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Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
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b: \(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
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Bài 3 :
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)
\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)
\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)
\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)
.....
\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)
\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left|\dfrac{x-1}{3}\right|=\dfrac{11}{5}\Rightarrow\dfrac{x-1}{3}=\pm\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x-1}{3}=\dfrac{11}{5}\\\dfrac{x-1}{3}=-\dfrac{11}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{33}{5}\\x-1=\dfrac{-33}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{38}{5}\\x=\dfrac{-28}{5}\end{matrix}\right.\)
\(a)\frac{11}{4}-2x=\frac{-1}{2}\)
\(2x=\frac{11}{4}-\left(\frac{-1}{2}\right)\)
\(2x=\frac{11}{4}+\frac{1}{2}\)
\(2x=\frac{11}{4}+\frac{2}{4}\)
\(2x=\frac{13}{4}\)
\(x=\frac{13}{4}:2\)
\(x=\frac{13}{8}\)
\(b)\left|\frac{3}{4}-\frac{1}{2x}\right| +\frac{1}{3}=\frac{5}{6}\)
\(\left|\frac{3}{4}-\frac{1}{2x}\right|=\frac{5}{6}-\frac{1}{3}\)
\(\left|\frac{3}{4}-\frac{1}{2x}\right|=\frac{5}{6}-\frac{2}{6}\)
\(\left|\frac{3}{4}-\frac{1}{2x}\right|=\frac{3}{6}\)
\(TH1:\)
\(\frac{3}{4}-\frac{1}{2x}=\frac{3}{6}\)
\(\frac{1}{2x}=\frac{3}{4}-\frac{3}{6}\)
\(\frac{1}{2x}=\frac{18}{24}-\frac{12}{24}\)
\(\frac{1}{2x}=\frac{6}{24}\)
\(\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow2x=4\)
\(x=4:2\)
\(x=2\)
\(TH2:\)
\(\frac{3}{4}-\frac{1}{2x}=\frac{-3}{6}\)
\(\frac{1}{2x}=\frac{3}{4}-\left(\frac{-3}{6}\right)\)
\(\frac{1}{2x}=\frac{3}{4}+\frac{3}{6}\)
\(\frac{1}{2x}=\frac{18}{24}+\frac{12}{24}\)
\(\frac{1}{2x}=\frac{30}{24}\)
\(\frac{1}{2x}=\frac{5}{4}\)
\(\Rightarrow1:2x=5:4\)
\(1:2x=1,25\)
\(2x=1:1,25\)
\(2x=0,8\)
\(x=0,8:2\)
\(\)\(x=0,4\)