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Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
/x/=-3
không có giá trĩ nào thảo mãn vì/x/ lớn hơn bằng o với mọi x
1) \(\left|x-\frac{3}{5}\right|< \frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}< \frac{1}{3}\\x-\frac{3}{5}< -\frac{1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{1}{3}+\frac{3}{5}\\x< \frac{-1}{3}+\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x< \frac{5}{15}+\frac{9}{15}\\x< \frac{-5}{15}+\frac{9}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)
vay \(\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)
2) \(\left|x+\frac{11}{2}\right|>\left|-5,5\right|\)
\(\left|x+\frac{11}{2}\right|>5,5\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{11}{2}>\frac{11}{2}\\x+\frac{11}{2}>-\frac{11}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{11}{2}-\frac{11}{2}\\x>\frac{-11}{2}-\frac{11}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)
vay \(\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)
3) \(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Rightarrow\left|x-\frac{7}{5}\right|>\frac{2}{5}\) va \(\left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{7}{5}>\frac{2}{5}\\x-\frac{7}{5}>\frac{-2}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{2}{5}+\frac{7}{5}\\x>\frac{-2}{5}+\frac{7}{5}\end{cases}}\)va \(\orbr{\begin{cases}x-\frac{7}{5}< \frac{3}{5}\\x-\frac{7}{5}< \frac{-3}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{3}{5}+\frac{7}{5}\\x< \frac{-3}{5}+\frac{7}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>\frac{9}{5}\\x>1\end{cases}}\)va \(\orbr{\begin{cases}x< 2\\x< \frac{4}{5}\end{cases}}\)
vay ....
\(\left|x\right|=7\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
Vậy \(x\in\left\{\pm7\right\}\)
\(\frac{x-3}{5}+\frac{x-3}{6}+\frac{x-3}{7}=0\)
\(\left(x-3\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)=0\)
mà \(\frac{1}{5}>0;\frac{1}{6}>0;\frac{1}{7}>0\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\ne0\)
=> x - 3= 0
x = 3