^2-3x+4\)
\(\Leftrightarrow x^2-4x-x^2+3x\le7+4\)
\(\Leftrightarrow-x\le11\)
\(\Leftrightarrow x\le-11\)
Nhìn sơ qua thì thấy bài 3, b thay -2 vào x rồi giải bình thường tìm m
Bài 2:
a) \(x+x^2=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=0-1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}\)
b) \(0x-3=0\)
\(\Leftrightarrow0x=3\)
\(\Rightarrow vonghiem\)
c) \(3y=0\)
\(\Leftrightarrow y=0\)
1. \(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)
\(\Rightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8=27x^3+27x^2+9x+1\)
\(\Rightarrow-18x^3-33x^2+9x+6=0\)\(\Rightarrow\left(x+2\right)\left(-18x^2+3x+3\right)=0\)
\(\Rightarrow\left(x+2\right)\left(2x-1\right)\left(-9x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{2};x=-\frac{1}{3}\end{cases}}\)
Vậy \(x=-2;x=\frac{1}{2};x=-\frac{1}{3}\)
2. \(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)
\(\Rightarrow\left(\frac{x-1988}{15}-1\right)+\left(\frac{x-1969}{17}-2\right)+\left(\frac{x-1946}{19}-3\right)+\left(\frac{x-1919}{21}-4\right)=0\)
\(\Rightarrow\frac{x-2003}{15}+\frac{x-2003}{17}+\frac{x-2003}{19}+\frac{x-2003}{21}=0\)
\(\Rightarrow x-2003=0\)do \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\ne0\)
Vậy \(x=2003\)
3. Đặt \(\hept{\begin{cases}2009-x=a\\x-2010=b\end{cases}}\)
\(\Rightarrow\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Rightarrow49a^2+49ab+49b^2=19a^2-19ab+19b^2\)
\(\Rightarrow30a^2+68ab+30b^2=0\Rightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5a=-3b\\3a=-5b\end{cases}}\)
Với \(5a=-3b\Rightarrow5\left(2009-x\right)=-3\left(x-2010\right)\)
\(\Rightarrow-2x=-4015\Rightarrow x=\frac{4015}{2}\)
Với \(3a=-5b\Rightarrow3\left(2009-x\right)=-5\left(x-2010\right)\)
\(\Rightarrow2x=4023\Rightarrow x=\frac{4023}{2}\)
Vậy \(x=\frac{4023}{2}\)hoặc \(x=\frac{4015}{2}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{\left(x+3\right)-x}{x\left(x+3\right)}=\frac{3}{10}\)
\(\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)
\(\Rightarrow x\left(x+3\right)=10=2.\left(2+3\right)\)
\(\Rightarrow x=2\)
pt <=> \(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
a.) \(\frac{1}{x+1}-\frac{5}{x+2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow\)\(\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{-5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\) \(x^2-2x-x+2-x^2-2x=-5x-2\)
\(\Leftrightarrow\) \(-2x-x-2x+5x=-2-2\)
\(\Leftrightarrow\) \(0x=-4\)(pt vô nghiệm )
\(\rightarrow\)S= \(\Phi\)
B) \(\frac{3x+2}{2}-\frac{3x+1}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow\frac{3\left(3x+2\right)}{6}-\frac{3x+1}{6}=6.2x+\frac{2.5}{6}\)
\(\Leftrightarrow9x+6-3x+1=12x+10\)
\(\Leftrightarrow9x-3x-12x=10-6-1\)
\(\Leftrightarrow-6x=3\)
\(\Leftrightarrow x=\frac{-1}{2}\)
\(\rightarrow S=\)\(\frac{-1}{2}\)
Câu b bạn sai r nhé. QUên đổi dấu r.
\(\frac{3x+2}{2}-\frac{3x+1}{6}=2x+\frac{5}{3}\)
\(\Rightarrow3\left(3x+2\right)-\left(3x+1\right)=12x+10\)
\(\Leftrightarrow9x+6-3x-1-12x=10\)
\(\Leftrightarrow-6x=10-6+1\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\frac{5}{6}\)
Vậy \(x=-\frac{5}{6}\)là nghiệm của phương trình
Cho x,y,z là các sô dương.Chứng minh rằng x/2x+y+z+y/2y+z+x+z/2z+x+y<=3/4
ĐKXĐ: \(x\ne0\)
Ta có: \(\frac{10}{x}+2\left(\frac{1}{80}-\frac{1}{x}\right)=\frac{2}{15}\)
\(\Leftrightarrow\frac{8}{x}+\frac{1}{40}-\frac{2}{15}=0\)
\(\Leftrightarrow\frac{960}{120x}+\frac{3x}{120x}-\frac{16x}{120x}=0\)
\(\Leftrightarrow960+3x-16x=0\)
\(\Leftrightarrow960-13x=0\)
\(\Leftrightarrow13x=960\)
hay \(x=\frac{960}{13}\)(tm)
Vậy: \(x=\frac{960}{13}\)