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1) \(\left|\dfrac{1}{2}x-\dfrac{1}{6}\right|=\dfrac{1}{3}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{6}=\dfrac{1}{3}\\\dfrac{1}{2}x-\dfrac{1}{6}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{1}{2}\\\dfrac{1}{2}x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(---\)
2) \(\left|\dfrac{1}{2}x+\dfrac{3}{5}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+\dfrac{3}{5}=\dfrac{1}{2}\\\dfrac{1}{2}x+\dfrac{3}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{1}{10}\\\dfrac{1}{2}x=-\dfrac{11}{10}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{11}{5}\end{matrix}\right.\)
\(---\)
3) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\left|\dfrac{-3}{4}\right|\)
\(\Rightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{4}\\\dfrac{3}{4}x-\dfrac{3}{4}=-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{3}{2}\\\dfrac{3}{4}x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
\(---\)
4) \(14-\left|\dfrac{3x}{2}-1\right|=9\)
\(\Rightarrow\left|\dfrac{3x}{2}-1\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3x}{2}-1=5\\\dfrac{3x}{2}-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3x}{2}=6\\\dfrac{3x}{2}=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\3x=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(---\)
5) \(17-\left|\dfrac{2}{3}-4x\right|=9\)
\(\Rightarrow\left|\dfrac{2}{3}-4x\right|=8\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}-4x=8\\\dfrac{2}{3}-4x=-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{22}{3}\\4x=\dfrac{26}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{6}\\x=\dfrac{13}{6}\end{matrix}\right.\)
\(---\)
6) \(5-\left|2x-3\right|=\dfrac{1}{2}\)
\(\Rightarrow\left|2x-3\right|=\dfrac{9}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
#\(Toru\)
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Ta có M(x) = x4 + 9/2 . x2 + 2/2 . x2 + x + 6 ( tách 11/2 . x2)
=> M(x) = x4 + 9/2.x2 + x2 + x + 6
Ta xét x2 + x + 6
= x2 + 1/2.x + 1/2.x + 1/4 + 23/4 (tách x và tách 6)
= x(x + 1/2) + 1/2(x + 1/2) + 23/4 (phân phối)
= (x + 1/2).(x + 1/2) + 23/4 (phân phối tiếp)
= (x + 1/2)2 + 23/4
Ghép kết quả trên vào M(x) ta đc:
M(x)= x4 + 9/2.x2 + (x + 1/2)2 + 23/4
Vì x4 >= 0, mọi x
9/2.x2 >= 0, mọi x.
(x + 1/2)2 >= 0, mọi x
Suy ra x4 + 9/2.x2 + (x + 1/2)2 >= 0, mọi x
Suy ra x4 + 9/2.x2 + (x + 1/2)2 + 23/4 > 0, mọi x
Vậy đa thức M(x) vô nghiệm
ko tránh khỏi thiếu sót, nếu làm sai ai đó sửa lại nhé
_Hết_
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`@` `\text {Ans}`
`\downarrow`
Ta có: \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\)
\(\Rightarrow\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\)`=`\(\dfrac{\left(5z-25\right)-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\)
`=`\(\dfrac{5z-25-3x+3-4y-12}{8}\)
`=`\(\dfrac{\left(5z-3x-4y\right)+\left(-25+3-12\right)}{8}\)
`=`\(\dfrac{50-34}{8}\)`=`\(\dfrac{16}{8}=2\)
`=>`\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=2\)
`=>`\(\left\{{}\begin{matrix}x=2\cdot2+1=5\\y=2\cdot4-3=5\\z=2\cdot6+5=17\end{matrix}\right.\)
Vậy, `x,y,z` lần lượt là `5; 5; 17.`
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\(-2x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(-2x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
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Không ghi lại đề
Ta có : \(-4\frac{3}{5}×2\frac{4}{23}\)
\(=-\frac{23}{5}×\frac{50}{23}\)
\(=-\frac{50}{5}=-10\)
\(-2\frac{3}{5}\div1\frac{6}{15}\)
\(=-\frac{13}{5}\div\frac{7}{5}\)
\(=-\frac{13}{5}×\frac{5}{7}\)
\(=-\frac{13}{7}\)
\(\Rightarrow\) \(-10\le x\le\frac{-13}{7}\)
\(\Rightarrow\) \(\frac{-70}{7}\le x\le\frac{-13}{7}\)
\(\Rightarrow x\in\left\{\frac{-70}{7};\frac{-69}{7};...;-2;\frac{-13}{7}\right\}\)
Vậy...
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\(2022-\left(\dfrac{1}{4}\right)^2\cdot4^2=2022-\left(\dfrac{1}{4}\cdot4\right)^2=2022-1^2=2021\)
2022 - (1/4)^2x=4^2
(1/4)^2x=2022 - 4^2
(1/4)^2x=2006
x=2006/(1/4)
x=8024
tri tuyet doi cua x-2=2
suy ra x-2=-2
x-2=2
suy ra x=0
x=4
vậy x={0,4}
| x - 2 | + 4 = 6
| x - 2 | = 6 - 4
| x - 2 | = 2
| x - 2 | = 2 <=> | x - 2 | = -2
x = 2 + 2 x = -2 + 2
x = 4 x = 0
Vậy ..