Giai cac phuong trinh :
a\(\frac{12}{x-1}-\frac{8}{x+1}=1\)
b\(\frac{x^3+7x^2+6x-30}{x^3-1}=\frac{x^2-x+16}{x^2+x+1}\)
Giai phuong trinh
\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
\(c,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(d,\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-10x=3-15\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)
KL :....
\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\) ĐKXĐ : \(x\ne0;2\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=2-2\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
KL ::
\(c,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\) ĐKXĐ : \(x\ne\pm2\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x^2+2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x+x+2+x^2-2x-x+2=2x^2+4\)
\(\Leftrightarrow0x=0\)
KL : PT vô số nghiệm
Giai phuong trinh sau: \(\frac{x^2+2x+2}{x+1}+\frac{x^2+8x+20}{x+4}=\frac{x^2+4x+6}{x+2}+\frac{x^2+6x+12}{x+3}\)
giai phuong trinh
a) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3x-24}\)
b) \(\frac{1}{3-x}+\frac{14}{x^2-9}=\frac{x-4}{3+x}+\frac{7}{3+x}\)
a) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3x-24}\) \(ĐK:x\ne8\)
\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3\left(x-8\right)}\)
\(\Leftrightarrow\frac{3.3}{6.\left(x-8\right)}+\frac{6.\left(3x-20\right)}{6\left(x-8\right)}-\frac{2\left(3x-102\right)}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow\frac{9+18x-120-6x+204}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow\frac{12x+93}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow8\left(12x+93\right)=-6\left(x-8\right)\)
\(\Leftrightarrow96x+744=-6x+48\)
\(\Leftrightarrow102x=-696\)
\(\Leftrightarrow x=\frac{-116}{17}\) (nhận)
Vậy .....
b) \(\frac{1}{3-x}+\frac{14}{x^2-9}=\frac{x-4}{3+x}+\frac{7}{3+x}\) \(ĐK:x\ne\pm3\)
\(\Leftrightarrow\frac{1}{3-x}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{x-4}{3+x}+\frac{7}{3+x}\)
\(\Leftrightarrow-\frac{3+x}{\left(x-3\right)\left(3+x\right)}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{-3-x+14}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)
\(\Leftrightarrow-3-x+14=x^2-3x-4x+12+7x-21\)
\(\Leftrightarrow x=-5\) (nhận)
Vậy ....
giai cac phuong trinh sau:
a.\(\frac{6}{\left(x+1\right)\left(x+2\right)}+\frac{8}{\left(x-1\right)\left(x+4\right)}=1\)
b.\(x^3+\frac{1}{x^3}=13\left(x+\frac{1}{x}\right)\)
\(\frac{x^2-8}{x^2-16}=\frac{1}{x+4}+\frac{1}{x-4}\)
giai phuong trinh
\(\frac{x^2-8}{x^2-16}=\frac{1}{x+4}+\frac{1}{x-4}\)
\(\Rightarrow\frac{x^2-8}{\left(x+4\right)\left(x-4\right)}=\frac{x-4}{\left(x+4\right)\left(x-4\right)}+\frac{x+4}{\left(x-4\right)\left(x+4\right)}\)
\(\Rightarrow x^2-8=x-4+x+4\)
\(\Rightarrow x^2-8=2x\)
\(\Rightarrow x^2-2x-8=0\)
\(\Delta=b^2-4ac=\left(-2\right)^2-4.1.\left(-8\right)=4+32=36>0\)
phương trình có 2 nghiệm phân biệt : \(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{2+\sqrt{36}}{2}=\frac{2+6}{2}=\frac{8}{2}=4\)
\(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{2-\sqrt{36}}{2}=\frac{2-6}{2}=\frac{-4}{2}=\left(-2\right)\)
giai phuong trinh
a.\(\frac{5x-3}{x^2-9}-\frac{x}{x-3}=\frac{2x-1}{x+3}\)
b.\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
b) \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
=> \(\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
=> \(\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
=> x - 100 = 0
=> x = 100
Giai cac phuong trinh:
a/ (x - 1)3 - x(x + 1)2 = 5x(2 - x) - 11(x + 2)
b/ \(\frac{\left(2x+1\right)^2}{5}\) _ \(\frac{\left(x-1\right)^2}{3}\) = \(\frac{7x^2-14x-5}{15}\)
c/ \(\frac{x^2-10x-29}{1971}\) + \(\frac{x^2-10x-27}{1973}\) = \(\frac{x^2-10x-1971}{29}\) + \(\frac{x^2-10x-1973}{27}\)
cho phuong trinh an x: \(\frac{x+a}{x+2}+\frac{x-2}{x-a}=2\)
a) giai phuong trinh vs a=4
b)Tim cac gtri cua a sao cho phuong trinh nhan x=-1 lam nghiem
a) Ta có: \(\frac{x+a}{x+2}+\frac{x-2}{x-a}=2\left(1\right)\)
Với a = 4
Thay vào phương trình (t) ta được:
\(\frac{x+2}{x+2}+\frac{x-2}{x-2}=2\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow x^2-4+x^2-4=2\left(x^2-4\right)\)
\(\Leftrightarrow2x^2=2x^2-8\)
\(\Leftrightarrow0x=-8\)
Vậy phương trình vô nghiệm
b) Nếu x = -1
\(\Rightarrow\frac{-1+a}{-1+2}+\frac{-1-2}{-1-a}=2\)
\(\Leftrightarrow\frac{-1+a}{1}+\frac{-3}{-1-a}=2\)
\(\Leftrightarrow\frac{\left(-1+a\right)\left(-1-a\right)}{-1-a}+\frac{-3}{-1-a}=\frac{2\left(-1-a\right)}{-1-a}\)
\(\Leftrightarrow1+a-a-a^2-3=-2-2a\)
\(\Leftrightarrow-a^2+2a=-2-1+3\)
\(\Leftrightarrow a\left(2-a\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=0\\2-a=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)
Vậy a = {0;2}
NĂM MỚI VUI VẺ
\(a,\frac{x+4}{x+2}+\frac{x-2}{x-4}=2\)
\(\frac{x+2+2}{x+2}+\frac{x-4+2}{x-4}=2\)
=> \(1+\frac{2}{x+2}+1+\frac{2}{x-4}=2\)
=>\(2\left(\frac{x-4+x+2}{\left(x+2\right)\left(x-4\right)}\right)=0\)
=> x=1 (t/m \(x\ne-2\) và \(x\ne4\))
Giai Phuong Trinh :
1+\(\frac{1}{x+2}\)=\(\frac{12}{8-x^3}\)