Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(C1:=3+1-3y\)
\(=4-3y\)
\(C2:\)
\(a.=3x\left(2y-1\right)\)
\(b.=\left(x-y\right)\left(x+y\right)+4\left(x+y\right)\)
\(=\left(x-y+4\right)\left(x+y\right)\)
\(C3:\)
\(a.6x^2+2x+12x-6x^2=7\)
\(14x=7\)
\(x=\frac{1}{2}\)
\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)
\(\frac{26}{5}x=-\frac{13}{2}\)
\(x=-\frac{13}{2}\times\frac{5}{26}\)
\(x=-\frac{5}{4}\)
Bạn Moon làm kiểu gì vậy ?
1) \(\left(3x^2y^2+x^2y^2\right):\left(x^2y^2\right)-3y\)
\(=\left[\left(x^2y^2\right)\left(3+1\right)\right]:\left(x^2y^2\right)-3y\)
\(=4-3y\)
2) a, \(6xy-3x=\left(3x\right)\left(2y-1\right)\)
b, \(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+4\right)\)
3) a, \(2x\left(3x+1\right)+\left(4-2x\right)3x=7\)
\(< =>6x^2+2x+12x-6x^2=7\)
\(< =>14x=7< =>x=\frac{7}{14}\)
b, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)
\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{26x}{5}=\frac{-13}{2}\)
\(< =>26x.2=\left(-13\right).5\)
\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne3\\x\ne\pm2\end{cases}}\)
b) \(D=\left(\frac{2+x}{2-x}-\frac{2-x}{2+x}-\frac{4x^2}{x^2-4}\right)\div\left(\frac{x-3}{2-x}\right)\)
\(\Leftrightarrow D=\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2-x}{x-3}\)
\(\Leftrightarrow D=\frac{4+4x+x^2-4+4x-x^2+4x^2}{\left(2+x\right)\left(x-3\right)}\)
\(\Leftrightarrow D=\frac{4x^2+8x}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow D=\frac{4x}{x-3}\)
c) Để D = 0
\(\Leftrightarrow\frac{4x}{x-3}=0\)
\(\Leftrightarrow4x=0\)
\(\Leftrightarrow x=0\)
Vậy để D = 0 \(\Leftrightarrow\)x = 0
d) Khi \(\left|2x-1\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=5\\1-2x=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\left(ktm\right)\\x=-2\left(ktm\right)\end{cases}}\)
Vậy khi \(\left|2x-1\right|=5\Leftrightarrow D\in\varnothing\)
a) Bình phương \(x+\frac{1}{x}=3\)
Kết quả: 7
b) Lập phương \(x+\frac{1}{x}=3\)
Kết quả: 18
c) Bình phương \(x^2+\frac{1}{x^2}\)
Kết quả: 47
\(\frac{1-3x}{2}-\frac{x+3}{2}\)
\(=\frac{1-3x-x-3}{2}\)
\(=\frac{-4x-2}{2}\)
\(=-2x-1\)
\(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)( ĐKXĐ : \(x\ne\pm2\))
\(=\frac{5\left(x+2\right)}{2\left(2x-4\right)}\cdot\frac{-\left(2x-4\right)}{x+2}\)
\(=\frac{-5\left(x+2\right)\left(2x-4\right)}{2\left(2x-4\right)\left(x+2\right)}\)
\(=-\frac{5}{2}\)
\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}\)( ĐKXĐ : \(x\ne-5;x\ne6\))
\(=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\frac{3}{-\left(x-6\right)}\)
\(=\frac{3\left(x-6\right)\left(x+6\right)}{-2\left(x+5\right)\left(x-6\right)}\)
\(=\frac{3\left(x+6\right)}{-2\left(x+5\right)}=\frac{3x+18}{-2x-10}=-\frac{3x+18}{2x+10}\)
a)
Điều kiện : \(\hept{\begin{cases}4x-8\ne0\\x+2\ne0\end{cases}}\)
\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
\(=\frac{5\left(x+2\right)}{-2\left(4-2x\right)}\cdot\frac{4-2x}{x+2}\)
\(=\frac{-5}{2}\)
b)
Điều kiện : \(\hept{\begin{cases}2x+10\ne0\\6-x\ne0\end{cases}}\)
\(\hept{\begin{cases}x\ne-5\\x\ne6\end{cases}}\)
\(=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}\)
\(=\frac{-6\left(x+6\right)\cdot3}{2x+10}\)
\(=\frac{-9\left(x+6\right)}{x+5}\)
\(=\frac{-9x-54}{x+5}\)
\(=\frac{-9\left(x+5\right)-9}{x+5}\)
\(=-9-\frac{9}{x+5}\)
\(\frac{4-x}{x^3+2x}-\frac{x+5}{x^3-x^2+2x-2}\)( ĐKXĐ : \(x\ne1\))
\(=\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)
\(=\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x\left(x+5\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-x^2+5x-4}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x^2+5x}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-x^2+5x-4-\left(x^2+5x\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-x^2+5x-4-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-2x^2-4}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-2\left(x^2+2\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
\(=\frac{-2}{x\left(x-1\right)}=\frac{-2}{x\left(x-1\right)}\)
Đang đánh máy thì bấm gửi -..-
a,<=>\(\frac{\left(2x+1\right)^2}{4}\)+\(\frac{2\left(2x-1\right)^2}{4}\)≥\(\frac{12\left(x+5\right)^2}{4}\)
<=>4x2+4x+1+2(4x2-4x+1)≥12(x2+10x+25)
<=>4x2+4x+1+8x2-8x+2≥12x2+120x+300
<=>4x2+4x+1+8x2-8x+2-12x2-120x-300≥0
<=>-124x-297≥0
<=>124x+297≤0
<=>124x≤-297
<=>x≤\(\frac{-297}{124}\)
b, Tương tự câu a
c, |5−3x|=2+x
TH1: 5-3x=2+x
<=> -3x - x = 2 - 5
<=> -4x = -3
<=> x = 3/4
TH2: 5-3x = -2 - x
<=> -3x + x = -2 - 5
<=> -2x = -7
<=> x = 7/2
\(\frac{3\left(x-2\right)}{4}\div\frac{2-x}{2}=\frac{3\left(x-2\right)}{4}\times\frac{-2}{x-2}=\frac{-3}{2}\)
học tốt
Rút gọn nhé !
\(\frac{3}{4}.\left(x-2\right):\frac{1}{2}.\left(2-x\right)=\frac{3x-6}{4}.2.\left(2-x\right)\)
\(=\frac{3x-6}{4}.\left(4-2x\right)=\frac{\left(3x-6\right).\left(4-2x\right)}{4}\)
\(=\frac{\left(12x-24\right)-\left(6x^2+12x\right)}{4}=\frac{-24-6x^2}{4}\)
\(=\frac{-12-3x^2}{2}=\frac{-3.\left(4+x^2\right)}{2}\)