x/y=5/3 và x-y=6 giúp mk vsT_T
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a) \(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2.\left(-6\right)=13\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1^3-3.\left(-6\right).1=19\)
\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)=13.19-\left(-6\right)^2.1=211\)
b) \(x^2+y^2=\left(x-y\right)^2+2xy=1^1+2.6=13\)
\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+3.6.1=19\)
\(x^5-y^5=\left(x^2+y^2\right)\left(x^3-y^3\right)+x^2y^2\left(x-y\right)=13.19+6^2.1=283\)
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a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)
b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)
\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)
d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)
\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)
a) �2=�5=�7;�+�+�=562x=5y=7z;x+y+z=56
�2=�5=�7=�+�+�2+5+7=5614=42x=5y=7z=2+5+7x+y+z=1456=4
⇒{�=4.2=8�=4.5=20�=4.7=28⇒⎩⎨⎧x=4.2=8y=4.5=20z=4.7=28
b) �1,1=�1,3=�1,4(1);2�−�=5,51,1x=1,3y=1,4z(1);2x−y=5,5
(1)⇒2�−�1,1.2−1,3=5,50,9(1)⇒1,1.2−1,32x−y=0,95,5
⇒⎩⎨⎧x=1,1.0,95,5=0,96,05y=1,3.0,95,5=0,97,15z=1,11,4.x=1,11,4.0,96,05=0,998,47
d) �2=�3=�5;���=−302x=3x=5z;xyz=−30
�2=�3=�5=���2.3.5=−3030=−12x=3x=5z=2.3.5xyz=30−30=−1
⇒{�=2.(−1)=−2�=3.(−1)=−3�=5.(−1)=−5⇒⎩⎨⎧x=2.(−1)=−2y=3.(−1)=−3z=5.(−1)=−5
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a, (x+3)*(y+2)=1
=> x+3 và y+2 là ước của 1
Ta có bảng sau:
x+3 | -1 | 1 |
x | -4 | 2 |
y+2 | -1 | 1 |
y | -3 | 1 |
Vậy...
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Theo đề bài ta có\(\frac{3+x}{7+y}=\frac{3}{7}\)
<=>\(\frac{3.a}{7.a}=\frac{3+x}{7+y}=\frac{3}{7}\)
x=20:(3+7).3=6
y=20:(3+7).7=14
vậy x= 6 ; y = 14
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Bài 1:
Ta có:
\(y-x=25\Rightarrow y=25+x\)
Mà \(7x=4y\Rightarrow7x=4\cdot\left(25+x\right)\)
\(7x=100+4x\)
\(\Rightarrow7x-4x=100\)
\(3x=100\)
\(x=\frac{100}{3}\)
bài 1 :
Ta có: 7x=4y ⇔ x/4=y/7
áp dụng tính chất dãy tỉ số bằng nhau ta có
x/4=y/7=(y-x)/(7-4)=100/3
⇒x= 4 x 100/3=400/3 ; y = 7 x 100/3=700/3
bài 2
ta có x/5 = y/6 ⇔ x/20=y/24
y/8 = z/7 ⇔ y/24=z/21
⇒x/20=y/24=z/21
ADTCDTSBN(bài 1 có)
x/20=y/24=z/21=(x+y)/(20+24)=69/48=23/16
⇒x= 20 x 23/16 = 115/4
y= 24x 23/16=138/2
z=21x23/16=483/16
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Ta có x + y = 5
<=> (x + y)2 = 25
<=> x2 + 2xy + y2 = 25
<=> x2 - 2xy + y2 = 1
<=> (x - y)2 = 1
<=> \(\orbr{\begin{cases}x-y=1\\x-y=-1\end{cases}}\)
Khi x - y = 1 => x = 3 ; y = 2
Khi x - y = -1 => x = 2 ; y = 3
Khi x = 3; y = 2 thì x3 + y3 = 33 + 23 = 35
Khi x = 2 ; y = 3 thì x3 + y3 = 23 + 33 = 35
Vậy x3 + y3 = 35
\(x^3+y^3\)
\(=x^3+3x^2y+3xy^2+y^3-3x^2y-3xy^2\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(=5^3-3\cdot6\cdot5\)
\(=125-90\)
\(=35\)
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a)
\(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{20}\)
\(\frac{y}{5}=\frac{z}{6}\Rightarrow\frac{y}{20}=\frac{z}{24}\)
=> \(\frac{x}{15}=\frac{y}{20}=\frac{z}{24}\)
Đến đây dễ rồi
b)
\(\left(\frac{x}{3}\right)^2=\frac{x}{3}\cdot\frac{x}{3}=\frac{x}{3}\cdot\frac{y}{4}=\frac{xy}{3\cdot4}=\frac{48}{12}=4=\left(\pm2\right)^2\)
TH1 : \(\frac{x}{3}=\frac{y}{4}=2\)
Sau đó tìm x và y
TH2 : \(\frac{x}{3}=\frac{y}{4}=-2\)
Sau đó lại tìm x và y
Sau cùng kết luận
Học tốt
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áp dụng dãy tí số bằng nhau ta có:
x/5 = y/-6 = z/7 = x+y-z/5+6-7 = 32/4 = 8
từ x/5=8 -> x=40
y/-6 =8 -> y= -48
z/7=8 -> z=56
Theo tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{x+y-z}{5+\left(-6\right)-7}=\frac{32}{-8}=-4\)
\(\Rightarrow\hept{\begin{cases}x=-4.5=-20\\y=-4.\left(-6\right)=24\\z=-4.7=-28\end{cases}}\)
`x/y=5/3`
`=>x/5=y/3`
Mà `x-y=6` nên áp dụng tính chất dãy tỉ số bằng nhau ta có
`x/5=y/3=(x-y)/(5-3)=6/2=3`
\(=>\left\{{}\begin{matrix}\dfrac{x}{5}=3=>x=15\\\dfrac{y}{3}=3=>y=9\end{matrix}\right.\)