\(\dfrac{x+\left(x-10\right)}{2}.8=160\)
\(x-5=20\)
\(x=25\)
\(\dfrac{x+\left(x-10\right)}{2}.8=160\\ \dfrac{x+\left(x-10\right)}{2}=160:8\\\dfrac{x+x-10}{2}=20\\ 2x-10=20.2\\ 2x-10=40\\ 2x=40+10\\ 2x=50\\ x=50:2\\ x=25 \)
\(\dfrac{x+\left(x-10\right)}{2}.8=160\)
\(x-5=20\)
\(x=25\)
\(\dfrac{x+\left(x-10\right)}{2}.8=160\\ \dfrac{x+\left(x-10\right)}{2}=160:8\\\dfrac{x+x-10}{2}=20\\ 2x-10=20.2\\ 2x-10=40\\ 2x=40+10\\ 2x=50\\ x=50:2\\ x=25 \)
\(\dfrac{1}{x+3}+\dfrac{8}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x}{x^2-2x-3}\) giải pt
giải pt:
\(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}=\dfrac{-2x}{\left(3-x\right)\left(x+1\right)}\)
\(x^2+\dfrac{1}{x^2}=x+\dfrac{1}{x}\)
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
e,\(\left(x^2+x+1\right)^2-2x^2-2x=5\)
Giải pt
giải pt sau: \(\dfrac{1}{x-5}\)-\(\dfrac{4}{\left(x-5\right)\left(x-1\right)}\)=\(\dfrac{5}{x-1}\)
CẦN GẤP
Giải pt sau:
\(\frac{x-1}{99}+\frac{\left(x-2\right)}{49}+\frac{\left(x-7\right)}{31}+\frac{\left(x-8\right)}{23}=10\)
giải pt:
\(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
Giải phương trình:
a) \(\dfrac{1}{x-2}+3=\dfrac{x-3}{2-x}\)
b) \(\dfrac{3}{\left(x-1\right)\left(x-2\right)}+\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{\left(x-2\right)\left(x-3\right)}\)
c) \(1+\dfrac{1}{x+2}=\dfrac{12}{8+x^3}\)
P=\(\left(\dfrac{3\left(x+2\right)}{2x^2+8}-\dfrac{2x^2-x-10}{\left(x+1\right)\left[\left(x+1\right)^2-2x\right]}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{x-1}\right)\cdot\dfrac{2}{x-1}\)
a) rút gọn P
b)tìm tất cả các giá trị nguyên của x để P có giá trị là bội của 4
Giải các pt sau:
a, \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
b,\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
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