\(A=x^2-2x+2\)
\(A=x^2-2x+1+1\)
\(A=\left(x-1\right)^2+1\ge1\)
\(MinA=1\Leftrightarrow x-1=0\Leftrightarrow x=1\)
(Nhớ k cho mình với nhoa!)
\(A=x^2-2.1.x+1+2\Rightarrow A=\left(x-1\right)^2+1\ge1\)
\(\Rightarrow A_{min}=1\Leftrightarrow x-1=0\Rightarrow x=1\)
t i c k cho mình nha cảm ơn
=> A = x2 - 2x + 1 + 1
=> A = (x - 1)2 + 1 >= 1
=> A MIN = 1 => x - 1 = 0 => x = 1
t i c k nhé!! 45465657678568578796898798796976965
\(A=x^2-2x+2\)
\(\Leftrightarrow A=x^2-2x+1+1\)
\(\Leftrightarrow A=\left(x-1\right)^2+1\)
Vì \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+1\ge1\)
Biểu thức A đạt GTNN khi:
(x-1)2+1=1
<=>(x-1)2=0
<=>x-1=0
<=>x=1
Vậy A đạt GTNN là 1 khi x=1
e) \(\frac{6}{11}x=\frac{9}{2}y\Rightarrow\frac{x}{\frac{9}{2}}=\frac{y}{\frac{6}{11}}\Rightarrow\frac{-x}{-\frac{81}{5}}=\frac{y}{\frac{108}{55}}\) (1)
\(\frac{9}{2}y=\frac{18}{5}f\Rightarrow\frac{y}{\frac{18}{5}}=\frac{f}{\frac{9}{2}}\Rightarrow\frac{y}{\frac{108}{55}}=\frac{f}{\frac{27}{11}}\) (2)
Từ (1) và (2) => \(\frac{-x}{-\frac{81}{5}}=\frac{y}{\frac{108}{55}}=\frac{f}{\frac{27}{11}}\)
Áp dụng t/c của dãy tỉ số bằng nhau:
\(\frac{-x}{-\frac{81}{5}}=\frac{y}{\frac{108}{55}}=\frac{f}{\frac{27}{11}}=\frac{-x+y+f}{-\frac{81}{5}+\frac{108}{55}+\frac{27}{11}}=\frac{-120}{-\frac{648}{55}}=\frac{275}{27}\)
\(\Rightarrow x=\frac{275}{27}\cdot\frac{81}{5}=165\)
\(y=\frac{275}{27}\cdot\frac{108}{55}=20\)
\(f=\frac{275}{27}\cdot\frac{27}{11}=25\)
f) Đặt \(\frac{x}{12}=\frac{y}{9}=\frac{f}{5}=k\)
\(\Rightarrow x=12k\)
\(y=9k\)
\(f=5k\)
=> xyf = 20
<=> 12k * 9k * 5k = 20
540* k^3 = 20
k^3 = 1/27
k= 1/3
=> x= 12k = 12* 1/3 = 4
y= 9k = 9 * 1/3 = 3
f= 5k = 5* 1/3 = 5/3
h) Ta có: 2x^3 - 1 = 15
2x^3 = 16
x^3 = 8
x = 2
Thay x= 2 vào x+ 16/9 = y- 25/16 :
2+ 16/9 = y- 25/16
34/9 = y- 25/16
y= 769/144
Thay x= 2 vào x+ 16/9 = f+ 9/25 :
2+ 16/9 = f+ 9/25
34/9 = f+ 9/25
f= 769/225
Vậy x= 2, y = 769/144, f= 769/225
i) Đặt \(\frac{3}{5}x=\frac{2}{3}y=k\)
\(\Rightarrow x=k:\frac{3}{5}=\frac{5}{3}k\)
\(y=k:\frac{2}{3}=\frac{3}{2}k\)
\(\Rightarrow x^2-y^2=38\)
\(\Leftrightarrow\left(\frac{5}{3}k\right)^2-\left(\frac{3}{2}k\right)^2=38\)
\(\frac{25}{9}k^2\cdot\frac{9}{4}k^2=38\)
\(\frac{25}{4}k^4=38\)
\(k^4=\frac{152}{25}\)
\(k=\)
Rồi xong....tính ko ra.