tim y, biet:cau a 24*y=480
cau b 12852/y-296=61
Tìm y, biết:
a) 24 x y : 20 = 480
b) 12852 : y - 296 = 61
a) 24 × y : 20 = 480
24 × y = 480 × 20
24 × y = 9600
y = 9600 : 24
y = 400
b) 12862 : y - 296 = 61
12852 : y = 61 + 296
12852 : y = 357
y = 12852 : 357
y = 36
tim x,y,z (cau a ) , tim x , y (cau b )
a, x : y : z = 3 : 4 : 5 va 5z2 - 3x2 - 2y2 = 594
b, z + y = x : y = 3.(x - y )
a) Đặt: x3=y4=z5=Kx3=y4=z5=K
=> x= 3K ; y = 4K ; z = 5K
Theo đề bài ta có: 5z2−3x2−2y2=5945z2−3x2−2y2=594
Hay: 5×(5K)2−3×(3K)2−2×(4K)2=5945×(5K)2−3×(3K)2−2×(4K)2=594
5 * 25K2 - 3* 9K2 - 2* 16K2 = 594
125K2 - 27K2 - 32K2 = 594
66K2 = 594
=> K2 = 594 : 66 = 9
=> K= căn của 9 = ±3±3
Với K = 3
=> x = 3 * 3 = 9
y = 4 * 3 = 12
z = 5 * 3 = 15
Với K = - 3
=> x = 3 * (- 3) = - 9
y = 4 * (- 3) = - 12
z = 5 * (- 3)= - 15
Vậy x = ±9±9 ; y = ±12±12 ; z = ±15
a,\(\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
b, 10x = 6y va 2x2 - y2 = -28
Tim x,y,z(cau a)
tim x,y ( cau b)
\(a)\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+x+x+z+2+x+y-3}{x+y+z}\)
\(=\dfrac{\left(x+y+z\right)+\left(x+y+z\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Lại có: \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
\(\Rightarrow2=\dfrac{1}{x+y+z}\Rightarrow2\left(x+y+z\right)=1\Rightarrow x+y+z=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y+z+x+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+2=3y\\\dfrac{1}{2}-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1+\dfrac{1}{2}}{3}\\y=\dfrac{\dfrac{1}{2}+2}{3}\\z=\dfrac{\dfrac{1}{2}-3}{3}\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)
Chúc bạn học tốt!
Giúp mình với :
a) Vẽ đồ thị (P) y= 0,25x2
b) Trên (P) tim diem M (x;y) sao cho y = x-1
Giup minh cau b thoi , cau a minh lam duoc roi
Cau 1; cho\(\frac{x}{2}=\frac{y}{5}\)vaxy=90. So cap (x;y) thoa man la
Cau 2 : Cho a+d=b+c va \(a^2+d^2=b^2+c^2\)(b,d khac 0).Khi do 4 so lap thanh ti le thuc nao
Cau 3 :GTLN cua phan so \(\frac{7n-8}{2n-3}\)
Cau 4: Cho A=\(\frac{12}{x-15}\) dieu kien de 0<A<1 va A>1
Cau 5 ; tim x biet /-x-5//x=5=10
Cau 6: tap hop cac so nguyen cua x thoa man (3x^2-51)^2014=(-24)^2014
Cau 7: tap hop cac so thoa man /x-y/+/y+9/25/=0
Tim cac so nguyen x,y biet
Cau a, x /10-1/y=3/10
\(\frac{x}{10}-\frac{1}{y}=\frac{3}{10}\)
Ta có: \(\frac{1}{y}=\frac{x}{10}-\frac{3}{10}\)
\(\Rightarrow\frac{1}{y}=\frac{x-3}{10}\)
\(\Rightarrow y.\left(x-3\right)=1.10\)
\(\Rightarrow y.\left(x-3\right)=10\)
\(\Rightarrow x-3\)thuộc \(Ư\left(10\right)\)\(=\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
Ta có bảng sau:
\(x-3\) | \(1\) | \(-1\) | \(2\) | \(-2\) | \(5\) | \(-5\) | \(10\) | \(-10\) |
\(x\) | \(4\) | \(2\) | \(5\) | \(1\) | \(8\) | \(-2\) | \(13\) | \(-7\) |
\(y\) | \(10\) | \(-10\) | \(5\) | \(-5\) | \(2\) | \(-2\) | \(1\) | \(-1\) |
thõa mãn | thõa mãn | thõa mãn | thõa mãn | thõa mãn | thõa mãn | thõa mãn | thõa mãn |
Vậy \(\left(x;y\right)\)thuộc \(\left\{\left(4;10\right),\left(2;-10\right),\left(5;5\right),\left(1;-5\right),\left(8;2\right),\left(-2;-2\right),\left(13;1\right),\left(-7;-1\right)\right\}\)
cau 1,(n^2-7)\(⋮\)n+3
cau 2 tim x;y;z
bietx-y=-9
y-z=-10
z+x=11
xin loi ban voi qua chua nhap cau hoi/
tim n ban nhe
cau 5 : tim y biet : 1000 - 345 + 345 : y = 655 + 345 : 5
1000-345+345:y=655+345:5
1000-345+345:y=724
345+345:y=1000-724
345:y=276-345
y=345-131
y=214
Đúng ghi Đ, sai ghi S.
Cho các số : 405; 480, 296; 324.
a, Những số chia hết cho 2 và 5 là: 480; 296.
b, Những số chia hết cho 3 và 9 là: 405; 324.