Tìm y biết:
a) y + 10,5 = 25,5 - 3,5
b) y x 10 = 1,5 : 0,01
So sánh 2 số x, y sau đây:
x = ( 42 - 4,2 . 10 + 76.7,6 ) : (0,01 . 0,1)
y = ( 689,7 + 0,3 ) : ( 7,4 : 0,2 - 2,2 - 1,5 )
So sánh hai số x, y sau đây:
x= ( 42 - 4,2 . 10 + 76 : 7,6 ) : ( 0,01 . 0,1 )
y= ( 689,7 + 0,3 ) : ( 7,4 : 0,2 - 2,2 - 1,5 )
Ta có:
x= ( 42 - 4,2 . 10 + 76 : 7,6 ) : ( 0,01 . 0,1 )
=(42-42+10):(0,001)
=10:0,001=10000
y= ( 689,7 + 0,3 ) : ( 7,4 : 0,2 - 2,2 - 1,5 )
=690:33,3=20,(720)
Vì 10000 >20,(720) nên x>y
Vậy x>y
Ta có : \(x=\left(42-4,2.10+76:7,6\right):\left(0,01.0,1\right)\)
\(=10:0,001=10000\)
\(y=\left(689,7+0,3\right):\left(7,4:0,2-2,2-1,5\right)\)
\(=20,\left(720\right)\)
Vì \(10000>20,\left(720\right)\)
\(\Rightarrow x>y\)
Ta có :
X=(42-4,2.10+76:7,6):(0,01.0,1)
X=(42-42+10):0,1
X=(0+10):0,1
X=10:0,1
X=100
Y=(689,7+0,3):(7,4:0,2-2,2-1,5)
Y=690:(37-2,2-1,5)
Y=690:(34,8-1,5)
Y=690:33,3
Y=20,(720)
(720) nghĩa là chu kì lặp đi lặp lại
Mà 100>20,(720)
Nên X>Y
So sánh 2 số x, y sau đây:
x = ( 42 - 4,2 . 10 + 76.7,6 ) : (0,01 . 0,1)
y = ( 689,7 + 0,3 ) : ( 7,4 : 0,2 - 2,2 - 1,5 )
\(x=\left(42-4,2.10+76.7,6\right):\left(0,01.0,1\right)\)
\(y=\left(689,7+0,3\right):\left(7,4:0,2-2,2-1,5\right)\)
\(Ta\) \(có:\)
\(x=\left(42-4,2.10+76.7,6\right):\left(0,01.0,1\right)\)
\(x=\left(42-42+76.7,6\right):\left(0,01.0,1\right)\)
\(x=\left(0+76.7,6\right):\left(0,01.0,1\right)\)
\(x=\left(577,6\right):\left(0,001\right)\)
\(x=577600\)
\(y=\left(689,7+0,3\right):\left(7,4:0,2-2,2-1,5\right)\)
\(y=\left(689,7+0,3\right):\left(37-3,7\right)\)
\(y=\left(690\right):\left(33,3\right)\)
\(y=20,\overline{720}\)
*** So sánh :
Vì \(577600>20,\overline{720}\) nên \(x>y\)
Hay \(x=\left(42-4,2.10+76.7,6\right):\left(0,01.0,1\right)\) \(>\) \(y=\left(689,7+0,3\right):\left(7,4:0,2-2,2-1,5\right)\)
Tìm giá trị của x và y biết:
a/ x+y = 10 và x=y
b/ 2x + 3y = 180 và x=y
c/ 3x + 5y = 13 và y = 2x
Lời giải:
a. Thay $x=y$ vào điều kiện ban đầu thì:
$x+x=10$
$2x=10$
$x=5$
$\Rightarrow y=x=5$
Vậy $(x,y)=(5,5)$
b. Thay $x=y$ vào điều kiện đầu:
$2x+3x=180$
$5x=180$
$x=36$
$y=x=36$
Vậy $(x,y)=(36,36)$
c. Thay $y=2x$ vào điều kiện đầu thì:
$3x+5.2x=13$
$13x=13$
$x=1$
$y=2x=2$
Vậy $(x,y)=(1,2)$
a) Ta có: x=y
mà x+y=10
nên \(x=y=\dfrac{10}{2}=5\)
b) Ta có: \(\left\{{}\begin{matrix}2x+3y=180\\x=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y+3y=180\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y=180\\x=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=36\\x=36\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}3x+5y=13\\y=2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+10x=13\\y=2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}13x=13\\y=2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Tìm số nguyên x và y biết:
a) \(\dfrac{x}{6}-\dfrac{3}{y}=\dfrac{1}{12}\)
b) (x - 3)( x+10) ≤ 0
Tìm số nguyên x và y biết:
a) \(\dfrac{x}{6}-\dfrac{3}{y}=\dfrac{1}{12}\)
b) (x - 3)( x+10) ≤ 0
tìm x,y biết:
a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
b) \(\left(\dfrac{1}{2}.x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\)( do \(x^2\ge0,\left(y-\dfrac{1}{10}\right)^4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
b) \(\left(\dfrac{1}{2}.x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\)( do \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0,\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
\(a,\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\\ b,\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\end{matrix}\right.\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
Mà \(x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x;y\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=0\\\left(y-\dfrac{1}{10}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(0;\dfrac{1}{10}\right)\)
b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)
Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\forall x;y\)
\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}=0\\\left(y^2-\dfrac{1}{4}\right)^{10}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(10;\dfrac{1}{2}\right);\left(10;-\dfrac{1}{2}\right)\right\}\)
Tìm các số x, y, z biết:
a) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và x - 2y + 3z = 33
b) x : y : z = 10 : 6 : 21 và y + 5x - 2z = -42
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-2y+3z}{2-2\cdot3+3\cdot5}=\dfrac{33}{11}=3\)
Do đó: x=6; y=9; z=15
Tìm x,y,z biết:a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{10}\)và y-x=6
Tìm x,y,z biết:b) \(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{7}\)và x-2y+z=18
a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)
⇒\(\dfrac{y-x}{5-2}=\dfrac{6}{3}=2\)
\(\dfrac{x}{2}=2\Rightarrow x=4\)
\(\dfrac{y}{5}=2\Rightarrow y=10\)
\(\dfrac{z}{10}=2\Rightarrow z=20\)
b) Ta có: \(\dfrac{x}{8}=\dfrac{2y}{6}=\dfrac{z}{7}\)
\(\dfrac{x-2y+z}{8-6+7}=\dfrac{18}{9}=2\)
\(\dfrac{x}{8}=2\Rightarrow x=16\)
\(\dfrac{y}{3}=2\Rightarrow y=6\)
\(\dfrac{z}{7}=2\Rightarrow z=14\)