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\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{4}\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{4}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy ..
\(\dfrac{1}{2}\)| \(\dfrac{1}{3}x\)- \(\dfrac{1}{4}\)| - \(\dfrac{1}{5}\)= \(\dfrac{1}{6}\)
=> \(\dfrac{1}{2}\)| \(\dfrac{1}{3}x\) - \(\dfrac{1}{4}\)| = \(\dfrac{11}{30}\)
=> | \(\dfrac{1}{3}x\)- \(\dfrac{1}{4}\)| = \(\dfrac{11}{15}\)
=> \(\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{4}=\dfrac{11}{15}\\\dfrac{1}{3}x-\dfrac{1}{4}=\dfrac{-11}{15}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{59}{60}\\\dfrac{1}{3}x=\dfrac{-29}{60}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{59}{20}\\x=\dfrac{-29}{20}\end{matrix}\right.\)
Chúc bạn học tốt !
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)
Do đó: x=18; y=12; z=9
a) Thay x + 3y - 2z vào biểu thức ta có:
\(\dfrac{x - 1}{3} = \dfrac{3(y + 2)}{3 . 4} = \dfrac{2(z - 2)}{2 . 3}\) = \(\dfrac{x - 1}{3} = \dfrac{3x + 6}{12} = \dfrac{2z - 4}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhua ta có:
\(\dfrac{x - 1}{3} = \dfrac{3y + 6}{12} = \dfrac{2z - 4}{6} = \dfrac{x - 1}{3}+ \dfrac{3y + 6}{12} -\dfrac{2z - 4}{6}\)
=\(\dfrac{x - 1 + 3y + 6 - 2z + 4}{3 + 12 -6} \) = \(\dfrac{(x + 3y - 2z) + ( -1 + 6 +4)}{3 + 12 - 6} \)
=\(\dfrac{36 + 9}{9}\) = 5
=> \(\dfrac{x - 1}{3} =\) 5 => x - 1 = 5.3 =15 => x = 5+1 = 6
=>
=>
Vậy ...
(Bạn dựa theo cách này và lm những bài tiếp nhé!)
\(\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{5}\right|\ge0\\\left|x+\dfrac{1}{15}\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+\dfrac{1}{3}+x+\dfrac{1}{5}+x+\dfrac{1}{15}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\)
Vậy ..
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)
Do đó: x=18; y=12; z=9
\(A=\left(3x^4y^2-\dfrac{1}{2}x^4y^2\right)+\left(\dfrac{1}{2}xy^5+5xy^5\right)+\left(\dfrac{-3}{4}x^2y^3-\dfrac{1}{4}x^2y^3\right)=\dfrac{11}{4}x^4y^2+\dfrac{26}{5}xy^5-x^2y^3\)
Bậc là 6
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{6}\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)
\(\Rightarrow3x+1=4x\)
\(\Rightarrow x=1\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}>0\\x+\dfrac{1}{3}>0\\x+\dfrac{1}{6}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|=x+\dfrac{1}{2}\\\left|x+\dfrac{1}{3}\right|=x+\dfrac{1}{3}\\\left|x+\dfrac{1}{6}\right|=x+\dfrac{1}{6}\end{matrix}\right.\)
Thay vào ta được:
\(x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)
\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=1\)
Vậy...................
Chúc bạn học tốt!!!
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Theo tính chất của dãy tỉ số bằng nhau, có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8x+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x=8y\\6z=12x\\8y=6z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{12}\\\dfrac{x}{6}=\dfrac{z}{12}\\\dfrac{y}{6}=\dfrac{z}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
Kết luận ...
a) \(A=2x^2-\dfrac{1}{3}y\)
A= \(\left(2-\dfrac{1}{3}\right)\)\(x^2y\)
A=\(\dfrac{5}{3}\)\(x^2y\)
Tại \(x=2;y=9\) ta có
A=\(\dfrac{5}{3}\).(2)\(^2\).9 = \(\dfrac{5}{3}\).4 .9 = 60
Vậy tại \(x=2;y=9\) biểu thức A= 60
b) P=\(2x^2+3xy+y^2\) (\(y^2\) là 1\(y^2\) nha bạn)
P=\(\left(2+3+1\right)\left(x^2.x\right)\left(y.y^2\right)\)
P= 6\(x^3y^3\)
Tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) ta có
P= 6.\(\left(-\dfrac{1}{2}\right)^3.\left(\dfrac{2}{3}\right)^3\) = 6.\(\left(-\dfrac{1}{8}\right).\dfrac{8}{27}\) = \(-\dfrac{2}{9}\)
Vậy tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) biểu thức P= \(-\dfrac{2}{9}\)
c)\(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)
=\(\left((-\dfrac{1}{2}).\dfrac{2}{3}\right)\left(x.x^3\right).y^2\)
=\(-\dfrac{1}{3}\)\(x^4y^2\)
Tại \(x=2;y=\dfrac{1}{4}\)ta có
\(-\dfrac{1}{3}\).\(\left(2\right)^4.\left(\dfrac{1}{4}\right)^2=-\dfrac{1}{3}.16.\dfrac{1}{16}=-\dfrac{1}{3}\)
\(\)Vậy \(x=2;y=\dfrac{1}{4}\) biểu thức \(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)= \(-\dfrac{1}{3}\)
CHÚC BẠN HỌC TỐT NHA
\(\Leftrightarrow\left[{}\begin{matrix}\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|-3=-4\\\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|-3=4\end{matrix}\right.\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{4}=7\\\dfrac{1}{2}x-\dfrac{1}{4}=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{29}{4}\\\dfrac{1}{2}x=-\dfrac{27}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{2}\\x=-\dfrac{27}{2}\end{matrix}\right.\)