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1) Tìm x, biết :
a) \(\dfrac{x-1}{3}=\dfrac{x+1}{5}\)
=> \(5\left(x-1\right)=3\left(x+1\right)\)
=> \(5x-5=3x+3\)
=> \(5x-5-3=3x\)
=> \(5x-8=3x\)
=> \(8=5x-3x\)
=> \(8=2x\)
=> x = 8 : 2
=> x = 4
Bài 2:
\(\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2\ge0\\\left(y+\dfrac{1}{2}\right)^2\ge0\\\left(z-\dfrac{1}{3}\right)^2\ge0\end{matrix}\right.\Rightarrow\left(2x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{3}\right)^2\ge0\)Mà \(\left(2x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{3}\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\\\left(z-\dfrac{1}{3}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{-1}{2}\\z=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{4},y=\dfrac{-1}{2},z=\dfrac{1}{3}\)
1)
a) \(2x+\dfrac{5}{2}=\dfrac{7}{2}\)
\(\Leftrightarrow2x=\dfrac{7}{2}-\dfrac{5}{2}\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
b) \(\left|5-\dfrac{1}{2}x\right|=\left|-\dfrac{1}{5}\right|\)
\(\Leftrightarrow\left|5-\dfrac{1}{2}x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}5-\dfrac{1}{2}x=\dfrac{1}{5}\\5-\dfrac{1}{2}x=-\dfrac{1}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{48}{5}\\x=\dfrac{52}{5}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{48}{5};x_2=\dfrac{52}{5}\)
1.
Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)
\(\Rightarrow x^2-y^2=\left(5k\right)^2-\left(4k\right)^2=25k^2-16k^2=9k^2=4\)
\(\Rightarrow k^2=\dfrac{4}{9}\Rightarrow k=\pm\dfrac{2}{3}\)
\(\circledast k=\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{8}{3}\end{matrix}\right.\)
\(\circledast k=-\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=-\dfrac{8}{3}\end{matrix}\right.\)
2.
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\Rightarrow x=2\)
\(\Rightarrow y=\dfrac{\dfrac{2\cdot2+1}{5}\cdot7+2}{3}=3\)
3.
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-\left(z-3\right)}{4+9-4}=\dfrac{95-8+3}{9}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10\cdot4+2}{2}=21\\y=\dfrac{10\cdot9+6}{3}=32\\z=10\cdot4+3=43\end{matrix}\right.\)
Mình k chép lại đề nha!
Ap dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-4}{4}=\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{2x+3y-z-4}{2+3-4}=46\)
Suy ra; x-1/2 => x-1=92 => x=93
y-2/3 => y-2=138 => y=140
z-4/4=46 => z-4= 184 => z=188
Vậy x=93
y=140
z=188
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-4}{4}\)
\(\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-4}{4}\)
Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(=\dfrac{2x-2+3y-6-z+4}{4+9-4}=\dfrac{\left(2x+3y-z\right)-2-6+4}{9}=\dfrac{54}{9}=6\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=6\Rightarrow x-1=12\Rightarrow x=13\\\dfrac{y-2}{3}=6\Rightarrow y-2=18\Rightarrow y=20\\\dfrac{z-4}{4}=6\Rightarrow z-4=24\Rightarrow z=28\end{matrix}\right.\)
b) áp dụng giống.
\(2\) )
\(B=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{x}{z}\right)\left(1+\dfrac{z}{4}\right)\)
\(B=\dfrac{2y}{x}.\dfrac{x+z}{z}.\dfrac{4+z}{4}\)
\(B=\dfrac{2y\left(x+z\right)\left(4+z\right)}{4xz}\)
\(B=\dfrac{\left(2xy+2yz\right)\left(4+z\right)}{4xz}\)
\(B=\dfrac{8xy+2xyz+8yz+2yz^2}{4xz}\)
\(\dfrac{x-2}{4}=\dfrac{y+1}{5}=\dfrac{z+3}{7}\)
\(\Rightarrow\dfrac{2\left(x-2\right)}{8}=\dfrac{y+1}{5}=\dfrac{2\left(z+3\right)}{14}\)
\(\Rightarrow\dfrac{2x-4}{8}=\dfrac{y+1}{5}=\dfrac{2z+6}{14}\)
Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(=\dfrac{2x-4+y+1-2z-6}{8+5-14}\)
\(=\dfrac{2x+y-2z-9}{-1}\)
\(=\dfrac{7-9}{-1}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-2}{4}=2\Rightarrow x-2=8\Rightarrow x=10\\\dfrac{y+1}{5}=2\Rightarrow y+1=10\Rightarrow y=9\\\dfrac{z+3}{7}=2\Rightarrow z+3=14\Rightarrow z=11\end{matrix}\right.\)
Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x
=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x
=> 2x+3y-1 / 12 = 2x+3y-1 / 6x
=> 12 = 6x => x =2
Bài 1:
| x-1 | +2x =2015 (*)
TH1: |x-1| = x-1 khi x-1 \(\ge\) 0 \(\Rightarrow x\ge1\)
\(\Rightarrow\) (*) có dạng:
(x-1) +2x=2015
\(\Rightarrow\) 3x-1=2015
\(\Rightarrow\) 3x=2016
\(\Rightarrow\) x=672 > 1 ( thỏa mãn )
TH2: |x-1| = 1-x khi x-1 < 0 \(\Rightarrow\) x<1
\(\Rightarrow\) (*) có dạng:
(1-x) +2x =2015
\(\Rightarrow\) 1+x =2015
\(\Rightarrow\) x=2014 > 1 ( ko thỏa mãn)
Vậy x= 672 thỏa mãn đề bài.
Bài 2:
\(2x+\dfrac{1}{7}=\dfrac{1}{y}\)
\(\Rightarrow2x=\dfrac{1}{y}-\dfrac{1}{7}\)
\(\Rightarrow2x=\dfrac{7-y}{7y}\)
\(\Rightarrow x=\dfrac{7-y}{7y}:2=\dfrac{7-y}{14y}\)
\(\Rightarrow14xy=7-y\)
\(\Rightarrow14xy+y=7\)
\(\Rightarrow y\left(14x+1\right)=7\)
\(\Rightarrow y;14x+1\inƯ\left(7\right)\)
Ta có bảng sau :
Vậy cặp ( x ; y ) nguyên thỏa mãn đề bài là : x = 0 và y = 7