Giải các phương trình :
\(a,\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(b,\left(3x-2\right)\left(4x+5\right)=0\)
Giải các phương trình :
\(a,\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(b,\left(3x-2\right)\left(4x+5\right)=0\)
a. \(x^2+9x+20=\left(x^2+4x\right)+\left(5x+20\right)\)
\(=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)
Tương tự: \(x^2+11x+30=\left(x+5\right)\left(x+6\right)\)
\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)
\(\Rightarrow PT=\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(=\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(=\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(=18\left(x+7\right)-18\left(x+4\right)=\left(x+7\right)\left(x+4\right)\)
\(=x^2+11x+28=54\)
\(=x^2+11x-26=0\)
\(=\left(x^2-2x\right)+\left(13x-26\right)=0\)
\(=x\left(x-2\right)+13\left(x-2\right)=0\)
\(=\left(x+13\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)
b. \(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-\frac{5}{4}\end{matrix}\right.\)
1.Giải phương trình: \(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
2.Giải phương trình: \(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
Giải các phương trình sau :
a, \(\left(6x+8\right)\left(6x+6\right)\left(6x+7\right)^2=72\)
b,\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
a/ Đặt \(6x+7=a\Rightarrow\left\{{}\begin{matrix}6x+8=a+1\\6x+6=a-1\end{matrix}\right.\)
\(\Rightarrow\left(a-1\right)\left(a+1\right)a^2-72=0\)
\(\Leftrightarrow\left(a^2-1\right)a^2-72=0\)
\(\Leftrightarrow a^4-a^2-72=0\)
\(\Leftrightarrow\left(a^2-9\right)\left(a^2+8\right)=0\)
\(\Leftrightarrow a^2=9\) (do \(a^2+8>0\))
\(\Rightarrow\left[{}\begin{matrix}a=3\\a=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}6x+7=3\\6x+7=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne-4;-5;-6;-7\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow x^2+11x-26=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
Giải các phương trình,bất phương trình:
c,\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
d,\(\frac{4}{-25x^2+20x-3}=\frac{3}{5x-1}-\frac{2}{5x-3}\)
e,\(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}-\frac{2}{x^2-4x+3}=0\)
g,\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)
h,\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
i,\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
k,\(\left(3x^2+10x-8\right)^2=\left(5x^2-2x+10\right)^2\)
l,\(\left(x^2-2x+1\right)-4=0\)
m,\(4x^2+4x++1=x^2\)
Xin đáy ai giúp mình đi
Giaỉ các phương trình sau:
a) \(\left(x^2+11x+12\right)\left(x^2+9x+20\right)\left(x^2+13x+42\right)=36\left(x^2+11x+30\right)\left(x^2+11x+31\right)\)
b) \(20\left(\frac{x-2}{x+1}\right)^2-5\left(\frac{x+2}{x-1}\right)^2+48\cdot\frac{x^2-4}{x^2-1}=0\)
Giải các phương trình sau:
c)\(\frac{x-1}{2x^2-4x}-\frac{7}{8x}=\frac{5-x}{4x^2-8x}-\frac{1}{8x-16}\)
d) \(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{3}{54}\)
\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x+28=54\)
\(\Leftrightarrow x^2+11x-26=0\)
Ta có \(\Delta=11^2+4.26=225,\sqrt{\Delta}=15\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-11+15}{2}=2\\x=\frac{-11-15}{2}=-13\end{cases}}\)
Vậy tập nghiệm S = {2;-13}
Giải các phương trình:
\(a.\left(x^2+1\right)\left(x^2-4x+4\right)=0\)
\(b.\left(3x-2\right)\left(\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right)=0\)
\(c.\left(3,3-11x\right)\left(\frac{7x+2}{5}+\frac{2\left(1-3x\right)}{3}\right)=0\)
a)\(\left(x^2+1\right)\left(x^2-4x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\\left(x-2\right)^2=0\end{cases}\Rightarrow}x=2}\)
b)\(\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\\ \Rightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)
c)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)
GIẢI PHƯƠNG TRÌNH:
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
phân tích mẫu thành nhân tử r áp dụng \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) sau đó rút gọn quy đồng
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\) \(\left(ĐKXĐ:x\ne0;x\ne-4;x\ne-5;x\ne-6;x\ne-7\right)\)
\(\Leftrightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{\left(x+6\right)\left(x+7\right)+\left(x+4\right)\left(x+7\right)+\left(x+4\right)\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{\left(x^2+13x+42\right)+\left(x^2+11x+28\right)+\left(x^2+9x+20\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{x^2+13x+42+x^2+11x+28+x^2+9x+20}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3x^2+33x+90}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3\left(x^2+11x+30\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=18.3\left(x^2+11x+30\right)\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=54\left(x+5\right)\left(x+6\right)\)
\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x+28-54=0\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow x^2+13x-2x-26=0\)
\(\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\)
\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+13=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-13\left(tm\right)\\x=2\left(tm\right)\end{cases}}\)
rần Duy Thanh:gà dữ đến dấu <=> thứ 3 r` mà còn quy đồng. What is your favourite food :v
giải phương trình
\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)^{ }\)
\(\frac{2a-9}{2a-5}+\frac{3a}{3a-2}=2\)
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\)
\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
\(\frac{8x^23}{3\left(1-4x^2\right)}=\frac{2x}{6x-3}-\frac{1+8x}{4+8x}\)
\(\frac{x-3}{x-2}+\frac{x-2}{x-4}=-1\)
\(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)
\(\frac{5x-2}{2-2x}+\frac{2x-1}{2}=1-\frac{x^2+x-3}{1-x}\)