4/x+2 + (-3)/x-2 + 12/x^2-4 giai chi tiet ho mk !
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\(\frac{3}{4}+\frac{1}{4}\cdot x+x-\frac{7}{6}\cdot x=\frac{5}{12}\)
\(\frac{3}{4}+\frac{1}{4}\cdot x-\frac{7}{6}\cdot x+x\cdot1=\frac{5}{12}\)
\(\frac{3}{4}+x\left(\frac{1}{4}-\frac{7}{6}+1\right)=\frac{5}{12}\)
\(\frac{3}{4}+x\cdot\frac{1}{12}=\frac{5}{12}\)
\(x\cdot\frac{1}{12}=\frac{5}{12}-\frac{3}{4}\)
\(x\cdot\frac{1}{12}=\frac{5}{12}-\frac{9}{12}\)
\(x\cdot\frac{1}{12}=\frac{-1}{3}\)
\(x=\frac{-1}{3}\text{ : }\frac{1}{12}\)
\(x=\frac{-1}{3}\cdot12\)
\(x=\frac{-12}{3}\)
\(x=-4\)
\(\text{b, }0,25\cdot x-\frac{2}{3}\cdot x=-1\frac{1}{6}\)
\(\frac{1}{4}\cdot x-\frac{2}{3}\cdot x=\frac{-7}{6}\)
\(x\cdot\left(\frac{1}{4}-\frac{2}{3}\right)=\frac{-7}{6}\)
\(x\cdot\frac{-5}{12}=\frac{-7}{6}\)
\(x=\frac{-7}{6}\text{ : }\frac{-5}{12}\)
\(x=\frac{-7}{6}\cdot\frac{12}{-5}\)
\(x=\frac{-14}{-5}\)
\(\frac{4}{7}x-\frac{2}{3}=\frac{1}{5}\)
\(\Leftrightarrow\frac{4}{7}x=\frac{1}{5}+\frac{2}{3}\)
\(\Leftrightarrow\frac{4}{7}x=\frac{3}{15}+\frac{10}{15}\)
\(\Leftrightarrow\frac{4}{7}x=\frac{13}{15}\)
\(\Leftrightarrow x=\frac{13}{15}:\frac{4}{7}=\frac{13}{15}\cdot\frac{7}{4}=\frac{91}{60}\)
\(x^4+x^2-20=0\)
\(\Leftrightarrow x^4-4x^2+5x^2-20=0\)
\(\Leftrightarrow x^2\left(x^2-4\right)+5\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x^2+5=0\end{cases}}\)loại \(x^2+5=0\)vì giải trên tập số thực nên x^2+5>0
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Vậy \(S=\left\{2;-2\right\}\)
x ^ 4 + x ^ 2 - 20 = 0
(x ^ 2 + 5) (x ^ 2 - 4) = 0
(x ^ 2 + 5) (x + 2) (x - 2) = 0
x ^ 2 + 5 = 0
x ^ 2 = -5
x = ± √-5
x = ± i√5
x + 2 = 0
x = -2
x - 2 = 0
x = 2
x = {-i√5, i√5, -2, 2}
\(x^5-x^4+3x^3+3x^2-x+1=0\)
\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^4-2x^3+5x^2-2x+1=0\left(#\right)\end{cases}}\)
\(\Leftrightarrow x=-1\)(vì biểu thức # vô nghiệm) (cái này bạn tự cm)
vậy....
:D ko bt khó không nhưng nhì so qua thì áp dụng
|a|+|b|>=|a+b|
|a|=|-a|
Theo đầu bài ta có:
\(\frac{1}{5}\cdot a+2+\frac{1}{2}\cdot a+7=a\)
\(\Rightarrow2+7=a-\frac{1}{2}\cdot a-\frac{1}{5}\cdot a\)
\(\Rightarrow a\cdot\frac{3}{10}=9\)
\(\Rightarrow a=30\)
\(\frac{1}{5}a+2+\frac{1}{2}a+7=a\left(\frac{1}{5}+\frac{1}{2}\right)+2+7=\frac{7}{10}a+10=\frac{7a}{10}+10\)
4(x-1)2=x2
<=>2(x-1)=x hoặc 2(x-1)=-x
<=>2x-2=x hoặc 2x-2=-x
<=>x=2 hoặc x=2/3
Vậy trung bình cộng các giá trị x là: (2+2/3):2=4/3
\(\frac{4}{x+2}+\frac{-3}{x-2}+\frac{12}{x^2-4}.\)
\(=\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4x-8-3x-6+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x-4}{x^2-4}\)
\(\frac{4}{x+2}+\frac{\left(-2\right)}{x-2}+\frac{12}{x^2-4}\)
\(=\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4\left(x-2\right)-3\left(x+2\right)+12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-2}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{1}{x+2}\)