Tìm x biết \(|^1_2x+3|=\dfrac{2}{5}\)
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Ta có : (P) \(y=x^2\)
(d) y=\(\dfrac{-1}{2}x+\dfrac{3}{2}\)
Xét phương trình hoành độ giao điểm của (P) và (d):
\(x^2=\dfrac{-1}{2}x+\dfrac{3}{2}\)
\(x^2+\dfrac{1}{2}x-\dfrac{3}{2}=0\\ \Leftrightarrow x^2+\dfrac{3}{2}x-x-\dfrac{3}{2}=0\\\Leftrightarrow x\left(x+\dfrac{3}{2}\right)-\left(x+\dfrac{3}{2}\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+\dfrac{3}{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy hoành độ giao điểm của (P) và (d) là 1 và \(\dfrac{-3}{2}\)
a, \(\left(-2xy^2z^3\right)^3.\left(\dfrac{5}{2}xy^3\right)^2.\left(\dfrac{-4}{125}xy\right)\)
\(=\left(-2\right)^3.x^3.\left(y^2\right)^3.z^3.\left(\dfrac{5}{2}\right)^2.x^2.\left(y^3\right)^2.\dfrac{-4}{125}.x.y\)
\(=\left(-2\right)^3.\left(\dfrac{5}{2}\right)^2.\dfrac{-4}{125}.\left(x^3.x^2.x\right).\left(y^6.y^6.y\right).z^3\)
\(=\left(-8\right).\dfrac{25}{4}.\dfrac{-4}{125}.x^6.y^{13}.z^3\)
\(=1,6.x^6.y^{13}.z^3\)
a, \(\left(-2xy^2z^3\right).\left(\dfrac{5}{2}xy^3\right)^2.\left(\dfrac{-4}{125}xy\right)\)
= \(\left(-5x^2y^5z^3\right)^5.\left(\dfrac{-4}{125}xy\right)\)
= \(\left(\dfrac{4}{25}x^3y^6z^3\right)^5\)
b, \(2\dfrac{1}{3}x^2y^5-3\dfrac{2}{5}x^3y-1\dfrac{1}{2}x^2y^5+2\dfrac{2}{3}x^3y\)
= \(\dfrac{7}{3}x^2y^5-\dfrac{17}{5}x^3y-\dfrac{3}{2}x^2y^5+\dfrac{8}{3}x^3y\)
= \(\dfrac{5}{6}x^2y^5-\dfrac{11}{15}x^3y\)
b) Để \(\frac{2}{3\left|x-1\right|+4}\)đạt GTLN
=> 3|x - 1|+ 4 đạt giá trị nhỏ nhất
mà 3|x - 1| \(\ge0\forall x\)
=> 3|x - 1| + 4 \(\ge\)4
Dấu "=" xảy ra <=> x - 1 = 0
=> x = 1
Vậy GTLN của \(B=\frac{1}{2}\Leftrightarrow x=1\)
a) Áp dụng tính chất dãy tỉ số bằng nhau ta được:
X/3 = y/4 = x/3 + y/4 = 28/7 = 4
=> x = 4 × 3 = 12
=> y = 4 × 4 = 16
Vậy x = 12, y = 16
B) Áp dụng tính chất dãy tỉ số bằng nhau ta được:
X/2 = y/(-5) = x/2 - y/(-5) = (-7)/7 = -1
=> x = -1 × 2 = -2
=> y = -1 × -5 = 5
Vậy x = -2, y = 5
C) làm tương tự như bài a, b
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
x8=y12=z15=x+y−z8+12−15=105=2x8=y12=z15=x+y−z8+12−15=105=2
Do đó: x=16; y=24; z=30
\(1.x-\dfrac{2}{3}\times\left(x+9\right)=1\)
\(x-\dfrac{2}{3}\times x-6=1\)
\(x\times\left(1-\dfrac{2}{3}\right)=7\)
\(x\times\dfrac{1}{3}=7\)
\(x=21\)
\(2.x-\dfrac{11}{15}=\dfrac{3+x}{5}\)
\(\dfrac{15x}{15}-\dfrac{11}{15}=\dfrac{9+3x}{15}\)
\(15x-11=9+3x\)
\(12x=20\)
\(x=\dfrac{5}{3}\)
a, - \(\dfrac{2}{5}\) + \(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)
\(\dfrac{4}{5}\).\(x\) = \(\dfrac{3}{5}\)+ \(\dfrac{2}{5}\)
\(\dfrac{4}{5}\).\(x\) = 1
\(x\) = \(\dfrac{5}{4}\)
b, - \(\dfrac{3}{7}\) - \(\dfrac{4}{7}\): \(x\) = \(\dfrac{2}{5}\)
\(\dfrac{4}{7}\): \(x\) = - \(\dfrac{3}{7}\) - \(\dfrac{2}{5}\)
\(\dfrac{4}{7}\): \(x\) = - \(\dfrac{29}{35}\)
\(x\) = \(\dfrac{4}{7}\): (- \(\dfrac{29}{35}\) )
\(x\) = - \(\dfrac{20}{29}\)
c, \(\dfrac{4}{7}\).\(x\) + \(\dfrac{2}{3}\) = - \(\dfrac{1}{5}\)
