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Bài 4:
b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)
a)\(\left|\dfrac{x-1}{3}\right|=\dfrac{11}{5}\Rightarrow\dfrac{x-1}{3}=\pm\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x-1}{3}=\dfrac{11}{5}\\\dfrac{x-1}{3}=-\dfrac{11}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{33}{5}\\x-1=\dfrac{-33}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{38}{5}\\x=\dfrac{-28}{5}\end{matrix}\right.\)
1.
$(3^2-2^3)x+3^2.2^2=4^2.3$
$\Leftrightarrow x+36=48$
$\Leftrightarrow x=48-36=12$
2.
$x^5-x^3=0$
$\Leftrightarrow x^3(x^2-1)=0$
$\Leftrightarrow x^3(x-1)(x+1)=0$
$\Leftrightarrow x^3=0$ hoặc $x-1=0$ hoặc $x+1=0$
$\Leftrightarrow x=0$ hoặc $x=\pm 1$
3.
$(x-1)^2+(-3)^2=5^2(-1)^{100}$
$\Leftrightarrow (x-1)^2+9=25$
$\Leftrightarrow (x-1)^2=25-9=16=4^2=(-4)^2$
$\Rightarrow x-1=4$ hoặc $x-1=-4$
$\Leftrightarrow x=5$ hoặc $x=-3$
4.
$(2x-1)^2-(2x-1)=0$
$\Leftrightarrow (2x-1)(2x-1-1)=0$
$\Leftrightarrow (2x-1)(2x-2)=0$
$\Leftrightarrow 2x-1=0$ hoặc $2x-2=0$
$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=1$
$\Lef
`@` `\text {Ans}`
`\downarrow`
\((3^2-2^3)x+3^2.2^2=4^2.3\)
`=> x + (3*2)^2 = 48`
`=> x+6^2 = 48`
`=> x + 36 = 48`
`=> x = 48 - 36`
`=> x=12`
Vậy, `x=12`
\(x^5-x^3=0\)
`=> x^3(x^2 - 1)=0`
`=>`\(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
Vậy, `x \in {0; +- 1 }`
\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\)
`=> (x-1)^2 + 9 = 25*1`
`=> (x-1)^2 + 9 = 25`
`=> (x-1)^2 = 25 - 9`
`=> (x-1)^2 = 16`
`=> (x-1)^2 = (+-4)^2`
`=>`\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
Vậy, `x \in {5; -3}`
\((2x-1)^2-(2x-1)=0\)
`=> (2x-1)(2x-1) - (2x-1)=0`
`=> (2x-1)(2x-1-1)=0`
`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy, `x \in {1; 1/2}`
a)
\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)
c)
\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
d)
\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
e)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)
f)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)
a) x - 3/97 + x - 2/98 = x - 1/99 + x/100
<=> x + 1/99 + 1 + x + 2/98 + 1 + x + 3/97 + 1 + (x + 4/96 + 1 + x + 5/95 + 1 + x + 10/90 + 1) = 0
<=> x + 100/99 + x + 100/98 + x + 100/97 + (x + 100/96 + x + 100/95 + x + 100/90) = 0
<=> (x + 100)(1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90) = 0
Mà 1/99 + 1/98 + 1/97 + 1/96 + 1/95 + 1/90 khác 0
=> x + 100 = 0
=> x = -100
c) (1/1.2 + 1/2.3 + ... + 1/99.100) - 2x = 1/2
<=> (1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100) - 2x = 1/2
<=> (1 - 1/100) - 2x = 1/2
<=> 99/100 - 2x = 1/2
<=> -2x = 1/2 - 99/100
<=> -2x = -49/100
<=> x = 49/200
=> x = 49/200
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Rightarrow\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Dễ thấy \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}>0\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
