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`@` `\text {Ans}`
`\downarrow`
`(4x - 1)^2 = (1 - 4x)^4`
`\Rightarrow (4x - 1)^2 - (1 - 4x)^4 = 0`
`\Rightarrow (4x - 1)^2 - (4x - 1)^4 = 0`
`\Rightarrow (4x - 1)^2. [1 - (4x - 1)^2] = 0`
`\Rightarrow`\(\left[{}\begin{matrix}\left(4x-1\right)^2=0\\1-\left(4x-1\right)^2=0\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}4x-1=0\\\left(4x-1\right)^2=1\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}4x=1\\4x-1=1\\4x-1=-1\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=\dfrac{1}{4}\\4x=2\\4x=0\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
Vậy, `x \in`\(\left\{0;\dfrac{1}{4};\dfrac{1}{2}\right\}\)
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a, |x - 1,7| = 2,3
=> x - 1,7 = 2,3 hoặc x - 1,7 = -2,3
=> x = 4 hoặc x = -0,6
câu b tương tự câu a
c, |x - 1| = 2x - 3
=> x - 1 = 2x - 3 hoặc x - 1 = 3 - 2x
=> x - 2x = -3 + 1 hoặc x + 2x = 3 + 1
=> -x = -2 hoặc 3x = 4
=> x = 2 hoặc x = 4/3
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a) (x + 3) . (x - \(\frac{1}{2}\)) = 0
=> \(\hept{\begin{cases}x+3=0\\x-\frac{1}{2}=0\end{cases}}\)
=> \(\hept{\begin{cases}x=-3\\x=\frac{1}{2}\end{cases}}\)
ok nha!! 5756758769723414657765887805674765756568678568
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a) Ta có: P(x) = 2x5 + 2 - 6x2 - 3x3 + 4x2 - 2x + x3 + 4x5
= (2x5 + 4x5) + 2 - (6x2 - 4x2) - (3x3 - x3) - 2x
= 6x5 + 2 - 2x2 - 2x3 - 2x
b) P(x) = 6x5 - 2x3 - 2x2 - 2x + 2
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Vì | 2x+1 | và | x+8 | luôn lớn hơn hoặc bằng 0
=> | 2x + 1 | + | x + 8 | = 4x luôn lớn hơn hoặc bằng 0
=> 4x luôn lớn hơn hoặc bằng 0
=> x luôn lớn hơn hoặc bằng 0
=> 2x + 1 + x + 8 = 4x
=> 3x + 9 = 4x
=> 9 = 4x - 3x
=> x = 9
Vậy,.......
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\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12};\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\)
\(\dfrac{x}{8}=\dfrac{-28}{19}\Rightarrow x=-\dfrac{224}{19}\\ \dfrac{y}{12}=\dfrac{-28}{19}\Rightarrow y=-\dfrac{336}{19}\\ \dfrac{z}{15}=\dfrac{-28}{19}\Rightarrow z=-\dfrac{420}{19}\)