Tìm x , y biết
x/3 = y/4 và xy = 300
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21 tháng 12 2022 lúc 20:30
Câu 14: tìm x,y biết
x/2 = y/3 = z/4 và 2x + 3y - z = 27
ta có : `x/2 = y/3 = z/4=> (2x)/4 =(3y)/9 = z/4`
`=> (2x)/4 =(3y)/9 = z/4` và `2x + 3y - z = 27`
Áp dụng t/c dãy tỉ số bằng nhau ta có:
`(2x)/4 =(3y)/9 = z/4 =(2x + 3y - z)/(4+9-4)=27/9=3`
`=>x/2=3=>x=3.2=6`
`=>y/3=3=>x=3.3=9`
`=>z/4=3=>z=3.4=12`
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tìm quan hệ x và y biết
x\(^4\)=y\(^4\)
x\(^5\)=y\(^5\)
Ta có:
\(x^4=y^4\)
\(\Rightarrow x^4-y^4=0\)
\(\Rightarrow\left(x^2\right)^2-\left(y^2\right)^2=0\)
\(\Rightarrow\left(x^2-y^2\right)\left(x^2+y^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-y^2=0\\x^2+y^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-y=0\\x+y=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
_______________
Ta có:
\(x^5=y^5\)
\(\Rightarrow x^5-y^5=0\)
\(\Rightarrow x-y=0\)
\(\Rightarrow x=y\)
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tìm x,y biết
x^3+y^3/6=x^3-2y^3 và x^6 .y^6
tìm hai số tự nhiên x và y,biết
x< 17/4<y
tìm 2 số x và y biết
x phần 3 = x cộng y = 20
\(\dfrac{x}{3}=x+y=20\Rightarrow x=60\Rightarrow60+y=20\Rightarrow y=-40\)
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Ta có:
\(\dfrac{x}{3}=20\)
\(\Rightarrow\)\(x=60\)
Lại có:
\(x+y=20\)
\(\Rightarrow\)\(y=20-60\)
\(\Rightarrow\)\(y=-40\)
Vây x = 60 và y = - 40
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\(\dfrac{x}{3}=x+y\)
\(\Leftrightarrow x-\dfrac{1}{3}x=-y\)
\(\Leftrightarrow y=-\dfrac{2}{3}x\)
Ta có: x+y=20
\(\Leftrightarrow x\cdot\dfrac{1}{3}=20\)
hay x=60
=> y=40
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Xem thêm câu trả lời
Tìm x,y biết
x/2=y/3 và 2x-3y=54
áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y}{4-9}=-\dfrac{54}{5}\)
\(\dfrac{x}{2}=-\dfrac{54}{5}\Rightarrow x=-\dfrac{54}{5}.2=-\dfrac{108}{5}\)
\(\dfrac{y}{3}=-\dfrac{54}{5}\Rightarrow y=-\dfrac{54}{5}.3=-\dfrac{162}{5}\)
Vậy \(x=-\dfrac{108}{5};y=-\dfrac{162}{5}\)
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Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)
nên \(\dfrac{2x}{4}=\dfrac{3y}{9}\)
mà 2x-3y=54
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y}{4-9}=\dfrac{-54}{5}\)
Do đó: \(x=-\dfrac{108}{5};y=-\dfrac{162}{5}\)
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Tìm các số nguyên x,y biết
X+3y+xy=5
Giúp mình với , ai đúng mình tick cho nha , cảm ơn mn nhiều
tìm x,y thuộc z biết
x/7=9/y và x>y
\(\Leftrightarrow xy=63\)
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;63\right);\left(3;21\right);\left(7;9\right);\left(-63;-1\right);\left(-21;-3\right);\left(-9;-7\right)\right\}\)
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Tìm các cặp số nguyên (x,y) biết
x/3-1/y+1=1/6
\(\dfrac{x}{3}-\dfrac{1}{y+1}=\dfrac{1}{6}\)
=>\(\dfrac{xy+x-3}{3\left(y+1\right)}=\dfrac{1}{6}\)
=>\(2\left(xy+x-3\right)=1\)
=>2xy+2x-6=1
=>2xy+2x=7
=>2x(y+1)=7
=>x(y+1)=7/2
mà x,y nguyên
nên \(\left(x,y\right)\in\varnothing\)
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1/y+1=x/3-1/6
1/y+1=2x/6-1/6
1/y+1= 2x-1/6
=> 1.6=(y+1).(2x-1)
ta có bảng
y+1 6 1
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y 5 0
x 1 7/2
loại
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