\(\dfrac{4}{7}\).\(x\) = -\(\dfrac{1}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{4}{7}\).\(x\) = - \(\dfrac{13}{15}\)
\(x\) = - \(\dfrac{13}{15}\): \(\dfrac{4}{7}\)
\(x\) = - \(\dfrac{91}{60}\)
\(a,\dfrac{2}{3}x-\dfrac{2}{5}=\dfrac{1}{2}x-\dfrac{1}{3}\\ \Rightarrow\dfrac{2}{3}x-\dfrac{1}{2}x-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\left(\dfrac{2}{3}-\dfrac{1}{2}\right)-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\dfrac{1}{6}=-\dfrac{11}{15}\\ \Rightarrow x=-\dfrac{22}{5}\\ b,\dfrac{1}{3}x+\dfrac{2}{5}.\left(x+1\right)=0\\ \Rightarrow\dfrac{1}{3}x+\left(x+1\right)=-\dfrac{2}{5}\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{2}{5}-\left(x+1\right)\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{7}{5}-x\\ \Rightarrow\dfrac{1}{3}.2x=-\dfrac{7}{5}\\ \Rightarrow2x=-\dfrac{21}{5}\\ \Rightarrow x=-\dfrac{21}{10}.\)
a, \(\dfrac{3}{7}\)\(x\)- \(\dfrac{2}{3}\)\(x\) = \(\dfrac{10}{21}\)
(\(\dfrac{3}{7}\) - \(\dfrac{2}{3}\)) \(\times\) \(x\) = \(\dfrac{10}{21}\)
- \(\dfrac{5}{21}\) \(\times\) \(x\) = \(\dfrac{10}{21}\)
\(x\) = \(\dfrac{10}{21}\) : (-\(\dfrac{5}{21}\))
\(x\) = -2
b, \(\dfrac{7}{35}\) : (\(x-\dfrac{1}{3}\)) = - \(\dfrac{2}{25}\)
\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{7}{35}\) : (- \(\dfrac{2}{25}\))
\(x\) - \(\dfrac{1}{3}\) = - \(\dfrac{5}{2}\)
\(x\) = - \(\dfrac{5}{2}\) + \(\dfrac{1}{3}\)
\(x\) = - \(\dfrac{13}{6}\)
c, 3.(\(x\) - \(\dfrac{1}{2}\)) - 5.(\(x\) + \(\dfrac{3}{5}\)) = - \(x\)+ \(\dfrac{1}{5}\)
3\(x\) - \(\dfrac{3}{2}\) - 5\(x\) - 3 = - \(x\) + \(\dfrac{1}{5}\)
- \(x\) + 5\(x\) - 3\(x\) = - \(\dfrac{3}{2}\) - 3 - \(\dfrac{1}{5}\)
\(x\) = - \(\dfrac{47}{10}\)
\(a,\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow x\left(\dfrac{3}{7}-\dfrac{2}{3}\right)=\dfrac{10}{21}\\ \Rightarrow x.-\dfrac{5}{21}=\dfrac{10}{21}\\ \Rightarrow x=-2\\ b,\dfrac{7}{35}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow\dfrac{1}{5}:\left(x-\dfrac{1}{3}\right)=-\dfrac{2}{25}\\ \Rightarrow x-\dfrac{1}{3}=-\dfrac{5}{2}\\ \Rightarrow x=-\dfrac{13}{6}\\ c,3.\left(x-\dfrac{1}{2}\right)-5.\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\\ \Rightarrow3x-\dfrac{3}{2}-5x+5=-x+\dfrac{1}{5}\)
\(\Rightarrow x\left(3-5\right)-\dfrac{3}{2}+5=-x+\dfrac{1}{5}\\ \Rightarrow-2x-\dfrac{13}{2}=-x+\dfrac{1}{5}\\ \Rightarrow-x-\dfrac{13}{5}=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{1}{5}-\dfrac{13}{5}\\ \Rightarrow x=-\dfrac{12}{5}.\)
\(\left|\dfrac{1}{2}x+3\right|=\dfrac{2}{5}\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+3=\dfrac{2}{5}\\\dfrac{1}{2}x+3=-\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{13}{5}\\\dfrac{1}{2}x=-\dfrac{17}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{26}{5}\\x=-\dfrac{34}{5}\end{matrix}\right.\\ \Rightarrow S=\left\{-\dfrac{26}{5};-\dfrac{34}{5}\right\}\)
`|1/2x+3|=2/5`
`+)1/2x+3=2/5`
`<=>1/2x=-13/5`
`<=>x=-26/5`
`+)1/2x+3=-2/5`
`<=>1/2x=-17/5`
`<=>x=-34/5`
Vậy `x=-26/5` hoặc `x=-34/5`