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\(\dfrac{1}{5}\times x-\dfrac{2}{3}=\dfrac{1}{10}\times x+\dfrac{5}{6}\)
\(\dfrac{1}{5}x-\dfrac{2}{3}-\dfrac{1}{10}x-\dfrac{5}{6}=0\)
\(\dfrac{1}{5}x-\dfrac{1}{10}x-\dfrac{2}{3}-\dfrac{5}{6}=0\)
\(\dfrac{1}{10}x-\dfrac{3}{2}=0\)
\(\dfrac{1}{10}x=\dfrac{3}{2}\)
\(x=15\)
\(\dfrac{1}{5}\).x - \(\dfrac{2}{3}\) = \(\dfrac{1}{10}\).x + \(\dfrac{5}{6}\)
⇒ \(\dfrac{1}{5}\).x - \(\dfrac{1}{10}\).x = \(\dfrac{5}{6}\) + \(\dfrac{2}{3}\)
⇒ \(\dfrac{2}{10}\).x - \(\dfrac{1}{10}\).x = \(\dfrac{5}{6}\) + \(\dfrac{4}{6}\)
⇒ \(\dfrac{1}{10}\).x = \(\dfrac{9}{6}\)
⇒ x = \(\dfrac{9}{6}\) : \(\dfrac{1}{10}\)
⇒ x = \(\dfrac{9}{6}\) . 10
⇒ x = \(\dfrac{90}{6}\)
⇒ x = 15
Vậy x = 15
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`7(x-1/2)^2=9`
`(x-1/2)^2=9/7`
\(=>\left[{}\begin{matrix}x-\dfrac{1}{2}=\sqrt{\dfrac{9}{7}}\\x-\dfrac{1}{2}=-\sqrt{\dfrac{9}{7}}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{\sqrt{7}}+\dfrac{1}{2}\\x=-\dfrac{3}{\sqrt{7}}+\dfrac{1}{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{6+\sqrt{7}}{2\sqrt{7}}\\x=\dfrac{-6+\sqrt{7}}{2\sqrt{7}}\end{matrix}\right.\)
7.(x-\(\dfrac{1}{2}\))2=9
7.x+\(\dfrac{1}{4}\) =9
7.x=\(\dfrac{37}{4}\)
x=\(\dfrac{37}{28}\)
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a: Ta có: \(\dfrac{x+1}{2}=\dfrac{2}{x+1}\)
\(\Leftrightarrow\left(x+1\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
b: Ta có: \(\dfrac{\left(x-2\right)^2}{7}=\dfrac{49}{\left(x-2\right)}\)
\(\Leftrightarrow x-2=7\)
hay x=9
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Với x=0
\(\Rightarrow3.f\left(0\right)-f\left(1\right)=0+1=1\)
\(f\left(0\right)-f\left(1\right)=\frac{1}{3}\)(1)
Với x=1
\(\Rightarrow3.f\left(1\right)-f\left(0\right)=1+1=2\)
\(f\left(1\right)-f\left(0\right)=\frac{2}{3}\)(2)
Với x=-1
\(3.f\left(-1\right)-f\left(2\right)=1+1=2\)
\(\Rightarrow f\left(-1\right)-f\left(2\right)=\frac{2}{3}\)(3)
Kết hợp (1);(2);(3) tính nhé
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ÁP dụng cái bất đẳng thức j j đó
mk có xem làm ở đâu rùi nhưng chưa học nên ko bt giải
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\(\Leftrightarrow\left|x^2+2021\right|=2019\\ \Leftrightarrow x\in\varnothing\left(x^2+2021>2019\Leftrightarrow\left|x^2+2021\right|>2019\right)\)
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24 - 16(x - 1/2) = 23
=> 16(x - 1/2) = 24 - 23
=> 16(x - 1/2) = 1
=> x - 1/2 = 1/16
=> x = 1/16 + 1/2
=> x = 9/16
\(24-16(x-\frac{1}{2})=23\)
\(16(x-\frac{1}{2})=24-23\)
\(16(x-\frac{1}{2})=1\)
\(x-\frac{1}{2}=\frac{1}{16}\)
\(x=\frac{1}{16}+\frac{1}{2}\)
\(x=\frac{9}{16}\)
Vậy số thực x cần tìm là \(\frac{9}{16}\)
Chúc bạn hok tốt ~
(x - 1)x + 2 = (x - 1)2
<=>(x - 1)x + 2 - (x - 1)2 = 0
<=> (x - 1)2[(x- 1)x - 1] = 0
<=> \(\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^x-1=0\end{cases}}\)
Khi (x - 1)x - 1 = 0
Nếu x = 0
=> (x - 1)x - 1 = 0 (tm)
Nếu x > 0
Khi đó (x - 1)x - 1 = 0
<=> \(\left(x-1\right)^x=1^x\)
<=> \(x-1=1\)
<=> x = 2
Khi (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
Vậy x \(\in\left\{0;1;2\right\}\)