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Bài 1 :
Đặt :
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=3k\\3y=4k\\4z=5k\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3k}{2}\\y=\dfrac{4k}{3}\\z=\dfrac{5k}{4}\end{matrix}\right.\)
Thay vào \(x+y+z=49\) ta được :
\(\dfrac{3k}{2}=\dfrac{4k}{3}=\dfrac{5k}{4}=49\)
\(\Leftrightarrow\dfrac{18k+16k+15k}{12}=\dfrac{588}{12}\)
\(\Leftrightarrow49k=588\)
\(\Leftrightarrow k=12\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3.12}{2}=18\\y=\dfrac{4.12}{3}=16\\z=\dfrac{5.12}{4}=15\end{matrix}\right.\)
Vậy ....
Bài1:
Từ \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{x}{90}=\dfrac{y}{80}=\dfrac{z}{75}\)
Áp dụng t/c của dãy tỉ số bằng nhau,ta có:
\(\dfrac{x}{90}=\dfrac{y}{80}=\dfrac{z}{75}=\dfrac{x+y+z}{90+80+75}=\dfrac{49}{245}=\dfrac{1}{5}\)
=>x=18;b=16;c=15
Vậy...
a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)
b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)
\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)
\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)
\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)
\(x=\)-\(\dfrac{27}{10}\)
\(x:\dfrac{2}{9}=\dfrac{9}{2}\)
\(x.\dfrac{9}{2}=\dfrac{9}{2}\)
\(x=\dfrac{9}{2}:\dfrac{9}{2}\)
\(x=1\)
\(\dfrac{x}{9}=\dfrac{5}{3}\)
\(x.3=5.9\)
\(x.3=45\)
\(x=45:3=15\)
\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)
\(x:\dfrac{8}{125}=\dfrac{125}{8}\)
\(x.\dfrac{125}{8}=\dfrac{125}{8}\)
\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)
\(\left(-\dfrac{2}{5}\right)^2\cdot\left|\dfrac{1}{3}-\dfrac{3}{5}\right|-\dfrac{2}{5}\cdot\sqrt{\dfrac{1}{25}}+\dfrac{4}{3}\)
\(=\dfrac{4}{25}\cdot\dfrac{4}{15}-\dfrac{2}{5}\cdot\dfrac{1}{5}+\dfrac{4}{3}\)
\(=\dfrac{16}{375}-\dfrac{2}{25}+\dfrac{4}{3}\)
\(=\dfrac{16}{375}-\dfrac{30}{375}+\dfrac{500}{375}\)
\(=\dfrac{486}{375}=\dfrac{162}{125}\)
a)
ta có \(\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{1}{13}.\dfrac{1}{14}=\dfrac{3}{7}.9.\dfrac{1}{2}.\dfrac{1}{13}-\dfrac{1}{13}.\dfrac{1}{14}\)\(=\dfrac{1}{13}.\left(\dfrac{3}{7}.\dfrac{9}{2}-\dfrac{1}{14}\right)=\dfrac{1}{13}.\dfrac{26}{14}=\dfrac{1.26}{13.14}\)\(=\dfrac{1.13.2}{13.7.2}=\dfrac{1}{7}\)
b)\(x-\left(\dfrac{5}{2}+2x\right)=x-\dfrac{5}{2}-2x=-x-\dfrac{5}{2}=\dfrac{7}{4}\)
\(\Rightarrow-x=\dfrac{7}{4}+\dfrac{5}{2}=\dfrac{17}{4}\)
\(\Rightarrow x=-\dfrac{17}{4}\)(vì -x là số đối của x)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x+y}{4+5}=\dfrac{18}{9}=2\)
Do đó: x=8; y=10
Ta có :
\(\dfrac{x}{10}=\dfrac{y}{5}\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{10}\)
\(\dfrac{y}{2}=\dfrac{z}{3}\Leftrightarrow\dfrac{y}{10}=\dfrac{z}{15}\)
\(\Leftrightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\)
\(\Leftrightarrow\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}=\dfrac{2x-3y+4z}{40-30+60}=\dfrac{330}{70}=\dfrac{33}{7}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{33}{7}\Leftrightarrow x=\dfrac{660}{7}\\\dfrac{y}{10}=\dfrac{33}{7}\Leftrightarrow y=\dfrac{330}{7}\\\dfrac{z}{15}=\dfrac{33}{7}\Leftrightarrow z=\dfrac{495}{7}\end{matrix}\right.\)
Vậy .....
Bài 1:
a.
$|x+\frac{7}{4}|=\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)
b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$
$|2x+1|=\frac{11}{15}$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)
c.
$3x(x+\frac{2}{3})=0$
\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)
d.
$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$
$\Leftrightarrow x=\frac{2}{5}$
Nguyễn Quý Trung:
\(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
Bạn bớt 2 vế đi 1/3 thì \(x=\dfrac{2}{5}\)
2) \(\dfrac{x}{y}=\left(\dfrac{x}{y}\right)^2\)
\(\Rightarrow\left(\dfrac{x}{y}\right)^2-\dfrac{x}{y}=0\)
\(\Rightarrow\dfrac{x}{y}\left(\dfrac{x}{y}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{y}=0\Rightarrow x=0;y\in R\\\dfrac{x}{y}-1=0\Rightarrow\dfrac{x}{y}=1\Rightarrow x=y\end{matrix}\right.\)
3) \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}.2^5+2^{15}.1=2^{15}.33⋮33\rightarrowđpcm\)
4)\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\\\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\end{matrix}\right.\)
\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\\\left(x-4-y\right)^{200}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-12+y\right)^{200}+\left(x-y-4\right)^{200}\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-y-4\right)^{200}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-12+y=0\Rightarrow x+y=12\\x-y-4=0\Rightarrow x-y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y\right)+\left(x-y\right)=12+4\Rightarrow x+y+x-y=16\Rightarrow2x=16\Rightarrow x=8\\y=8-4=4\end{matrix}\right.\)