1) giải phương trình:
a) \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x+5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
b) \(\frac{7x+10}{x+1}\left(x^2-x-2\right)-\frac{7x+10}{x+1}\left(2x^2-3x-5\right)=0\)
c) \(\frac{2x+5}{x+3}+1=\frac{4}{x^2+2x-3}-\frac{3x-1}{1-x}\)
d) \(\frac{13}{2x^2+x-21}+\frac{1}{2x+7}+\frac{6}{9-x^2}=0\)
e) \(\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
f) \(\frac{1+\frac{x}{x+3}}{1-\frac{x}{x+3}}=3\)
\(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)0 \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\) \(\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)\) giải dùm minh đi
1. \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)
2 . \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)
3 . \(4\left(3x-2\right)-3\left(x-4\right)=7x+10\)
4. \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
giải pt \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\left(x+3\right)^4+\left(x+5\right)^4=2\)
\(\frac{x^2+x+4}{2}+\frac{x^2+x+7}{3}=\frac{x^2+x+13}{5}+\frac{x^2+x+16}{6}\)
\(\frac{x+19}{3}+\frac{x+13}{5}=\frac{x+7}{7}+\frac{x+1}{9}\)
giúp vs mình cần gấp
Giải phương trinh:
1,x8-x5+x2-x+1=0
2,\(\frac{1}{\left(2x-1\right)^2}-\frac{1}{\left(3x+1\right)^2}=\frac{4}{4\left(x+2\right)^2}\)
3, \(\frac{x\left(x^2-36\right)}{4-7x}-\frac{21x+1}{x^3+2}=4\)
4,\(\frac{4x^2+16}{x^2+6}-\frac{3}{x^2+1}=\frac{5}{x^2+3}+\frac{7}{x^2+5}\)
Tìm điều kiện xác định rồi giải các phương trình sau:
a) \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
b) \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
c) \(\frac{8x^2}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)
Help me!
Ta có: \(a^3+b^3+3\left(a^2+b^2\right)+4\left(a+b\right)+4=0\)
\(\Leftrightarrow a^3+b^3+3a^2+3b^2+4a+4b+4=0\)
\(\Leftrightarrow\left(a+1\right)^3+\left(b+1\right)^3+a+b+2=0\)
\(\Leftrightarrow\left(a+b+2\right)\left[\left(a+1\right)^2-\left(a+1\right)\left(b+1\right)+\left(b+1\right)^2\right]+\left(a+b+2\right)=0\)
\(\Leftrightarrow\left(a+b+2\right)\left[\left(a+1\right)^2-\left(a+1\right)\left(b+1\right)+\left(b+1\right)^2+1\right]=0\left(1\right)\)
Đặt \(a+1=x;b+1=y\)
Xét \(x^2-xy+y^2+1=x^2-2.x.\frac{y}{2}+\frac{y^2}{4}-\frac{y^2}{4}+y^2+1\)
\(=\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2+1\)
Mà \(\hept{\begin{cases}\left(x-\frac{y}{2}\right)^2\ge0;\forall x,y\\\frac{3}{4}y^2\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2\ge0;\forall x,y\)
\(\Rightarrow\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2+1\ge1>0;\forall x,y\)
Hay \(\left(a+1\right)^2-\left(a+1\right)\left(b+1\right)+\left(b+1\right)^2+1>0\)
Từ đó\(\left(1\right)\)xảy ra \(\Leftrightarrow a+b+2=0\)
\(\Leftrightarrow a+b=-2\)Thay vào biểu thức M ta đuợc:
\(M=2018.\left(-2\right)^2=8072\)
Vậy ...
Giải phương trình:
1.\(\frac{x-5}{x-5}+\frac{x-6}{x-5}+\frac{x-7}{x-5}+...+\frac{1}{x-5}=4\left(x\in N\right)\)
2.\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
3.\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{x\left(x+2\right)}\right)=\frac{31}{16}\left(x\in N\right)\)
4.\(8\left(x^2+\frac{1}{x^2}\right)-34\left(x+\frac{1}{x}\right)+51=0\)
5.\(6x^4-5x^3-38x^2-5x+6=0\)
g) \(|9-7x|=5x-3\)
Vì \(|9-7x|\ge0;\forall x\)
\(\Rightarrow5x-3\ge0\)
\(\Rightarrow x\ge\frac{3}{5}\)
Ta có: \(|9-7x|=5x-3\)
\(\Leftrightarrow\orbr{\begin{cases}9-7x=5x-3\\9-7x=3-5x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-7x-5x=-3-9\\-7x+5x=3-9\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-12x=-12\\-2x=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1>\frac{3}{5}\left(chon\right)\\x=3>\frac{3}{5}\left(chon\right)\end{cases}}\)
Vậy \(x\in\left\{1;3\right\}\)
h) \(8x-|4x+1|=x+2\)
\(\Leftrightarrow|4x+1|=7x+2\)
Vì \(|4x+1|\ge0;\forall x\)
\(\Rightarrow7x+2\ge0\)
\(\Rightarrow x\ge\frac{-2}{7}\)
Ta có: \(|4x+1|=7x+2\)
\(\Leftrightarrow\orbr{\begin{cases}4x+1=7x+2\\4x+1=-7x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-3x=1\\11x=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}< \frac{-2}{7}\left(loai\right)\\x=\frac{-3}{11}>\frac{-2}{7}\left(chon\right)\end{cases}}\)
Vậy \(x=\frac{-3}{11}\)